Encuentra elementos comunes en tres arrays ordenadas

Dadas tres arrays ordenadas en orden no decreciente, imprima todos los elementos comunes en estas arrays.

Ejemplos: 

C++

// C++ program to print common elements in three arrays
#include <bits/stdc++.h>
using namespace std;
 
// This function prints common elements in ar1
void findCommon(int ar1[], int ar2[], int ar3[], int n1,
                int n2, int n3)
{
    // Initialize starting indexes for ar1[], ar2[] and
    // ar3[]
    int i = 0, j = 0, k = 0;
 
    // Iterate through three arrays while all arrays have
    // elements
    while (i < n1 && j < n2 && k < n3) {
        // If x = y and y = z, print any of them and move
        // ahead in all arrays
        if (ar1[i] == ar2[j] && ar2[j] == ar3[k]) {
            cout << ar1[i] << " ";
            i++;
            j++;
            k++;
        }
 
        // x < y
        else if (ar1[i] < ar2[j])
            i++;
 
        // y < z
        else if (ar2[j] < ar3[k])
            j++;
 
        // We reach here when x > y and z < y, i.e., z is
        // smallest
        else
            k++;
    }
}
 
// Driver code
int main()
{
    int ar1[] = { 1, 5, 10, 20, 40, 80 };
    int ar2[] = { 6, 7, 20, 80, 100 };
    int ar3[] = { 3, 4, 15, 20, 30, 70, 80, 120 };
    int n1 = sizeof(ar1) / sizeof(ar1[0]);
    int n2 = sizeof(ar2) / sizeof(ar2[0]);
    int n3 = sizeof(ar3) / sizeof(ar3[0]);
 
    cout << "Common Elements are ";
    findCommon(ar1, ar2, ar3, n1, n2, n3);
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta (kriSania804)

C

// C program to print common elements in three arrays
#include <stdio.h>
 
// This function prints common elements in ar1
void findCommon(int ar1[], int ar2[], int ar3[], int n1,
                int n2, int n3)
{
    // Initialize starting indexes for ar1[], ar2[] and
    // ar3[]
    int i = 0, j = 0, k = 0;
 
    // Iterate through three arrays while all arrays have
    // elements
    while (i < n1 && j < n2 && k < n3) {
        // If x = y and y = z, print any of them and move
        // ahead in all arrays
        if (ar1[i] == ar2[j] && ar2[j] == ar3[k]) {
            printf("%d ", ar1[i]);
            i++;
            j++;
            k++;
        }
 
        // x < y
        else if (ar1[i] < ar2[j])
            i++;
 
        // y < z
        else if (ar2[j] < ar3[k])
            j++;
 
        // We reach here when x > y and z < y, i.e., z is
        // smallest
        else
            k++;
    }
}
 
// Driver code
int main()
{
    int ar1[] = { 1, 5, 10, 20, 40, 80 };
    int ar2[] = { 6, 7, 20, 80, 100 };
    int ar3[] = { 3, 4, 15, 20, 30, 70, 80, 120 };
    int n1 = sizeof(ar1) / sizeof(ar1[0]);
    int n2 = sizeof(ar2) / sizeof(ar2[0]);
    int n3 = sizeof(ar3) / sizeof(ar3[0]);
 
    printf("Common Elements are ");
    findCommon(ar1, ar2, ar3, n1, n2, n3);
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta
// (kriSania804)

Java

// Java program to find common elements in three arrays
class FindCommon {
    // This function prints common elements in ar1
    void findCommon(int ar1[], int ar2[], int ar3[])
    {
        // Initialize starting indexes for ar1[], ar2[] and
        // ar3[]
        int i = 0, j = 0, k = 0;
 
        // Iterate through three arrays while all arrays
        // have elements
        while (i < ar1.length && j < ar2.length
               && k < ar3.length) {
            // If x = y and y = z, print any of them and
            // move ahead in all arrays
            if (ar1[i] == ar2[j] && ar2[j] == ar3[k]) {
                System.out.print(ar1[i] + " ");
                i++;
                j++;
                k++;
            }
 
            // x < y
            else if (ar1[i] < ar2[j])
                i++;
 
            // y < z
            else if (ar2[j] < ar3[k])
                j++;
 
            // We reach here when x > y and z < y, i.e., z
            // is smallest
            else
                k++;
        }
    }
 
    // Driver code to test above
    public static void main(String args[])
    {
        FindCommon ob = new FindCommon();
 
        int ar1[] = { 1, 5, 10, 20, 40, 80 };
        int ar2[] = { 6, 7, 20, 80, 100 };
        int ar3[] = { 3, 4, 15, 20, 30, 70, 80, 120 };
 
        System.out.print("Common elements are ");
        ob.findCommon(ar1, ar2, ar3);
    }
}
 
/*This code is contributed by Rajat Mishra */

Python

# Python function to print common elements in three sorted arrays
def findCommon(ar1, ar2, ar3, n1, n2, n3):
 
    # Initialize starting indexes for ar1[], ar2[] and ar3[]
    i, j, k = 0, 0, 0
 
    # Iterate through three arrays while all arrays have elements
    while (i < n1 and j < n2 and k < n3):
 
        # If x = y and y = z, print any of them and move ahead
        # in all arrays
        if (ar1[i] == ar2[j] and ar2[j] == ar3[k]):
            print ar1[i],
            i += 1
            j += 1
            k += 1
 
        # x < y
        elif ar1[i] < ar2[j]:
            i += 1
 
        # y < z
        elif ar2[j] < ar3[k]:
            j += 1
 
        # We reach here when x > y and z < y, i.e., z is smallest
        else:
            k += 1
 
 
# Driver program to check above function
ar1 = [1, 5, 10, 20, 40, 80]
ar2 = [6, 7, 20, 80, 100]
ar3 = [3, 4, 15, 20, 30, 70, 80, 120]
n1 = len(ar1)
n2 = len(ar2)
n3 = len(ar3)
print "Common elements are",
findCommon(ar1, ar2, ar3, n1, n2, n3)
 
# This code is contributed by __Devesh Agrawal__

C#

// C# program to find common elements in
// three arrays
using System;
 
class GFG {
 
    // This function prints common element
    // s in ar1
    static void findCommon(int[] ar1, int[] ar2, int[] ar3)
    {
 
        // Initialize starting indexes for
        // ar1[], ar2[] and ar3[]
        int i = 0, j = 0, k = 0;
 
        // Iterate through three arrays while
        // all arrays have elements
        while (i < ar1.Length && j < ar2.Length
               && k < ar3.Length) {
 
            // If x = y and y = z, print any of
            // them and move ahead in all arrays
            if (ar1[i] == ar2[j] && ar2[j] == ar3[k]) {
                Console.Write(ar1[i] + " ");
                i++;
                j++;
                k++;
            }
 
            // x < y
            else if (ar1[i] < ar2[j])
                i++;
 
            // y < z
            else if (ar2[j] < ar3[k])
                j++;
 
            // We reach here when x > y and
            // z < y, i.e., z is smallest
            else
                k++;
        }
    }
 
    // Driver code to test above
    public static void Main()
    {
 
        int[] ar1 = { 1, 5, 10, 20, 40, 80 };
        int[] ar2 = { 6, 7, 20, 80, 100 };
        int[] ar3 = { 3, 4, 15, 20, 30, 70, 80, 120 };
 
        Console.Write("Common elements are ");
 
        findCommon(ar1, ar2, ar3);
    }
}
 
// This code is contributed by Sam007.

PHP

<?php
// PHP program to print common elements
// in three arrays
 
// This function prints common elements
// in ar1
function findCommon( $ar1, $ar2, $ar3,
                         $n1, $n2, $n3)
{
     
    // Initialize starting indexes for
    // ar1[], ar2[] and ar3[]
    $i = 0; $j = 0; $k = 0;
 
    // Iterate through three arrays while
    // all arrays have elements
    while ($i < $n1 && $j < $n2 && $k < $n3)
    {
         
        // If x = y and y = z, print any
        // of them and move ahead in all
        // arrays
        if ($ar1[$i] == $ar2[$j] &&
                      $ar2[$j] == $ar3[$k])
        {
            echo $ar1[$i] , " ";
            $i++;
            $j++;
            $k++;
        }
 
        // x < y
        else if ($ar1[$i] < $ar2[$j])
            $i++;
 
        // y < z
        else if ($ar2[$j] < $ar3[$k])
            $j++;
 
        // We reach here when x > y and
        // z < y, i.e., z is smallest
        else
            $k++;
    }
}
 
// Driver program to test above function
    $ar1 = array(1, 5, 10, 20, 40, 80);
    $ar2 = array(6, 7, 20, 80, 100);
    $ar3 = array(3, 4, 15, 20, 30, 70,
                                80, 120);
    $n1 = count($ar1);
    $n2 = count($ar2);
    $n3 = count($ar3);
 
    echo "Common Elements are ";
     
    findCommon($ar1, $ar2, $ar3,$n1, $n2, $n3);
     
// This code is contributed by anuj_67.
?>

Javascript

<script>
 
      // JavaScript program to print
      // common elements in three arrays
 
      // This function prints common elements in ar1
      function findCommon(ar1, ar2, ar3, n1, n2, n3)
      {
       
        // Initialize starting indexes
        // for ar1[], ar2[] and ar3[]
        var i = 0,
          j = 0,
          k = 0;
 
        // Iterate through three arrays
        // while all arrays have elements
        while (i < n1 && j < n2 && k < n3)
        {
         
          // If x = y and y = z, print any of them and move ahead
          // in all arrays
          if (ar1[i] == ar2[j] && ar2[j] == ar3[k])
          {
            document.write(ar1[i] + " ");
            i++;
            j++;
            k++;
          }
 
          // x < y
          else if (ar1[i] < ar2[j]) i++;
           
          // y < z
          else if (ar2[j] < ar3[k]) j++;
           
          // We reach here when x > y and z < y, i.e., z is smallest
          else k++;
        }
      }
 
      // Driver code
      var ar1 = [1, 5, 10, 20, 40, 80];
      var ar2 = [6, 7, 20, 80, 100];
      var ar3 = [3, 4, 15, 20, 30, 70, 80, 120];
      var n1 = ar1.length;
      var n2 = ar2.length;
      var n3 = ar3.length;
 
      document.write("Common Elements are ");
      findCommon(ar1, ar2, ar3, n1, n2, n3);
       
      // This code is contributed by rdtank.
       
</script>

C++

// C++ program to print common elements in three arrays
#include <bits/stdc++.h>
using namespace std;
 
// This function prints common elements in ar1
void findCommon(int ar1[], int ar2[], int ar3[], int n1, int n2, int n3)
{
    // Initialize starting indexes for ar1[], ar2[] and ar3[]
    int i = 0, j = 0, k = 0;
 
    // Declare three variables prev1, prev2, prev3 to track
    // previous element
    int prev1, prev2, prev3;
 
    // Initialize prev1, prev2, prev3 with INT_MIN
    prev1 = prev2 = prev3 = INT_MIN;
 
    // Iterate through three arrays while all arrays have
    // elements
    while (i < n1 && j < n2 && k < n3) {
 
        // If ar1[i] = prev1 and i < n1, keep incrementing i
        while (ar1[i] == prev1 && i < n1)
            i++;
 
        // If ar2[j] = prev2 and j < n2, keep incrementing j
        while (ar2[j] == prev2 && j < n2)
            j++;
 
        // If ar3[k] = prev3 and k < n3, keep incrementing k
        while (ar3[k] == prev3 && k < n3)
            k++;
 
        // If x = y and y = z, print any of them, update
        // prev1 prev2, prev3 and move ahead in each array
        if (ar1[i] == ar2[j] && ar2[j] == ar3[k]) {
            cout << ar1[i] << " ";
            prev1 = ar1[i++];
            prev2 = ar2[j++];
            prev3 = ar3[k++];
        }
 
        // If x < y, update prev1 and increment i
        else if (ar1[i] < ar2[j])
            prev1 = ar1[i++];
 
        // If y < z, update prev2 and increment j
        else if (ar2[j] < ar3[k])
            prev2 = ar2[j++];
 
        // We reach here when x > y and z < y, i.e., z is
        // smallest update prev3 and imcrement k
        else
            prev3 = ar3[k++];
    }
}
 
// Driver code
int main()
{
    int ar1[] = { 1, 5, 10, 20, 40, 80, 80 };
    int ar2[] = { 6, 7, 20, 80, 80, 100 };
    int ar3[] = { 3, 4, 15, 20, 30, 70, 80, 80, 120 };
    int n1 = sizeof(ar1) / sizeof(ar1[0]);
    int n2 = sizeof(ar2) / sizeof(ar2[0]);
    int n3 = sizeof(ar3) / sizeof(ar3[0]);
 
    cout << "Common Elements are ";
    findCommon(ar1, ar2, ar3, n1, n2, n3);
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

C

// C program to print common elements in three arrays
#include <limits.h>
#include <stdio.h>
 
// This function prints common elements in ar1
void findCommon(int ar1[], int ar2[], int ar3[], int n1, int n2, int n3)
{
    // Initialize starting indexes for ar1[], ar2[] and ar3[]
    int i = 0, j = 0, k = 0;
 
    // Declare three variables prev1, prev2, prev3 to track
    // previous element
    int prev1, prev2, prev3;
 
    // Initialize prev1, prev2, prev3 with INT_MIN
    prev1 = prev2 = prev3 = INT_MIN;
 
    // Iterate through three arrays while all arrays have
    // elements
    while (i < n1 && j < n2 && k < n3) {
 
        // If ar1[i] = prev1 and i < n1, keep incrementing i
        while (ar1[i] == prev1 && i < n1)
            i++;
 
        // If ar2[j] = prev2 and j < n2, keep incrementing j
        while (ar2[j] == prev2 && j < n2)
            j++;
 
        // If ar3[k] = prev3 and k < n3, keep incrementing k
        while (ar3[k] == prev3 && k < n3)
            k++;
 
        // If x = y and y = z, print any of them, update
        // prev1 prev2, prev3 and move ahead in each array
        if (ar1[i] == ar2[j] && ar2[j] == ar3[k]) {
            printf("%d ", ar1[i]);
            prev1 = ar1[i++];
            prev2 = ar2[j++];
            prev3 = ar3[k++];
        }
 
        // If x < y, update prev1 and increment i
        else if (ar1[i] < ar2[j])
            prev1 = ar1[i++];
 
        // If y < z, update prev2 and increment j
        else if (ar2[j] < ar3[k])
            prev2 = ar2[j++];
 
        // We reach here when x > y and z < y, i.e., z is
        // smallest update prev3 and imcrement k
        else
            prev3 = ar3[k++];
    }
}
 
// Driver code
int main()
{
    int ar1[] = { 1, 5, 10, 20, 40, 80, 80 };
    int ar2[] = { 6, 7, 20, 80, 80, 100 };
    int ar3[] = { 3, 4, 15, 20, 30, 70, 80, 80, 120 };
    int n1 = sizeof(ar1) / sizeof(ar1[0]);
    int n2 = sizeof(ar2) / sizeof(ar2[0]);
    int n3 = sizeof(ar3) / sizeof(ar3[0]);
 
    printf("Common Elements are ");
    findCommon(ar1, ar2, ar3, n1, n2, n3);
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java program to find common
// elements in three arrays
class FindCommon {
 
    // This function prints common elements in ar1
    void findCommon(int ar1[], int ar2[], int ar3[])
    {
 
        // Initialize starting indexes for ar1[],
        // ar2[] and ar3[]
        int i = 0, j = 0, k = 0;
        int n1 = ar1.length;
        int n2 = ar2.length;
        int n3 = ar3.length;
 
        // Declare three variables prev1,
        // prev2, prev3 to track previous
        // element
        int prev1, prev2, prev3;
 
        // Initialize prev1, prev2,
        // prev3 with INT_MIN
        prev1 = prev2 = prev3 = Integer.MIN_VALUE;
 
        while (i < n1 && j < n2 && k < n3) {
 
            // If ar1[i] = prev1 and i < n1,
            // keep incrementing i
            while (i < n1 && ar1[i] == prev1)
                i++;
 
            // If ar2[j] = prev2 and j < n2,
            // keep incrementing j
            while (j < n2 && ar2[j] == prev2)
                j++;
 
            // If ar3[k] = prev3 and k < n3,
            // keep incrementing k
            while (k < n3 && ar3[k] == prev3)
                k++;
 
            if (i < n1 && j < n2 && k < n3) {
 
                // If x = y and y = z, print any of
                // them, update prev1 prev2, prev3
                // and move ahead in each array
                if (ar1[i] == ar2[j] && ar2[j] == ar3[k]) {
                    System.out.print(ar1[i] + " ");
                    prev1 = ar1[i];
                    prev2 = ar2[j];
                    prev3 = ar3[k];
                    i++;
                    j++;
                    k++;
                }
 
                // If x < y, update prev1
                // and increment i
                else if (ar1[i] < ar2[j]) {
                    prev1 = ar1[i];
                    i++;
                }
 
                // If y < z, update prev2
                // and increment j
                else if (ar2[j] < ar3[k]) {
                    prev2 = ar2[j];
                    j++;
                }
 
                // We reach here when x > y
                // and z < y, i.e., z is
                // smallest update prev3
                // and imcrement k
                else {
                    prev3 = ar3[k];
                    k++;
                }
            }
        }
    }
 
    // Driver code
    public static void main(String args[])
    {
        FindCommon ob = new FindCommon();
 
        int ar1[] = { 1, 5, 10, 20, 40, 80, 80 };
        int ar2[] = { 6, 7, 20, 80, 80, 100 };
        int ar3[] = { 3, 4, 15, 20, 30, 70, 80, 80, 120 };
 
        System.out.print("Common elements are ");
 
        ob.findCommon(ar1, ar2, ar3);
    }
}
 
// This code is contributed by rajsanghavi9.

Python3

# Python 3 program for above approach
import sys
 
# This function prints
# common elements in ar1
 
 
def findCommon(ar1, ar2, ar3, n1,
               n2, n3):
 
    # Initialize starting indexes
    # for ar1[], ar2and
    # ar3[]
    i = 0
    j = 0
    k = 0
 
    # Declare three variables prev1,
    # prev2, prev3 to track
    # previous element
    # Initialize prev1, prev2,
    # prev3 with INT_MIN
    prev1 = prev2 = prev3 = -sys.maxsize - 1
 
    # Iterate through three arrays
    # while all arrays have
    # elements
    while (i < n1 and j < n2 and k < n3):
 
        # If ar1[i] = prev1 and i < n1,
        # keep incrementing i
        while (ar1[i] == prev1 and i < n1-1):
            i += 1
 
        # If ar2[j] = prev2 and j < n2,
        # keep incrementing j
        while (ar2[j] == prev2 and j < n2):
            j += 1
 
        # If ar3[k] = prev3 and k < n3,
        # keep incrementing k
        while (ar3[k] == prev3 and k < n3):
            k += 1
 
        # If x = y and y = z, pr
        # any of them, update
        # prev1 prev2, prev3 and move
        # ahead in each array
        if (ar1[i] == ar2[j] and ar2[j] == ar3[k]):
            print(ar1[i], end=" ")
            prev1 = ar1[i]
            prev2 = ar2[j]
            prev3 = ar3[k]
            i += 1
            j += 1
            k += 1
 
        # If x < y, update prev1
        # and increment i
        elif (ar1[i] < ar2[j]):
            prev1 = ar1[i]
            i += 1
 
        # If y < z, update prev2
        # and increment j
        elif (ar2[j] < ar3[k]):
            prev2 = ar2[j]
            j += 1
 
        # We reach here when x > y
        # and z < y, i.e., z is
        # smallest update prev3
        # and imcrement k
        else:
            prev3 = ar3[k]
            k += 1
 
 
# Driver code
ar1 = [1, 5, 10, 20, 40, 80, 80]
ar2 = [6, 7, 20, 80, 80, 100]
ar3 = [3, 4, 15, 20, 30, 70, 80, 80, 120]
n1 = len(ar1)
n2 = len(ar2)
n3 = len(ar3)
 
print("Common Elements are ")
findCommon(ar1, ar2, ar3, n1, n2, n3)
 
# This code is contributed by splevel62.

C#

// C# program to find common
// elements in three arrays
using System;
class GFG {
 
    // This function prints common elements in ar1
    static void findCommon(int[] ar1, int[] ar2, int[] ar3)
    {
 
        // Initialize starting indexes for ar1[],
        // ar2[] and ar3[]
        int i = 0, j = 0, k = 0;
        int n1 = ar1.Length;
        int n2 = ar2.Length;
        int n3 = ar3.Length;
 
        // Declare three variables prev1,
        // prev2, prev3 to track previous
        // element
        int prev1, prev2, prev3;
 
        // Initialize prev1, prev2,
        // prev3 with INT_MIN
        prev1 = prev2 = prev3 = Int32.MinValue;
 
        while (i < n1 && j < n2 && k < n3) {
 
            // If ar1[i] = prev1 and i < n1,
            // keep incrementing i
            while (i < n1 && ar1[i] == prev1)
                i++;
 
            // If ar2[j] = prev2 and j < n2,
            // keep incrementing j
            while (j < n2 && ar2[j] == prev2)
                j++;
 
            // If ar3[k] = prev3 and k < n3,
            // keep incrementing k
            while (k < n3 && ar3[k] == prev3)
                k++;
 
            if (i < n1 && j < n2 && k < n3) {
 
                // If x = y and y = z, print any of
                // them, update prev1 prev2, prev3
                // and move ahead in each array
                if (ar1[i] == ar2[j] && ar2[j] == ar3[k]) {
                    Console.Write(ar1[i] + " ");
                    prev1 = ar1[i];
                    prev2 = ar2[j];
                    prev3 = ar3[k];
                    i++;
                    j++;
                    k++;
                }
 
                // If x < y, update prev1
                // and increment i
                else if (ar1[i] < ar2[j]) {
                    prev1 = ar1[i];
                    i++;
                }
 
                // If y < z, update prev2
                // and increment j
                else if (ar2[j] < ar3[k]) {
                    prev2 = ar2[j];
                    j++;
                }
 
                // We reach here when x > y
                // and z < y, i.e., z is
                // smallest update prev3
                // and imcrement k
                else {
                    prev3 = ar3[k];
                    k++;
                }
            }
        }
    }
 
    // Driver code
    public static void Main()
    {
        // FindCommon ob = new FindCommon();
        int[] ar1 = { 1, 5, 10, 20, 40, 80, 80 };
        int[] ar2 = { 6, 7, 20, 80, 80, 100 };
        int[] ar3 = { 3, 4, 15, 20, 30, 70, 80, 80, 120 };
 
        Console.Write("Common elements are ");
 
        findCommon(ar1, ar2, ar3);
    }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
 
// Javascript program to print common
// elements in three arrays
 
// This function prints
// common elements in ar1
function findCommon(ar1, ar2, ar3, n1,
                    n2, n3)
{
     
    // Initialize starting indexes
    // for ar1[], ar2[] and
    // ar3[]
    var i = 0, j = 0, k = 0;
 
    // Declare three variables prev1,
    // prev2, prev3 to track
    // previous element
    var prev1, prev2, prev3;
 
    // Initialize prev1, prev2,
    // prev3 with INT_MIN
    prev1 = prev2 = prev3 = -1000000000;
 
    // Iterate through three arrays
    // while all arrays have
    // elements
    while (i < n1 && j < n2 && k < n3) {
     
        // If ar1[i] = prev1 and i < n1,
        // keep incrementing i
        while (ar1[i] == prev1 && i < n1)
            i++;
 
        // If ar2[j] = prev2 and j < n2,
        // keep incrementing j
        while (ar2[j] == prev2 && j < n2)
            j++;
 
        // If ar3[k] = prev3 and k < n3,
        // keep incrementing k
        while (ar3[k] == prev3 && k < n3)
            k++;
 
        // If x = y and y = z, print
        // any of them, update
        // prev1 prev2, prev3 and move
        //ahead in each array
        if (ar1[i] == ar2[j] && ar2[j] == ar3[k]) {
            document.write(ar1[i] + " ");
            prev1 = ar1[i];
            prev2 = ar2[j];
            prev3 = ar3[k];
            i++;
            j++;
            k++;
        }
 
        // If x < y, update prev1
        // and increment i
        else if (ar1[i] < ar2[j]) {
            prev1 = ar1[i];
            i++;
        }
 
        // If y < z, update prev2
        // and increment j
        else if (ar2[j] < ar3[k]) {
            prev2 = ar2[j];
            j++;
        }
 
        // We reach here when x > y
        // and z < y, i.e., z is
        // smallest update prev3
        // and imcrement k
        else {
            prev3 = ar3[k];
            k++;
        }
    }
}
 
// Driver code
var ar1 = [ 1, 5, 10, 20, 40, 80, 80 ];
var ar2 = [ 6, 7, 20, 80, 80, 100 ];
var ar3 = [ 3, 4, 15, 20, 30, 70, 80, 80, 120 ];
var n1 = ar1.length;
var n2 = ar2.length;
var n3 = ar3.length;
document.write("Common Elements are ");
findCommon(ar1, ar2, ar3, n1, n2, n3);
 
 
</script>

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
void commonElements(vector<int>arr1,vector<int>arr2,vector<int>arr3 ,int n1 ,int n2 ,int n3)
{
 
    // creating a max variable
    // for storing the maximum
    // value present in the all
    // the three array
    // this will be the size of
    // array for calculating the
    // frequency of each element
    // present in all the array
    int Max = INT_MIN;
 
    // deleting duplicates in linear time
    // for arr1
    int res1 = 1;
    for (int i = 1; i < n1; i++) {
        Max = max(arr1[i], Max);
        if (arr1[i] != arr1[res1 - 1]) {
            arr1[res1] = arr1[i];
            res1++;
        }
    }
 
    // deleting duplicates in linear time
    // for arr2
    int res2 = 1;
    for (int i = 1; i < n2; i++) {
        Max = max(arr2[i], Max);
        if (arr2[i] != arr2[res2 - 1]) {
            arr2[res2] = arr2[i];
            res2++;
        }
    }
 
        // deleting duplicates in linear time
        // for arr3
    int res3 = 1;
    for (int i = 1; i < n3; i++) {
        Max = max(arr3[i], Max);
        if (arr3[i] != arr3[res3 - 1]) {
            arr3[res3] = arr3[i];
            res3++;
        }
    }
 
    // creating an array for finding frequency
    vector<int>freq(Max + 1,0);
 
    // calculating the frequency of
    // all the elements present in
    // all the array
    for (int i = 0; i < res1; i++)
        freq[arr1[i]]++;
    for (int i = 0; i < res2; i++)
        freq[arr2[i]]++;
    for (int i = 0; i < res3; i++)
        freq[arr3[i]]++;
 
    // iterating till max and
    // whenever the frequency of element
    // will be three we print that element
    for (int i = 0; i <= Max; i++){
        if (freq[i] == 3){
            cout<<i<<" ";
        }
    }
}
 
// Driver Code
int main(){
 
vector<int>arr1 = { 1, 5, 10, 20, 40, 80 };
vector<int>arr2 = { 6, 7, 20, 80, 100 };
vector<int>arr3 = { 3, 4, 15, 20, 30, 70, 80, 120 };
 
commonElements(arr1, arr2, arr3, 6, 5, 8);
 
}
 
// This code is contributed by shinjanpatra

Java

// Java implementation of the above approach
 
class GFG {
    public static void commonElements(int[] arr1,
                                      int[] arr2,
                                      int[] arr3, int n1,
                                      int n2, int n3)
    {
        // creating a max variable
        // for storing the maximum
        // value present in the all
        // the three array
        // this will be the size of
        // array for calculating the
        // frequency of each element
        // present in all the array
        int max = Integer.MIN_VALUE;
 
        // deleting duplicates in linear time
        // for arr1
        int res1 = 1;
        for (int i = 1; i < n1; i++) {
            max = Math.max(arr1[i], max);
            if (arr1[i] != arr1[res1 - 1]) {
                arr1[res1] = arr1[i];
                res1++;
            }
        }
 
        // deleting duplicates in linear time
        // for arr2
        int res2 = 1;
        for (int i = 1; i < n2; i++) {
            max = Math.max(arr2[i], max);
            if (arr2[i] != arr2[res2 - 1]) {
                arr2[res2] = arr2[i];
                res2++;
            }
        }
 
        // deleting duplicates in linear time
        // for arr3
        int res3 = 1;
        for (int i = 1; i < n3; i++) {
            max = Math.max(arr3[i], max);
            if (arr3[i] != arr3[res3 - 1]) {
                arr3[res3] = arr3[i];
                res3++;
            }
        }
 
        // creating an array for finding frequency
        int[] freq = new int[max + 1];
 
        // calculating the frequency of
        // all the elements present in
        // all the array
        for (int i = 0; i < res1; i++)
            freq[arr1[i]]++;
        for (int i = 0; i < res2; i++)
            freq[arr2[i]]++;
        for (int i = 0; i < res3; i++)
            freq[arr3[i]]++;
 
        // iterating till max and
        // whenever the frequency of element
        // will be three we print that element
        for (int i = 0; i <= max; i++)
            if (freq[i] == 3)
                System.out.print(i + " ");
    }
 
    // Driver Code
    public static void main(String[] arg)
    {
 
        int arr1[] = { 1, 5, 10, 20, 40, 80 };
        int arr2[] = { 6, 7, 20, 80, 100 };
        int arr3[] = { 3, 4, 15, 20, 30, 70, 80, 120 };
 
        commonElements(arr1, arr2, arr3, 6, 5, 8);
    }
}

Python3

# Python implementation of the above approach   
import sys
def commonElements(arr1,  arr2,  arr3 , n1 , n2 , n3):
 
    # creating a max variable
    # for storing the maximum
    # value present in the all
    # the three array
    # this will be the size of
    # array for calculating the
    # frequency of each element
    # present in all the array
    Max = -sys.maxsize -1
 
    # deleting duplicates in linear time
    # for arr1
    res1 = 1
    for i in range(1, n1):
        Max = max(arr1[i], Max)
        if arr1[i] != arr1[res1 - 1]:
            arr1[res1] = arr1[i]
            res1 += 1
 
    # deleting duplicates in linear time
    # for arr2
    res2 = 1
    for i in range(1, n2):
        Max = max(arr2[i], Max)
        if (arr2[i] != arr2[res2 - 1]):
            arr2[res2] = arr2[i]
            res2 += 1
 
    # deleting duplicates in linear time
    # for arr3
    res3 = 1
    for i in range(1, n3):
        Max = max(arr3[i], Max)
        if (arr3[i] != arr3[res3 - 1]):
            arr3[res3] = arr3[i]
            res3 += 1
 
    # creating an array for finding frequency
    freq = [0 for i in range(Max + 1)]
 
    # calculating the frequency of
    # all the elements present in
    # all the array
    for i in range(res1):
        freq[arr1[i]] += 1
    for i in range(res2):
        freq[arr2[i]] += 1
    for i in range(res3):
        freq[arr3[i]] += 1
 
    # iterating till max and
    # whenever the frequency of element
    # will be three we print that element
    for i in range(Max + 1):
        if freq[i] == 3:
            print(i,end = " ")
 
# Driver Code
arr1 = [ 1, 5, 10, 20, 40, 80 ]
arr2 = [ 6, 7, 20, 80, 100 ]
arr3 = [ 3, 4, 15, 20, 30, 70, 80, 120 ]
 
commonElements(arr1, arr2, arr3, 6, 5, 8)
 
# This code is contributed by shinjanpatra

C#

// C# implementation of the above approach
using System;
class GFG {
    public static void commonElements(int[] arr1,
                                      int[] arr2,
                                      int[] arr3, int n1,
                                      int n2, int n3)
    {
        // creating a max variable
        // for storing the maximum
        // value present in the all
        // the three array
        // this will be the size of
        // array for calculating the
        // frequency of each element
        // present in all the array
        int max = int.MinValue;
 
        // deleting duplicates in linear time
        // for arr1
        int res1 = 1;
        for (int i = 1; i < n1; i++) {
            max = Math.Max(arr1[i], max);
            if (arr1[i] != arr1[res1 - 1]) {
                arr1[res1] = arr1[i];
                res1++;
            }
        }
 
        // deleting duplicates in linear time
        // for arr2
        int res2 = 1;
        for (int i = 1; i < n2; i++) {
            max = Math.Max(arr2[i], max);
            if (arr2[i] != arr2[res2 - 1]) {
                arr2[res2] = arr2[i];
                res2++;
            }
        }
 
        // deleting duplicates in linear time
        // for arr3
        int res3 = 1;
        for (int i = 1; i < n3; i++) {
            max = Math.Max(arr3[i], max);
            if (arr3[i] != arr3[res3 - 1]) {
                arr3[res3] = arr3[i];
                res3++;
            }
        }
 
        // creating an array for finding frequency
        int[] freq = new int[max + 1];
 
        // calculating the frequency of
        // all the elements present in
        // all the array
        for (int i = 0; i < res1; i++)
            freq[arr1[i]]++;
        for (int i = 0; i < res2; i++)
            freq[arr2[i]]++;
        for (int i = 0; i < res3; i++)
            freq[arr3[i]]++;
 
        // iterating till max and
        // whenever the frequency of element
        // will be three we print that element
        for (int i = 0; i <= max; i++)
            if (freq[i] == 3)
                Console.Write(i + " ");
    }
 
    // Driver Code
    public static void Main()
    {
 
        int[] arr1 = { 1, 5, 10, 20, 40, 80 };
        int[] arr2 = { 6, 7, 20, 80, 100 };
        int[] arr3 = { 3, 4, 15, 20, 30, 70, 80, 120 };
 
        commonElements(arr1, arr2, arr3, 6, 5, 8);
    }
}

Javascript

<script>
// javascript implementation of the above approach   
function commonElements(arr1,  arr2,  arr3 , n1 , n2 , n3)
{
 
        // creating a max variable
        // for storing the maximum
        // value present in the all
        // the three array
        // this will be the size of
        // array for calculating the
        // frequency of each element
        // present in all the array
        var max = Number.MIN_VALUE;
 
        // deleting duplicates in linear time
        // for arr1
        var res1 = 1;
        for (var i = 1; i < n1; i++) {
            max = Math.max(arr1[i], max);
            if (arr1[i] != arr1[res1 - 1]) {
                arr1[res1] = arr1[i];
                res1++;
            }
        }
 
        // deleting duplicates in linear time
        // for arr2
        var res2 = 1;
        for (var i = 1; i < n2; i++) {
            max = Math.max(arr2[i], max);
            if (arr2[i] != arr2[res2 - 1]) {
                arr2[res2] = arr2[i];
                res2++;
            }
        }
 
        // deleting duplicates in linear time
        // for arr3
        var res3 = 1;
        for (var i = 1; i < n3; i++) {
            max = Math.max(arr3[i], max);
            if (arr3[i] != arr3[res3 - 1]) {
                arr3[res3] = arr3[i];
                res3++;
            }
        }
 
        // creating an array for finding frequency
        var freq = Array(max + 1).fill(0);
 
        // calculating the frequency of
        // all the elements present in
        // all the array
        for (i = 0; i < res1; i++)
            freq[arr1[i]]++;
        for (i = 0; i < res2; i++)
            freq[arr2[i]]++;
        for (i = 0; i < res3; i++)
            freq[arr3[i]]++;
 
        // iterating till max and
        // whenever the frequency of element
        // will be three we print that element
        for (i = 0; i <= max; i++)
            if (freq[i] == 3)
                document.write(i + " ");
    }
 
    // Driver Code
 
        var arr1 = [ 1, 5, 10, 20, 40, 80 ];
        var arr2 = [ 6, 7, 20, 80, 100 ];
        var arr3 = [ 3, 4, 15, 20, 30, 70, 80, 120 ];
 
        commonElements(arr1, arr2, arr3, 6, 5, 8);
 
// This code is contributed by Rajput-Ji
</script>

C++

#include <bits/stdc++.h>
using namespace std;
 
void findCommon(int a[], int b[], int c[], int n1, int n2,
                int n3)
{
    // three sets to maintain frequency of elements
    unordered_set<int> uset, uset2, uset3;
    for (int i = 0; i < n1; i++) {
        uset.insert(a[i]);
    }
    for (int i = 0; i < n2; i++) {
        uset2.insert(b[i]);
    }
    // checking if elements of 3rd array are present in
    // first 2 sets
    for (int i = 0; i < n3; i++) {
        if (uset.find(c[i]) != uset.end()
            && uset2.find(c[i]) != uset.end()) {
            // using a 3rd set to prevent duplicates
            if (uset3.find(c[i]) == uset3.end())
                cout << c[i] << " ";
            uset3.insert(c[i]);
        }
    }
}
 
// Driver code
int main()
{
    int ar1[] = { 1, 5, 10, 20, 40, 80 };
    int ar2[] = { 6, 7, 20, 80, 100 };
    int ar3[] = { 3, 4, 15, 20, 30, 70, 80, 120 };
    int n1 = sizeof(ar1) / sizeof(ar1[0]);
    int n2 = sizeof(ar2) / sizeof(ar2[0]);
    int n3 = sizeof(ar3) / sizeof(ar3[0]);
 
    cout << "Common Elements are " << endl;
    findCommon(ar1, ar2, ar3, n1, n2, n3);
    return 0;
}

Java

/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
    static void findCommon(int a[], int b[], int c[], int n1, int n2,int n3)
{
    // three sets to maintain frequency of elements
    HashSet<Integer> uset = new HashSet<>();
    HashSet<Integer> uset2 = new HashSet<>();
    HashSet<Integer> uset3 = new HashSet<>();
    for (int i = 0; i < n1; i++) {
        uset.add(a[i]);
    }
    for (int i = 0; i < n2; i++) {
        uset2.add(b[i]);
    }
    // checking if elements of 3rd array are present in
    // first 2 sets
    for (int i = 0; i < n3; i++) {
        if (uset.contains(c[i]) && uset2.contains(c[i])) {
            // using a 3rd set to prevent duplicates
            if (uset3.contains(c[i]) == false)
                System.out.print(c[i]+" ");
            uset3.add(c[i]);
        }
    }
}
 
// Driver Code
public static void main(String args[])
{
    int ar1[] = { 1, 5, 10, 20, 40, 80 };
    int ar2[] = { 6, 7, 20, 80, 100 };
    int ar3[] = { 3, 4, 15, 20, 30, 70, 80, 120 };
    int n1 = ar1.length;
    int n2 = ar2.length;
    int n3 = ar3.length;
 
    System.out.println("Common Elements are ");
    findCommon(ar1, ar2, ar3, n1, n2, n3);
}
}
 
// This code is contributed by shinjanpatra

Python3

# Python implementation of the approach
def findCommon(a, b, c, n1, n2, n3):
 
    # three sets to maintain frequency of elements
    uset = set()
    uset2 = set()
    uset3 = set()
    for i in range(n1):
        uset.add(a[i])
 
    for i in range(n2):
        uset2.add(b[i])
 
    # checking if elements of 3rd array are present in first 2 sets
    for i in range(n3):
 
        if(c[i] in uset and c[i] in uset2):
 
            # using a 3rd set to prevent duplicates
            if c[i] not in uset3:
                print(c[i], end = " ")
            uset3.add(c[i])
 
# Driver code
ar1 = [ 1, 5, 10, 20, 40, 80 ]
ar2 = [ 6, 7, 20, 80, 100 ]
ar3 = [ 3, 4, 15, 20, 30, 70, 80, 120 ]
n1 = len(ar1)
n2 = len(ar2)
n3 = len(ar3)
 
print("Common Elements are ")
findCommon(ar1, ar2, ar3, n1, n2, n3)
 
# This code is contributed by shinjanpatra.

Javascript

<script>
 
function findCommon(a, b, c, n1, n2, n3)
{
 
    // three sets to maintain frequency of elements
    let uset = new Set();
    let uset2 = new Set();
    let uset3 = new Set();
    for(let i=0;i<n1;i++){
        uset.add(a[i]);
    }
     for(let i=0;i<n2;i++){
        uset2.add(b[i]);
    }
      // checking if elements of 3rd array are present in first 2 sets
    for(let i=0;i<n3;i++)
    {
        if(uset.has(c[i]) == true && uset2.has(c[i]) == true)
        {
            // using a 3rd set to prevent duplicates
            if(uset3.has(c[i]) == false)
                document.write(c[i]," ");
            uset3.add(c[i]);
        }
    }          
}
 
// Driver code
let ar1 = [ 1, 5, 10, 20, 40, 80 ];
let ar2 = [ 6, 7, 20, 80, 100 ];
let ar3 = [ 3, 4, 15, 20, 30, 70, 80, 120 ];
let n1 = ar1.length;
let n2 = ar2.length;
let n3 = ar3.length;
 
document.write("Common Elements are ","</br>");
findCommon(ar1, ar2, ar3, n1, n2, n3);
 
// This code is contributed by shinjanpatra.
</script>

C#

using System;
using System.Collections.Generic;
 
public class GFG {
    static void findCommon(int[] a, int[] b, int[] c,
                           int n1, int n2, int n3)
    {
        // three sets to maintain frequency of elements
        HashSet<int> uset = new HashSet<int>();
        HashSet<int> uset2 = new HashSet<int>();
        HashSet<int> uset3 = new HashSet<int>();
        for (int i = 0; i < n1; i++) {
            uset.Add(a[i]);
        }
        for (int i = 0; i < n2; i++) {
            uset2.Add(b[i]);
        }
        // checking if elements of 3rd array are present in
        // first 2 sets
        for (int i = 0; i < n3; i++) {
            if (uset.Contains(c[i])
                && uset2.Contains(c[i])) {
                // using a 3rd set to prevent duplicates
                if (!uset3.Contains(c[i])) {
                    Console.Write(c[i]);
                    Console.Write(" ");
                }
                uset3.Add(c[i]);
            }
        }
    }
 
    // Driver code
    public static void Main()
    {
        int[] ar1 = { 1, 5, 10, 20, 40, 80 };
        int[] ar2 = { 6, 7, 20, 80, 100 };
        int[] ar3 = { 3, 4, 15, 20, 30, 70, 80, 120 };
 
        int n1 = ar1.Length;
        int n2 = ar2.Length;
        int n3 = ar3.Length;
 
        Console.Write("Common Elements are ");
        Console.Write("\n");
        findCommon(ar1, ar2, ar3, n1, n2, n3);
    }
}
// This code is contributed by Aarti_Rathi

C++

#include <bits/stdc++.h>
using namespace std;
 
bool binary_search(int arr[], int n, int element)
{
    int l = 0, h = n - 1;
    while (l <= h) {
        int mid = (l + h) / 2;
        if (arr[mid] == element) {
            return true;
        }
        else if (arr[mid] > element) {
            h = mid - 1;
        }
        else {
            l = mid + 1;
        }
    }
    return false;
}
void findCommon(int a[], int b[], int c[], int n1, int n2,
                int n3)
{
    // Iterate on first array
    for (int j = 0; j < n1; j++) {
        if (j != 0 && a[j] == a[j - 1]) {
            continue;
        }
        // check if the element is present in 2nd and 3rd
        // array.
        if (binary_search(b, n2, a[j])
            && binary_search(c, n3, a[j])) {
            cout << a[j] << " ";
        }
    }
}
 
// Driver code
int main()
{
    int ar1[] = { 1, 5, 10, 20, 40, 80 };
    int ar2[] = { 6, 7, 20, 80, 100 };
    int ar3[] = { 3, 4, 15, 20, 30, 70, 80, 120 };
    int n1 = sizeof(ar1) / sizeof(ar1[0]);
    int n2 = sizeof(ar2) / sizeof(ar2[0]);
    int n3 = sizeof(ar3) / sizeof(ar3[0]);
 
    cout << "Common Elements are " << endl;
    findCommon(ar1, ar2, ar3, n1, n2, n3);
    return 0;
}
 
// This code is contributed by Anchal Agarwal

Java

// Java program to implement
// above approach
public class Main
{
 
    public static boolean binary_search(int arr[], int n, int element)
{
    int l = 0, h = n - 1;
    while (l <= h) {
        int mid = (l + h) / 2;
        if (arr[mid] == element) {
            return true;
        }
        else if (arr[mid] > element) {
            h = mid - 1;
        }
        else {
            l = mid + 1;
        }
    }
    return false;
 }
     
    public static void findCommon(int a[], int b[], int c[], int n1, int n2,int n3)
    {
        // Iterate on first array
        for (int j = 0; j < n1; j++)
        {
           if (j != 0 && a[j] == a[j - 1]) {
                continue;
            }
            // check if the element is present in 2nd and 3rd
            // array.
            if (binary_search(b, n2, a[j]) && binary_search(c, n3, a[j])) {
                 
                System.out.print(a[j] + " ");
                 
            }
        }
    }
 
    /* Driver code */
public static void main(String[] args)
    {
 
        int ar1[] = { 1, 5, 10, 20, 40, 80 };
        int ar2[] = { 6, 7, 20, 80, 100 };
        int ar3[] = { 3, 4, 15, 20, 30, 70, 80, 120 };
        int n1 = ar1.length;
        int n2 = ar2.length;
        int n3 = ar3.length;
         System.out.println("Common elements are ");
        // function calling
        findCommon(ar1, ar2, ar3, n1, n2, n3);
    }
}
 
//this code is contributed by Machhaliya Muhammad

Python3

# Python program to Find all range
# Having set bit sum X in array
 
 
def binary_search(arr, n, element):
 
    l,h = 0,n - 1
    while (l <= h):
        mid = (l + h) // 2
        if (arr[mid] == element):
            return True
 
        elif (arr[mid] > element):
            h = mid - 1
 
        else:
            l = mid + 1
 
    return False
 
def findCommon(a, b, c, n1, n2, n3):
 
    # Iterate on first array
    for j in range(n1):
        if (j != 0 and a[j] == a[j - 1]):
            continue
     
        # check if the element is present in 2nd and 3rd
        # array.
        if (binary_search(b, n2, a[j]) and binary_search(c, n3, a[j])):
            print(a[j],end=" ")
 
 
# Driver code
 
ar1 = [ 1, 5, 10, 20, 40, 80 ]
ar2 = [ 6, 7, 20, 80, 100 ]
ar3 = [ 3, 4, 15, 20, 30, 70, 80, 120 ]
n1 = len(ar1)
n2 = len(ar2)
n3 = len(ar3)
 
print("Common Elements are ")
findCommon(ar1, ar2, ar3, n1, n2, n3)
 
 
# This code is contributed by shinjanpatra

Javascript

<script>
function binary_search(arr, n, element)
{
    let l = 0, h = n - 1
    while (l <= h) {
        let mid = Math.floor((l + h) / 2)
        if (arr[mid] == element) {
            return true
        }
        else if (arr[mid] > element) {
            h = mid - 1
        }
        else {
            l = mid + 1
        }
    }
    return false
}
function findCommon(a, b, c, n1, n2, n3)
{
    // Iterate on first array
    for (let j = 0; j < n1; j++) {
        if (j != 0 && a[j] == a[j - 1]) {
            continue;
        }
        // check if the element is present in 2nd and 3rd
        // array.
        if (binary_search(b, n2, a[j])
            && binary_search(c, n3, a[j])) {
            document.write(a[j]," ");
        }
    }
}
 
// Driver code
 
let ar1 = [ 1, 5, 10, 20, 40, 80 ]
let ar2 = [ 6, 7, 20, 80, 100 ]
let ar3 = [ 3, 4, 15, 20, 30, 70, 80, 120 ]
let n1 = ar1.length
let n2 = ar2.length
let n3 = ar3.length
 
document.write("Common Elements are ","</br>")
findCommon(ar1, ar2, ar3, n1, n2, n3)
 
// This code is contributed by shinjanpatra
 
</script>

C#

// C# program to implement
// above approach
 
using System;
 
public class GFG
{
 
    public static bool binary_search(int[] arr, int n, int element)
    {
        int l = 0, h = n - 1;
        while (l <= h)
        {
            int mid = (l + h) / 2;
            if (arr[mid] == element)
            {
                return true;
            }
            else if (arr[mid] > element)
            {
                h = mid - 1;
            }
            else
            {
                l = mid + 1;
            }
        }
        return false;
    }
 
    public static void findCommon(int[] a, int[] b, int[] c, int n1, int n2, int n3)
    {
        // Iterate on first array
        for (int j = 0; j < n1; j++)
        {
            if (j != 0 && a[j] == a[j - 1])
            {
                continue;
            }
            // check if the element is present in 2nd and 3rd
            // array.
            if (binary_search(b, n2, a[j]) && binary_search(c, n3, a[j]))
            {
 
                Console.Write(a[j] + " ");
 
            }
        }
    }
 
    /* Driver code */
    public static void Main()
    {
 
        int[] ar1 = { 1, 5, 10, 20, 40, 80 };
        int[] ar2 = { 6, 7, 20, 80, 100 };
        int[] ar3 = { 3, 4, 15, 20, 30, 70, 80, 120 };
        int n1 = ar1.Length;
        int n2 = ar2.Length;
        int n3 = ar3.Length;
        Console.WriteLine("Common elements are ");
        // function calling
        findCommon(ar1, ar2, ar3, n1, n2, n3);
    }
}
 
//this code is contributed by Saurabh Jaiswal

Publicación traducida automáticamente

Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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