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