Encuentre números presentes en al menos dos de las tres arrays

Dados 3 arreglos , arr[] , brr[] y crr[] , la tarea es encontrar los elementos comunes en al menos 2 arreglos de los 3 arreglos dados
Ejemplos :

Entrada :arr[] = {1, 1, 3, 2, 4}, brr[] = {2, 3, 5}, crr[] = {3, 6}
Salida : {2, 3}
Explicación : Elementos 2 y 3 aparecen en al menos 2 arreglos

Entrada :arr[] = {3, 1}, brr[] = {2, 3}, crr[] = {1, 2}
Salida :{2, 3, 1}

Enfoque : la tarea se puede resolver con la ayuda de HashSet Siga los pasos a continuación para resolver el problema:

• Cree tres hashsets (s1, s2 y s3) para almacenar distintos elementos de las tres arrays y también un HashSet para almacenar distintos elementos de las tres arrays.
• Iterar en el conjunto y verificar los elementos comunes en al menos dos conjuntos (s1, s2), (s1, s3) y (s2, s3).
• Si un elemento cumple la condición anterior, imprima ese elemento.
   

A continuación se muestra la implementación del código anterior:

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
void get(vector<int>& p, vector<int>& c, vector<int>& m)
{
 
    // Store distinct elements of array arr[]
    set<int> s1;
 
    // Store distinct elements of array brr[]
    set<int> s2;
 
    // Store distinct elements of array crr[]
    set<int> s3;
 
    // Store the distinct elements
    // of all three arrays
    set<int> set;
    for (auto i : p) {
        s1.insert(i);
        set.insert(i);
    }
 
    for (auto i : c) {
        s2.insert(i);
        set.insert(i);
    }
 
    for (int i : m) {
        s3.insert(i);
        set.insert(i);
    }
 
    // Stores the common elements
    vector<int> result;
    for (auto i : set) {
        if (s1.count(i) && s2.count(i)
            || s2.count(i) && s3.count(i)
            || s1.count(i) && s3.count(i))
            cout << i << " ";
    }
}
 
// Driver Code
int main()
{
    vector<int> arr = { 1, 1, 3, 2, 4 };
    vector<int> brr = { 2, 3, 5 };
    vector<int> crr = { 3, 6 };
    get(arr, brr, crr);
 
    return 0;
}
 
    // This code is contributed by rakeshsahni

// Java implementation of above approach
 
import java.io.*;
import java.util.*;
 
class GFG {
    public static void get(
        int[] p, int[] c, int[] m)
    {
 
        // Store distinct elements of array arr[]
        Set<Integer> s1 = new HashSet<>();
 
        // Store distinct elements of array brr[]
        Set<Integer> s2 = new HashSet<>();
 
        // Store distinct elements of array crr[]
        Set<Integer> s3 = new HashSet<>();
 
        // Store the distinct elements
        // of all three arrays
        Set<Integer> set = new HashSet<>();
        for (int i : p) {
            s1.add(i);
            set.add(i);
        }
 
        for (int i : c) {
            s2.add(i);
            set.add(i);
        }
 
        for (int i : m) {
            s3.add(i);
            set.add(i);
        }
 
        // Stores the common elements
        List<Integer> result
            = new ArrayList<>();
        for (int i : set) {
            if (s1.contains(i)
                    && s2.contains(i)
                || s2.contains(i)
                       && s3.contains(i)
                || s1.contains(i)
                       && s3.contains(i))
                System.out.print(i + " ");
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 1, 1, 3, 2, 4 };
        int[] brr = { 2, 3, 5 };
        int[] crr = { 3, 6 };
        get(arr, brr, crr);
    }
}

# Python3 implementation of above approach
def get(p, c, m) :
 
    # Store distinct elements of array arr[]
    s1 = set();
 
    # Store distinct elements of array brr[]
    s2 = set();
 
    # Store distinct elements of array crr[]
    s3 = set();
 
    # Store the distinct elements
    # of all three arrays
    set_obj = set();
     
    for i in p :
        s1.add(i);
        set_obj.add(i);
 
    for i in c :
        s2.add(i);
        set_obj.add(i);
 
    for i in m :
        s3.add(i);
        set_obj.add(i);
 
    # Stores the common elements
    result = [];
    for i in set_obj :
        if (list(s1).count(i) and list(s2).count(i)
            or list(s2).count(i) and list(s3).count(i)
            or list(s1).count(i) and list(s3).count(i)) :
            print(i,end= " ");
   
# Driver Code
if __name__ == "__main__" :
 
    arr = [ 1, 1, 3, 2, 4 ];
    brr = [ 2, 3, 5 ];
    crr = [ 3, 6 ];
    get(arr, brr, crr);
 
    # This code is contributed by AnkThon

// C# implementation of above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG {
    static void get(
        int[] p, int[] c, int[] m)
    {
 
        // Store distinct elements of array arr[]
        HashSet<int> s1 = new HashSet<int>();
 
        // Store distinct elements of array brr[]
        HashSet<int> s2 = new HashSet<int>();
 
        // Store distinct elements of array crr[]
        HashSet<int> s3 = new HashSet<int>();
 
        // Store the distinct elements
        // of all three arrays
        HashSet<int> set = new HashSet<int>();
        foreach (int i in p) {
            s1.Add(i);
            set.Add(i);
        }
 
        foreach (int i in c) {
            s2.Add(i);
            set.Add(i);
        }
 
        foreach (int i in m) {
            s3.Add(i);
            set.Add(i);
        }
 
        // Stores the common elements
        ArrayList result
            = new ArrayList();
        foreach (int i in set) {
            if (s1.Contains(i)
                    && s2.Contains(i)
                || s2.Contains(i)
                       && s3.Contains(i)
                || s1.Contains(i)
                       && s3.Contains(i))
                Console.Write(i + " ");
        }
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 1, 1, 3, 2, 4 };
        int[] brr = { 2, 3, 5 };
        int[] crr = { 3, 6 };
        get(arr, brr, crr);
    }
}
 
// This code is contributed by Samim Hossain Mondal.

<script>
// Javascript implementation of above approach
function get(p, c, m)
{
 
  // Store distinct elements of array arr[]
  let s1 = new Set();
 
  // Store distinct elements of array brr[]
  let s2 = new Set();
 
  // Store distinct elements of array crr[]
  let s3 = new Set();
 
  // Store the distinct elements
  // of all three arrays
  let set = new Set();
  for (i of p) {
    s1.add(i);
    set.add(i);
  }
 
  for (i of c) {
    s2.add(i);
    set.add(i);
  }
 
  for (i of m) {
    s3.add(i);
    set.add(i);
  }
 
  // Stores the common elements
  let result = [];
  for (i of [...set].reverse()) {
    if (s1.has(i) && s2.has(i)
      || s2.has(i) && s3.has(i)
      || s1.has(i) && s3.has(i))
      document.write(i + " ");
  }
}
 
// Driver Code
let arr = [1, 1, 3, 2, 4];
let brr = [2, 3, 5];
let crr = [3, 6];
get(arr, brr, crr);
 
// This code is contributed by gfgking
</script>

2 3 

Complejidad de tiempo : O(n1+n2+n3) (donde n1, n2 y n3 son la longitud de las arrays dadas)
Espacio auxiliar : O(n1+n2+n3)