Given 3 arrays, arr[], brr[], and crr[], the task is to find the common elements in at least 2 arrays out of the given 3 arrays
Examples:
Input: arr[] = {1, 1, 3, 2, 4}, brr[] = {2, 3, 5}, crr[] = {3, 6}
Output: {2, 3}
Explanation: Elements 2 and 3 appear in atleast 2 arraysInput: arr[] = {3, 1}, brr[] = {2, 3}, crr[] = {1, 2}
Output: {2, 3, 1}
Approach: The task can be solved with the help of HashSet Follow the below steps to solve the problem:
- Create three hashsets (s1, s2, and s3) to store distinct elements of the three arrays and also one HashSet to store distinct elements of all three arrays.
- Iterate in set and check for the common elements in atleast two sets(s1, s2), (s1, s3) and (s2, s3).
- If an element satisfies the above condition, print that element.
Below is the implementation of the above code:
C++
// 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 Codeint 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
// Java implementation of above approachimport 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
# Python3 implementation of above approachdef 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 Codeif __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#
// C# implementation of above approachusing 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. |
Javascript
<script>// Javascript implementation of above approachfunction 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 Codelet 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> |
Output
2 3
Time Complexity: O(n1+n2+n3) (where n1, n2 and n3 are the length of given arrays)
Auxiliary Space: O(n1+n2+n3)
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