Intersection of Two Arrays with Distinct Elements

Last Updated : 04 Oct, 2024

Given two arrays a[] and b[] with distinct elements of size n and m respectively, the task is to find intersection (or common elements) of the two arrays. We can return the answer in any order.

Note: Intersection of two arrays can be defined as a set containing distinct common elements between the two arrays.

Examples:

Input: a[] = { 5, 6, 2, 1, 4 }, b[] = { 7, 9, 4, 2 }
Output: { 2, 4 }
Explanation: The only common elements in both arrays are 2 and 4.

Input: a[] = { 4, 5, 2, 3 } , b[] = { 1, 7 }
Output: { }
Explanation: There are no common elements in array a[] and b[]

Table of Content

[Naive Approach] Using nested loop - O(n*m) Time and O(1) Space

The idea is to traverse the first array a[] and for each element from a[], check whether it is present in array b[]. If present then add this element to result array.

C++

// C++ program for intersection of two arrays with // distinct elements using nested loops  #include <iostream> #include <vector> using namespace std;  vector<int> intersection(vector<int>& a, vector<int>& b) {     vector<int> res;      // Traverse through a[] and search every element     // a[i] in b[]     for (int i = 0; i < a.size(); i++) {          	for (int j = 0; j < b.size(); j++) {                        // If found in b[], then add this           	// element to result array             if (a[i] == b[j]) {                	res.push_back(a[i]);               	break;             }         }     }      return res; }  int main() {     vector<int> a = {5, 6, 2, 1, 4};      vector<int> b = {7, 9, 4, 2};      vector<int> res = intersection(a, b);        for (int i = 0; i < res.size(); i++)          cout << res[i] << " ";      return 0; }

// C program for intersection of two arrays with // distinct elements using nested loops  #include <stdio.h>  int* intersection(int a[], int n, int b[], int m, int* resSize) {     int* res = (int*)malloc(100 * sizeof(int));      *resSize = 0;         	     for (int i = 0; i < n; i++) {         for (int j = 0; j < m; j++) {             if (a[i] == b[j]) {                 res[(*resSize)++] = a[i];                 break;             }         }     }      return res; }  int main() {     int a[] = {5, 6, 2, 1, 4};     int b[] = {7, 9, 4, 2};     int resSize;          int* res = intersection(a, 5, b, 4, &resSize);      for (int i = 0; i < resSize; i++) {         printf("%d ", res[i]);     }      return 0; }

Java

// Java program for intersection of two arrays with // distinct elements using nested loops  import java.util.ArrayList;  class GfG {     static ArrayList<Integer> intersection(int[] a, int[] b) {         ArrayList<Integer> res = new ArrayList<>();          for (int i = 0; i < a.length; i++) {             for (int j = 0; j < b.length; j++) {                 if (a[i] == b[j]) {                     res.add(a[i]);                     break;                 }             }         }          return res;     }      public static void main(String[] args) {         int[] a = {5, 6, 2, 1, 4};         int[] b = {7, 9, 4, 2};          ArrayList<Integer> res = intersection(a, b);          for (int num : res) {             System.out.print(num + " ");         }     } }

Python

# Python program for intersection of two arrays with # distinct elements using nested loops  def intersection(a, b):     res = []      for i in range(len(a)):         for j in range(len(b)):             if a[i] == b[j]:                 res.append(a[i])                 break      return res  if __name__ == "__main__":     a = [5, 6, 2, 1, 4]     b = [7, 9, 4, 2]      res = intersection(a, b)      for num in res:         print(num, end=" ")

// C# program for intersection of two arrays with // distinct elements using nested loops  using System; using System.Collections.Generic;  class GfG {     static List<int> intersection(int[] a, int[] b) {         List<int> res = new List<int>();          for (int i = 0; i < a.Length; i++) {             for (int j = 0; j < b.Length; j++) {                 if (a[i] == b[j]) {                     res.Add(a[i]);                     break;                 }             }         }          return res;     }      static void Main() {         int[] a = {5, 6, 2, 1, 4};         int[] b = {7, 9, 4, 2};          List<int> res = intersection(a, b);          foreach (int num in res) {             Console.Write(num + " ");         }     } }

JavaScript

// JavaScript program for intersection of two arrays with // distinct elements using nested loops  function intersection(a, b) {     let res = [];      for (let i = 0; i < a.length; i++) {         for (let j = 0; j < b.length; j++) {             if (a[i] === b[j]) {                 res.push(a[i]);                 break;             }         }     }      return res; }  let a = [5, 6, 2, 1, 4]; let b = [7, 9, 4, 2];  let res = intersection(a, b);  console.log(res.join(" "));

Output

2 4

Time Complexity : O(n*m), where n and m are size of array a[] and b[] respectively.
Auxiliary Space : O(1)

[Better Approach] Using Sorting and Two Pointers - O(n*logm) Time and O(1) Space

The idea is to sort both the arrays and then maintain a pointer at the beginning of each array. By comparing the elements at both pointers, we can decide how to proceed:
If the element in the first array is smaller than the one in the second, move the pointer in the first array forward, because that element can't be part of the intersection.
If the element in the first array is greater, move the second pointer forward.
If the two elements are equal, you add that element to the result and move both pointers forward.
This continues until one of the pointers reaches the end of its array.
To know more about the implementation of this approach, please refer the post Intersection of Two Sorted Arrays with Distinct Elements.

[Expected Approach] Using Hash Set - O(n+m) Time and O(n) Space

The idea is to use a hash set to store the elements of array a[]. Then, go through array b[] and check if each element is present in the hash set. If an element is found in the hash set, add it to the result array since it is common in both the arrays.

C++

// C++ program for intersection of two arrays with // distinct elements using hash set  #include <iostream> #include <vector> #include <unordered_set> using namespace std;  vector<int> intersect(vector<int>& a, vector<int>& b) {        // Put all elements of a[] in hash set     unordered_set<int> st(a.begin(), a.end());       vector<int> res;                                 for (int i = 0; i < b.size(); i++) {                // If the element is in st         // then add it to result array         if (st.find(b[i]) != st.end()) {             res.push_back(b[i]);          }     }      return res; }  int main() {     vector<int> a = {5, 6, 2, 1, 4};      vector<int> b = {7, 9, 4, 2};      vector<int> res = intersect(a, b);     for (int i = 0; i < res.size(); i++)          cout << res[i] << " ";      return 0; }

Java

// Java program for intersection of two arrays with // distinct elements using hash set  import java.util.*;  class GfG {     static ArrayList<Integer> intersect(int[] a, int[] b) {       	         // Put all elements of a[] in hash set         HashSet<Integer> st = new HashSet<>();         for (int num : a) {             st.add(num);         }         ArrayList<Integer> res = new ArrayList<>();         for (int i = 0; i < b.length; i++) {                        // If the element is in st             // then add it to result array             if (st.contains(b[i])) {                 res.add(b[i]);             }         }          return res;     }      public static void main(String[] args) {         int[] a = {5, 6, 2, 1, 4};         int[] b = {7, 9, 4, 2};          ArrayList<Integer> res = intersect(a, b);         for (int num : res) {             System.out.print(num + " ");         }     } }

Python

# Python program for intersection of two arrays with # distinct elements using hash set  def intersect(a, b):        # Put all elements of a[] in hash set     st = set(a)     res = []     for i in range(len(b)):                # If the element is in st         # then add it to result array         if b[i] in st:             res.append(b[i])      return res  if __name__ == "__main__":     a = [5, 6, 2, 1, 4]     b = [7, 9, 4, 2]      res = intersect(a, b)     for num in res:         print(num, end=" ")

// C# program for intersection of two arrays with // distinct elements using hash set  using System; using System.Collections.Generic;  class GfG {     static List<int> intersect(int[] a, int[] b) {                // Put all elements of a[] in hash set         HashSet<int> st = new HashSet<int>(a);         List<int> res = new List<int>();         for (int i = 0; i < b.Length; i++) {                        // If the element is in st             // then add it to result array             if (st.Contains(b[i])) {                 res.Add(b[i]);             }         }          return res;     }      static void Main() {         int[] a = {5, 6, 2, 1, 4};         int[] b = {7, 9, 4, 2};          List<int> res = intersect(a, b);         foreach (int num in res) {             Console.Write(num + " ");         }     } }

JavaScript

// JavaScript program for intersection of two arrays with // distinct elements using hash set  function intersect(a, b) {        // Put all elements of a[] in hash set     let st = new Set(a);     let res = [];     for (let i = 0; i < b.length; i++) {                // If the element is in st         // then add it to result array         if (st.has(b[i])) {             res.push(b[i]);         }     }      return res; }  let a = [5, 6, 2, 1, 4]; let b = [7, 9, 4, 2];  let res = intersect(a, b); console.log(res.join(" "));

Output

4 2

Time Complexity: O(n + m), where n and m are size of array a[] and b[] respectively.
Auxiliary Space: O(n)

Related Articles:

Union of Two Arrays with Distinct Elements

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Practice Tags :

Intersection of Two Arrays with Distinct Elements

[Naive Approach] Using nested loop - O(n*m) Time and O(1) Space

[Better Approach] Using Sorting and Two Pointers - O(n*logm) Time and O(1) Space

[Expected Approach] Using Hash Set - O(n+m) Time and O(n) Space

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