Intersection of Two Arrays with Distinct Elements (original) (raw)

Last Updated : 23 Jul, 2025

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[]

Try It Yourself

Table of Content

[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

[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 #include using namespace std;

vector intersection(vector& a, vector& b) { vector 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 a = {5, 6, 2, 1, 4}; vector 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

// 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 intersection(int[] a, int[] b) { ArrayList 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#

// C# program for intersection of two arrays with // distinct elements using nested loops

using System; using System.Collections.Generic;

class GfG { static List intersection(int[] a, int[] b) { List res = new List();

    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(" "));

**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 #include #include using namespace std;

vector intersect(vector& a, vector& 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 a = {5, 6, 2, 1, 4}; vector 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 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#

// C# program for intersection of two arrays with // distinct elements using hash set

using System; using System.Collections.Generic;

class GfG { static List 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(" "));

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

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