Check if two arrays are equal or not

Last Updated : 26 Dec, 2024

Given two arrays, a and b of equal length. The task is to determine if the given arrays are equal or not. Two arrays are considered equal if:

Both arrays contain the same set of elements.
The arrangements (or permutations) of elements may be different.
If there are repeated elements, the counts of each element must be the same in both arrays.

Examples:

Input: a[] = [1, 2, 5, 4, 0], b[] = [2, 4, 5, 0, 1]
Output: true
Input: a[] = [1, 2, 5, 4, 0, 2, 1], b[] = [2, 4, 5, 0, 1, 1, 2]
Output: true
Input: a[] = [1, 7, 1], b[] = [7, 7, 1]
Output: false

Table of Content

[Naive Approach] Using Sorting - O(n*log n) Time and O(1) Space
[Expected Approach] Using Hashing - O(n) Time and O(n) Space

[Naive Approach] Using Sorting - O(n*log n) Time and O(1) Space

The basic idea is to sort the both arrays. Compare the arrays element by element one by one if all are same then it is equal array otherwise it is not.

C++

// C++ program to find given two array
// are equal or not
#include <bits/stdc++.h>
using namespace std;

bool checkEqual(vector<int>& a, vector<int>& b) {
    int n = a.size(), m = b.size();
  
    // If lengths of array are not equal means
    // array are not equal
    if (n != m) return false;
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
    for (int i = 0; i < n; i++)
        if (a[i] != b[i])
            return false;

    // If all elements were same.
    return true;
}

int main() {
    vector<int> a = { 3, 5, 2, 5, 2 };
    vector<int> b = { 2, 3, 5, 5, 2 };

    if (checkEqual(a, b))
        cout << "true";
    else
        cout << "false";
    return 0;
}

Java

import java.util.Arrays;

class GfG {
    static boolean checkEqual(int[] a, int[] b) {
      
        // If lengths of array are not equal means
        // array are not equal
        if (a.length != b.length) return false;
        Arrays.sort(a);
        Arrays.sort(b);
        for (int i = 0; i < a.length; i++)
            if (a[i] != b[i])
                return false;

        // If all elements were same.
        return true;
    }

    public static void main(String[] args) {
        int[] a = { 3, 5, 2, 5, 2 };
        int[] b = { 2, 3, 5, 5, 2 };
        if (checkEqual(a, b))
            System.out.println("true");
        else
            System.out.println("false");
    }
}

Python

def checkEqual(a, b):
  
    # If lengths of array are not equal means
    # array are not equal
    if len(a) != len(b):
        return False
    return sorted(a) == sorted(b)

if __name__ == '__main__':
    a = [3, 5, 2, 5, 2]
    b = [2, 3, 5, 5, 2]

    if checkEqual(a, b):
        print("true")
    else:
        print("false")

using System;
using System.Linq;

class GfG {
    static bool CheckEqual(int[] a, int[] b) {
      
        // If lengths of array are not equal means
        // array are not equal
        if (a.Length != b.Length) return false;
        Array.Sort(a);
        Array.Sort(b);
        for (int i = 0; i < a.Length; i++)
            if (a[i] != b[i])
                return false;

        // If all elements were same.
        return true;
    }

    static void Main() {
        int[] a = { 3, 5, 2, 5, 2 };
        int[] b = { 2, 3, 5, 5, 2 };

        if (CheckEqual(a, b))
            Console.WriteLine("true");
        else
            Console.WriteLine("false");
    }
}

JavaScript

function checkEqual(a, b) {
    // If lengths of array are not equal means
    // array are not equal
    if (a.length !== b.length) return false;
    a.sort();
    b.sort();
    for (let i = 0; i < a.length; i++)
        if (a[i] !== b[i])
            return false;

    // If all elements were same.
    return true;
}

//Driver Code
const a = [3, 5, 2, 5, 2];
const b = [2, 3, 5, 5, 2];
console.log(checkEqual(a, b) ? 'true' : 'false');

Output

true

Time Complexity: O(n*log n), since sorting is used
Auxiliary Space: O(1)

[Expected Approach] Using Hashing- O(n) Time and O(n) Space

Use of a hash map to count the occurrences of each element in one array and then verifying these counts against the second array.

C++

#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;

bool checkEqual(vector<int>& a, vector<int>& b) {
  	int n = a.size(), m = b.size();
    if (n != m)
        return false;

    unordered_map<int, int> mp;
    for (int i = 0; i < n; i++)
        mp[a[i]]++;

    for (int i = 0; i < n; i++) {
        if (mp.find(b[i]) == mp.end())
            return false;

        if (mp[b[i]] == 0)
            return false;
      
        mp[b[i]]--;
    }
    return true;
}

int main() {
    vector<int> a = { 3, 5, 2, 5, 2 };
    vector<int> b = { 2, 3, 5, 5, 2 };

    if (checkEqual(a, b))
        cout << "true";
    else
        cout << "false";
    return 0;
}

Java

import java.io.*;
import java.util.*;

class GfG {

    public static boolean checkEqual(int a[], int b[]) {
        int n = a.length, m = b.length;
        if (n != m)
            return false;

        Map<Integer, Integer> map
            = new HashMap<Integer, Integer>();
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (map.get(a[i]) == null)
                map.put(a[i], 1);
            else {
                count = map.get(a[i]);
                count++;
                map.put(a[i], count);
            }
        }

        for (int i = 0; i < n; i++) {

            if (!map.containsKey(b[i]))
                return false;

            if (map.get(b[i]) == 0)
                return false;

            count = map.get(b[i]);
            --count;
            map.put(b[i], count);
        }

        return true;
    }

    public static void main(String[] args) {
        int a[] = { 3, 5, 2, 5, 2 };
        int b[] = { 2, 3, 5, 5, 2 };

        if (checkEqual(a, b))
            System.out.println("true");
        else
            System.out.println("false");
    }
}

Python

def checkEqual(a, b):
    n, m = len(a), len(b)
    if n != m:
        return False

    mp = {}
    for num in a:
        mp[num] = mp.get(num, 0) + 1

    for num in b:
        if num not in mp:
            return False
        if mp[num] == 0:
            return False
        mp[num] -= 1
    return True

if __name__ == '__main__':
    a = [3, 5, 2, 5, 2]
    b = [2, 3, 5, 5, 2]

    if checkEqual(a, b):
        print("true")
    else:
        print("false")

using System;
using System.Collections.Generic;
using System.Linq;

class GfG {
    static bool CheckEqual(int[] a, int[] b) {
        int n = a.Length, m = b.Length;
        if (n != m)
            return false;

        var mp = new Dictionary<int, int>();
        foreach (var num in a) {
            if (mp.ContainsKey(num))
                mp[num]++;
            else
                mp[num] = 1;
        }

        foreach (var num in b) {
            if (!mp.ContainsKey(num) || mp[num] == 0)
                return false;
            mp[num]--;
        }
        return true;
    }

    static void Main() {
        int[] a = { 3, 5, 2, 5, 2 };
        int[] b = { 2, 3, 5, 5, 2 };

        Console.WriteLine(CheckEqual(a, b) ? "true" : "false");
    }
}

JavaScript

function checkEqual(a, b) {
    const n = a.length, m = b.length;
    if (n !== m)
        return false;

    const mp = {};
    for (let num of a) {
        mp[num] = (mp[num] || 0) + 1;
    }

    for (let num of b) {
        if (!(num in mp) || mp[num] === 0)
            return false;
        mp[num]--;
    }
    return true;
}

// Driver Code
const a = [3, 5, 2, 5, 2];
const b = [2, 3, 5, 5, 2];

console.log(checkEqual(a, b) ? 'true' : 'false');

Output

true

Time Complexity: O(n), where n is the length of given array
Auxiliary Space: O(n)

Find missing elements of a range

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Practice Tags :

Check if two arrays are equal or not

[Naive Approach] Using Sorting - O(n*log n) Time and O(1) Space

[Expected Approach] Using Hashing- O(n) Time and O(n) Space

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