Check if two arrays are equal or not
Last Updated :
26 Dec, 2024
Improve
Try it on GfG Practice 
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: trueInput: a[] = [1, 2, 5, 4, 0, 2, 1], b[] = [2, 4, 5, 0, 1, 1, 2]
Output: trueInput: 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
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++ 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;
}
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");
}
}
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");
}
}
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.
#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;
}
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");
}
}
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");
}
}
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)
