Check if two arrays are permutations of each other using Mathematical Operation
Last Updated :
02 May, 2025
Given two unsorted arrays of same size where arr[i] >= 0 for all i, the task is to check if two arrays are permutations of each other or not.
Examples:
Input: arr1[] = {2, 1, 3, 5, 4, 3, 2}
arr2[] = {3, 2, 2, 4, 5, 3, 1}
Output: Yes
Input: arr1[] = {2, 1, 3, 5}
arr2[] = {3, 2, 2, 4}
Output: No
It has been already discussed in Check if two arrays are permutations of each other using Sorting and Hashing. But in this post, a different approach is discussed.
Approach:
- Traverse the first array A, add and multiply all the elements and store them in variables as Sum1 and Mul1 respectively.
- Similarly, traverse the second array B, add and multiply all the elements and store them in variables as Sum2 and Mul2 respectively.
- Now, compare both sum1, sum2 and mul1, mul2. If Sum1 == Sum2 and Mul1 == Mul2, then both arrays are permutations of each other, else not.
Implementation:
C++
// CPP code to check if arrays
// are permutations of each other
#include <iostream>
using namespace std;
// Function to check if arrays
// are permutations of each other.
bool arePermutations(int a[], int b[], int n, int m)
{
int sum1 = 0, sum2 = 0, mul1 = 1, mul2 = 1;
// Calculating sum and multiply of first array
for (int i = 0; i < n; i++) {
sum1 += a[i];
mul1 *= a[i];
}
// Calculating sum and multiply of second array
for (int i = 0; i < m; i++) {
sum2 += b[i];
mul2 *= b[i];
}
// If sum and mul of both arrays are equal,
// return true, else return false.
return ((sum1 == sum2) && (mul1 == mul2));
}
// Driver code
int main()
{
int a[] = { 1, 3, 2 };
int b[] = { 3, 1, 2 };
int n = sizeof(a) / sizeof(int);
int m = sizeof(b) / sizeof(int);
if (arePermutations(a, b, n, m))
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
}
// Java code to check if arrays
// are permutations of each other
import java.io.*;
class GFG {
// Function to check if arrays
// are permutations of each other.
static boolean arePermutations(int a[], int b[], int n, int m)
{
int sum1 = 0, sum2 = 0, mul1 = 1, mul2 = 1;
// Calculating sum and multiply of first array
for (int i = 0; i < n; i++) {
sum1 += a[i];
mul1 *= a[i];
}
// Calculating sum and multiply of second array
for (int i = 0; i < m; i++) {
sum2 += b[i];
mul2 *= b[i];
}
// If sum and mul of both arrays are equal,
// return true, else return false.
return ((sum1 == sum2) && (mul1 == mul2));
}
// Driver code
public static void main (String[] args) {
int a[] = { 1, 3, 2 };
int b[] = { 3, 1, 2 };
int n = a.length;
int m = b.length;
if (arePermutations(a, b, n, m)==true)
System.out.println( "Yes");
else
System.out.println( "No");
}
}
// This code is contributed by inder_verma..
# Python 3 program to check if arrays
# are permutations of each other
# Function to check if arrays
# are permutations of each other
def arePermutations(a, b, n, m) :
sum1, sum2, mul1, mul2 = 0, 0, 1, 1
# Calculating sum and multiply of first array
for i in range(n) :
sum1 += a[i]
mul1 *= a[i]
# Calculating sum and multiply of second array
for i in range(m) :
sum2 += b[i]
mul2 *= b[i]
# If sum and mul of both arrays are equal,
# return true, else return false.
return((sum1 == sum2) and (mul1 == mul2))
# Driver code
if __name__ == "__main__" :
a = [ 1, 3, 2]
b = [ 3, 1, 2]
n = len(a)
m = len(b)
if arePermutations(a, b, n, m) :
print("Yes")
else :
print("No")
# This code is contributed by ANKITRAI1
// C# code to check if arrays
// are permutations of each other
using System;
class GFG
{
// Function to check if arrays
// are permutations of each other.
static bool arePermutations(int[] a, int[] b,
int n, int m)
{
int sum1 = 0, sum2 = 0,
mul1 = 1, mul2 = 1;
// Calculating sum and multiply
// of first array
for (int i = 0; i < n; i++)
{
sum1 += a[i];
mul1 *= a[i];
}
// Calculating sum and multiply
// of second array
for (int i = 0; i < m; i++)
{
sum2 += b[i];
mul2 *= b[i];
}
// If sum and mul of both arrays
// are equal, return true, else
// return false.
return ((sum1 == sum2) &&
(mul1 == mul2));
}
// Driver code
public static void Main ()
{
int[] a = { 1, 3, 2 };
int[] b = { 3, 1, 2 };
int n = a.Length;
int m = b.Length;
if (arePermutations(a, b, n, m) == true)
Console.Write( "Yes");
else
Console.Write( "No");
}
}
// This code is contributed
// by ChitraNayal
<script>
// JavaScript code to check if arrays
// are permutations of each other
// Function to check if arrays
// are permutations of each other.
function arePermutations(a,b,n,m)
{
let sum1 = 0, sum2 = 0, mul1 = 1, mul2 = 1;
// Calculating sum and multiply of first array
for (let i = 0; i < n; i++) {
sum1 += a[i];
mul1 *= a[i];
}
// Calculating sum and multiply of second array
for (let i = 0; i < m; i++) {
sum2 += b[i];
mul2 *= b[i];
}
// If sum and mul of both arrays are equal,
// return true, else return false.
return ((sum1 == sum2) && (mul1 == mul2));
}
// Driver code
let a=[1, 3, 2 ];
let b=[3, 1, 2];
let n = a.length;
let m = b.length;
if (arePermutations(a, b, n, m)==true)
document.write( "Yes");
else
document.write( "No");
// This code is contributed by rag2127
</script>
<?php
// PHP code to check if arrays
// are permutations of each other
// Function to check if arrays
// are permutations of each other.
function arePermutations($a, $b, $n, $m)
{
$sum1 = 0; $sum2 = 0;
$mul1 = 1; $mul2 = 1;
// Calculating sum and multiply
// of first array
for ($i = 0; $i < $n; $i++)
{
$sum1 += $a[$i];
$mul1 *= $a[$i];
}
// Calculating sum and multiply
// of second array
for ($i = 0; $i < $m; $i++)
{
$sum2 += $b[$i];
$mul2 *= $b[$i];
}
// If sum and mul of both arrays
// are equal, return true, else
// return false.
return (($sum1 == $sum2) &&
($mul1 == $mul2));
}
// Driver code
$a = array( 1, 3, 2 );
$b = array( 3, 1, 2 );
$n = sizeof($a);
$m = sizeof($b);
if (arePermutations($a, $b, $n, $m))
echo "Yes" . "\n";
else
echo "No" . "\n";
// This code is contributed
// by Akanksha Rai(Abby_akku)
Complexity Analysis:
- Time Complexity:O(n) where n is size of given array
- Auxiliary space: O(1) as it is using constant space
Efficient Approach:
To determine if two arrays are permutations of each other, we can use the following approach with O(1) time complexity:
- Check if the sizes of both arrays are equal. If not, return false.
- Calculate the XOR of all elements in both arrays.
- If the XOR result is 0, it means the arrays have the same elements (although the order may be different) and are permutations of each other. Return true.
- If the XOR result is not 0, return false.
Here's the modified code using this approach:
C++
#include <iostream>
using namespace std;
// Function to check if arrays are permutations of each other.
bool arePermutations(int a[], int b[], int n, int m) {
if (n != m) {
return false;
}
int xorResult = 0;
// Calculate XOR of all elements in both arrays
for (int i = 0; i < n; i++) {
xorResult ^= a[i];
xorResult ^= b[i];
}
// If XOR result is 0, arrays are permutations of each other
return (xorResult == 0);
}
// Driver code
int main() {
int a[] = {2,1,3,5,4,3,2};
int b[] = {3, 2,2,4,5,3,1};
int n = sizeof(a) / sizeof(int);
int m = sizeof(b) / sizeof(int);
if (arePermutations(a, b, n, m)) {
cout << "Yes" << endl;
} else {
cout << "No" << endl;
}
return 0;
}
import java.util.Arrays;
public class PermutationCheck {
// Function to check if arrays are permutations of each other.
public static boolean arePermutations(int[] a, int[] b, int n, int m) {
if (n != m) {
return false;
}
int xorResult = 0;
// Calculate XOR of all elements in both arrays
for (int i = 0; i < n; i++) {
xorResult ^= a[i];
xorResult ^= b[i];
}
// If XOR result is 0, arrays are permutations of each other
return (xorResult == 0);
}
// Driver code
public static void main(String[] args) {
int[] a = {2, 1, 3, 5, 4, 3, 2};
int[] b = {3, 2, 2, 4, 5, 3, 1};
int n = a.length;
int m = b.length;
if (arePermutations(a, b, n, m)) {
System.out.println("Yes");
} else {
System.out.println("No");
}
}
}
# Function to check if arrays are permutations of each other.
def are_permutations(a, b):
n = len(a)
m = len(b)
if n != m:
return False
xor_result = 0
# Calculate XOR of all elements in both arrays
for i in range(n):
xor_result ^= a[i]
xor_result ^= b[i]
# If XOR result is 0, arrays are permutations of each other
return (xor_result == 0)
# Driver code
a = [2, 1, 3, 5, 4, 3, 2]
b = [3, 2, 2, 4, 5, 3, 1]
if are_permutations(a, b):
print("Yes")
else:
print("No")
# This code is contributed by akshitaguprzj3
using System;
class Program
{
// Function to check if arrays are permutations of each other.
static bool ArePermutations(int[] a, int[] b, int n, int m)
{
if (n != m)
{
return false;
}
int xorResult = 0;
// Calculate XOR of all elements in both arrays
for (int i = 0; i < n; i++)
{
xorResult ^= a[i];
xorResult ^= b[i];
}
// If XOR result is 0, arrays are permutations of each other
return (xorResult == 0);
}
static void Main(string[] args)
{
int[] a = { 2, 1, 3, 5, 4, 3, 2 };
int[] b = { 3, 2, 2, 4, 5, 3, 1 };
int n = a.Length;
int m = b.Length;
if (ArePermutations(a, b, n, m))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by shivamgupta310570
// Function to check if arrays are permutations of each other.
function arePermutations(a, b, n, m) {
// If the arrays have different lengths, they can't be permutations.
if (n !== m) {
return false;
}
let xorResult = 0;
// Calculate XOR of all elements in both arrays.
for (let i = 0; i < n; i++) {
xorResult ^= a[i];
xorResult ^= b[i];
}
// If XOR result is 0, arrays are permutations of each other.
return xorResult === 0;
}
// Driver code
function main() {
const a = [2, 1, 3, 5, 4, 3, 2];
const b = [3, 2, 2, 4, 5, 3, 1];
const n = a.length;
const m = b.length;
if (arePermutations(a, b, n, m)) {
console.log("Yes");
} else {
console.log("No");
}
}
main();
Complexity Analysis:
Time Complexity: O(n) where n is size of given array.
Auxiliary space: O(1) as it is using constant space
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