Check if two arrays are permutations of each other using Mathematical Operation

Last Updated : 18 Oct, 2023

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: YesInput: 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:

1. Traverse the first array A, add and multiply all the elements and store them in variables as Sum1 and Mul1 respectively.
2. Similarly, traverse the second array B, add and multiply all the elements and store them in variables as Sum2 and Mul2 respectively.
3. 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 ` `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

 `// 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..`

Python3

 `# 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#

 `// 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`

Javascript

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PHP

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Output

```Yes

```

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 ` `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;` `}`

Java

 `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"``);` `        ``}` `    ``}` `}`

Python3

 `# 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`

C#

 `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`

Javascript

 `// 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();`

Output

```Yes

```

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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