XOR of All Elements Except Self

Last Updated : 4 Jul, 2026

Given an array arr[] of size n, construct another array res[] of size n such that res[i] is equal to the XOR of all elements of arr[] except arr[i].

Examples:

Input: arr[] = [2, 1, 5, 9]
Output: [13, 14, 10, 6]
Explanation:
res[0] = 1 ^ 5 ^ 9 = 13
res[1] = 2 ^ 5 ^ 9 = 14
res[2] = 2 ^ 1 ^ 9 = 10
res[3] = 2 ^ 1 ^ 5 = 6
Therefore, the resulting array is [13, 14, 10, 6].

Input: arr[] = [2, 1]
Output: [1, 2]
Explanation:
res[0] = 1
res[1] = 2
Since only one element remains after excluding the current element, each position contains that remaining element. Therefore, the resulting array is [1, 2].

Try It Yourself
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Table of Content

[Naive Approach] Simple Simulation - O(n ^ 2) Time and O(n) Space

For every index i, calculate the XOR of all elements except arr[i]. To do this, traverse the entire array again in an inner loop and XOR every element except the current one.

C++ 
#include <iostream>
#include <vector>
using namespace std;

vector<int> getXor(vector<int>& arr) {
    int n = arr.size();

    vector<int> res(n);

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

        for (int j = 0; j < n; j++) {
            if (i != j) {
                currXor ^= arr[j];
            }
        }

        res[i] = currXor;
    }

    return res;
}

int main() {
    vector<int> arr1 = {2, 1, 5, 9};
    vector<int> res1 = getXor(arr1);

    for (int x : res1) {
        cout << x << " ";
    }
    cout << endl;

    vector<int> arr2 = {2, 1};
    vector<int> res2 = getXor(arr2);

    for (int x : res2) {
        cout << x << " ";
    }

    return 0;
}
Java 
import java.util.ArrayList;
import java.util.Arrays;

public class Main {
    public static ArrayList<Integer> getXor(int[] arr) {
        int n = arr.length;

        ArrayList<Integer> res = new ArrayList<Integer>();

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

            for (int j = 0; j < n; j++) {
                if (i!= j) {
                    currXor ^= arr[j];
                }
            }

            res.add(currXor);
        }

        return res;
    }

    public static void main(String[] args) {
        int[] arr1 = {2, 1, 5, 9};
        ArrayList<Integer> res1 = getXor(arr1);

        System.out.println(res1);

        int[] arr2 = {2, 1};
        ArrayList<Integer> res2 = getXor(arr2);

        System.out.println(res2);
    }
}
Python 
def getXor(arr):
    n = len(arr)

    res = [0] * n

    for i in range(n):
        currXor = 0

        for j in range(n):
            if i != j:
                currXor ^= arr[j]

        res[i] = currXor

    return res


print(getXor([2, 1, 5, 9]))
print(getXor([2, 1]))
C# 
using System;
using System.Collections.Generic;

public class Program
{
    public static List<int> getXor(int[] arr)
    {
        int n = arr.Length;

        List<int> res = new List<int>();

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

            for (int j = 0; j < n; j++)
            {
                if (i!= j)
                {
                    currXor ^= arr[j];
                }
            }

            res.Add(currXor);
        }

        return res;
    }

    public static void Main()
    {
        int[] arr1 = {2, 1, 5, 9};
        List<int> res1 = getXor(arr1);

        Console.WriteLine(string.Join(" ", res1));

        int[] arr2 = {2, 1};
        List<int> res2 = getXor(arr2);

        Console.WriteLine(string.Join(" ", res2));
    }
}
JavaScript 
function getXor(arr) {
    let n = arr.length;

    let res = new Array(n);

    for (let i = 0; i < n; i++) {
        let currXor = 0;

        for (let j = 0; j < n; j++) {
            if (i !== j) {
                currXor ^= arr[j];
            }
        }

        res[i] = currXor;
    }

    return res;
}

console.log(getXor([2, 1, 5, 9]));
console.log(getXor([2, 1]));

Output
13 14 10 6 
1 2

[Better Approach] Prefix XOR and Suffix XOR - O(n) Time and O(n) Space

Instead of recalculating the XOR for every index, precompute:

  • prefixXor[i] = XOR of elements from index 0 to i
  • suffixXor[i] = XOR of elements from index i to n - 1

Then for every index i: result[i] = (XOR of elements before i) XOR (XOR of elements after i). which can be obtained using prefix and suffix arrays.

C++ 
#include <iostream>
#include <vector>
using namespace std;

vector<int> getXor(vector<int>& arr) {
    int n = arr.size();

    vector<int> prefix(n), suffix(n), res(n);

    prefix[0] = arr[0];

    for (int i = 1; i < n; i++) {
        prefix[i] = prefix[i - 1] ^ arr[i];
    }

    suffix[n - 1] = arr[n - 1];

    for (int i = n - 2; i >= 0; i--) {
        suffix[i] = suffix[i + 1] ^ arr[i];
    }

    for (int i = 0; i < n; i++) {
        int left = (i > 0) ? prefix[i - 1] : 0;
        int right = (i < n - 1) ? suffix[i + 1] : 0;

        res[i] = left ^ right;
    }

    return res;
}

int main() {
    vector<int> arr1 = {2, 1, 5, 9};

    for (int x : getXor(arr1)) {
        cout << x << " ";
    }
    cout << endl;

    vector<int> arr2 = {2, 1};

    for (int x : getXor(arr2)) {
        cout << x << " ";
    }

    return 0;
}
Java 
import java.util.ArrayList;

class GFG {
    static ArrayList<Integer> getXor(int[] arr) {
        int n = arr.length;

        int[] prefix = new int[n];
        int[] suffix = new int[n];

        prefix[0] = arr[0];

        for (int i = 1; i < n; i++) {
            prefix[i] = prefix[i - 1] ^ arr[i];
        }

        suffix[n - 1] = arr[n - 1];

        for (int i = n - 2; i >= 0; i--) {
            suffix[i] = suffix[i + 1] ^ arr[i];
        }

        ArrayList<Integer> res = new ArrayList<>();

        for (int i = 0; i < n; i++) {
            int left = (i > 0) ? prefix[i - 1] : 0;
            int right = (i < n - 1) ? suffix[i + 1] : 0;

            res.add(left ^ right);
        }

        return res;
    }

    public static void main(String[] args) {
        System.out.println(
            getXor(new int[]{2, 1, 5, 9})
        );

        System.out.println(
            getXor(new int[]{2, 1})
        );
    }
}
Python 
def getXor(arr):
    n = len(arr)

    prefix = [0] * n
    suffix = [0] * n

    prefix[0] = arr[0]

    for i in range(1, n):
        prefix[i] = prefix[i - 1] ^ arr[i]

    suffix[n - 1] = arr[n - 1]

    for i in range(n - 2, -1, -1):
        suffix[i] = suffix[i + 1] ^ arr[i]

    res = [0] * n

    for i in range(n):
        left = prefix[i - 1] if i > 0 else 0
        right = suffix[i + 1] if i < n - 1 else 0

        res[i] = left ^ right

    return res


print(getXor([2, 1, 5, 9]))
print(getXor([2, 1]))
C# 
using System;
using System.Collections.Generic;

class GFG {
    static List<int> getXor(int[] arr) {
        int n = arr.Length;

        int[] prefix = new int[n];
        int[] suffix = new int[n];

        prefix[0] = arr[0];

        for (int i = 1; i < n; i++) {
            prefix[i] = prefix[i - 1] ^ arr[i];
        }

        suffix[n - 1] = arr[n - 1];

        for (int i = n - 2; i >= 0; i--) {
            suffix[i] = suffix[i + 1] ^ arr[i];
        }

        List<int> res = new List<int>();

        for (int i = 0; i < n; i++) {
            int left = (i > 0) ? prefix[i - 1] : 0;
            int right = (i < n - 1) ? suffix[i + 1] : 0;

            res.Add(left ^ right);
        }

        return res;
    }

    static void Main() {
        Console.WriteLine(
            string.Join(" ", getXor(new int[] {2, 1, 5, 9}))
        );

        Console.WriteLine(
            string.Join(" ", getXor(new int[] {2, 1}))
        );
    }
}
JavaScript 
function getXor(arr) {
    let n = arr.length;

    let prefix = new Array(n);
    let suffix = new Array(n);
    let res = new Array(n);

    prefix[0] = arr[0];

    for (let i = 1; i < n; i++) {
        prefix[i] = prefix[i - 1] ^ arr[i];
    }

    suffix[n - 1] = arr[n - 1];

    for (let i = n - 2; i >= 0; i--) {
        suffix[i] = suffix[i + 1] ^ arr[i];
    }

    for (let i = 0; i < n; i++) {
        let left = (i > 0) ? prefix[i - 1] : 0;
        let right = (i < n - 1) ? suffix[i + 1] : 0;

        res[i] = left ^ right;
    }

    return res;
}

console.log(getXor([2, 1, 5, 9]));
console.log(getXor([2, 1]));

Output
13 14 10 6 
1 2

[Expected Approach] Using XOR Property - O(n) Time and O(1) Space

Observe that XOR has the following properties:

  • a ^ a = 0
  • a ^ 0 = a

First compute the XOR of all elements and store it in totalXor.

For any index i: totalXor = arr[0] ^ arr[1] ^ ... ^ arr[n-1], If we XOR totalXor with arr[i], arr[i] gets cancelled out. Therefore: result[i] = totalXor ^ arr[i]

C++ 
#include <iostream>
#include <vector>
using namespace std;

vector<int> getXor(vector<int>& arr) {
    int totalXor = 0;

    for (int x : arr) {
        totalXor ^= x;
    }

    vector<int> res(arr.size());

    for (int i = 0; i < arr.size(); i++) {
        res[i] = totalXor ^ arr[i];
    }

    return res;
}

int main() {
    vector<int> arr1 = {2, 1, 5, 9};

    for (int x : getXor(arr1)) {
        cout << x << " ";
    }
    cout << endl;

    vector<int> arr2 = {2, 1};

    for (int x : getXor(arr2)) {
        cout << x << " ";
    }

    return 0;
}
Java 
import java.util.ArrayList;

class GFG {
    static ArrayList<Integer> getXor(int[] arr) {
        int totalXor = 0;

        for (int x : arr) {
            totalXor ^= x;
        }

        ArrayList<Integer> res = new ArrayList<>();

        for (int x : arr) {
            res.add(totalXor ^ x);
        }

        return res;
    }

    public static void main(String[] args) {
        System.out.println(
            getXor(new int[] {2, 1, 5, 9})
        );

        System.out.println(
            getXor(new int[] {2, 1})
        );
    }
}
Python 
def getXor(arr):
    totalXor = 0

    for x in arr:
        totalXor ^= x

    res = []

    for x in arr:
        res.append(totalXor ^ x)

    return res


print(getXor([2, 1, 5, 9]))
print(getXor([2, 1]))
C# 
using System;
using System.Collections.Generic;

class GFG {
    static List<int> getXor(int[] arr) {
        int totalXor = 0;

        foreach (int x in arr) {
            totalXor ^= x;
        }

        List<int> res = new List<int>();

        foreach (int x in arr) {
            res.Add(totalXor ^ x);
        }

        return res;
    }

    static void Main() {
        Console.WriteLine(
            string.Join(" ", getXor(new int[] {2, 1, 5, 9}))
        );

        Console.WriteLine(
            string.Join(" ", getXor(new int[] {2, 1}))
        );
    }
}
JavaScript 
function getXor(arr) {
    let totalXor = 0;

    for (let x of arr) {
        totalXor ^= x;
    }

    let res = [];

    for (let x of arr) {
        res.push(totalXor ^ x);
    }

    return res;
}

console.log(getXor([2, 1, 5, 9]));
console.log(getXor([2, 1]));

Output
13 14 10 6 
1 2
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