Single Number

Medium Arrays

Every element appears twice except one. Find the single element in O(n) time and O(1) space using XOR.

Sample input

[4, 1, 2, 1, 2]

Sample output

Solution

from functools import reduce

nums = [4, 1, 2, 1, 2]
print(reduce(lambda a, b: a ^ b, nums))

const nums = [4, 1, 2, 1, 2];
console.log(nums.reduce((a, b) => a ^ b, 0));

public class Main {
    public static void main(String[] args) {
        int[] nums = {4, 1, 2, 1, 2};
        int result = 0;
        for (int num : nums) result ^= num;
        System.out.println(result);
    }
}

fun main() {
    val nums = intArrayOf(4, 1, 2, 1, 2)
    println(nums.reduce { a, b -> a xor b })
}

let nums = [4, 1, 2, 1, 2]
print(nums.reduce(0, ^))

void main() {
  final nums = [4, 1, 2, 1, 2];
  print(nums.reduce((a, b) => a ^ b));
}

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

int main() {
    vector<int> nums = {4, 1, 2, 1, 2};
    int result = 0;
    for (int num : nums) result ^= num;
    cout << result << endl;
    return 0;
}

#include <stdio.h>

int main() {
    int nums[] = {4, 1, 2, 1, 2};
    int n = 5, result = 0;
    for (int i = 0; i < n; i++) result ^= nums[i];
    printf("%d\n", result);
    return 0;
}