Find the XOR of Numbers Which Appear Twice

Find the XOR of Numbers Which Appear Twice - Problem

You are given an array nums, where each number in the array appears either once or twice.

Return the bitwise XOR of all the numbers that appear twice in the array, or 0 if no number appears twice.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,2,3,2]

› Output: 2

💡 Note: Only the number 2 appears twice in the array, so we return 2

Example 2 — Multiple Duplicates

$ Input: nums = [1,2,1,3]

› Output: 1

💡 Note: Only the number 1 appears twice, so we return 1

Example 3 — No Duplicates

$ Input: nums = [1,2,3]

› Output: 0

💡 Note: No number appears twice, so we return 0

Constraints

1 ≤ nums.length ≤ 50
1 ≤ nums[i] ≤ 50

Visualization

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Understanding the Visualization

Input Array

Array with numbers appearing once or twice: [1,2,3,2]

Find Duplicates

Identify which numbers appear exactly twice: 2

XOR Result

XOR all duplicate numbers together: 2

Key Takeaway

🎯 Key Insight: Use hash set to track seen numbers and XOR duplicates as they're found

Asked in

G Google 15 M Microsoft 12

The key insight is to identify numbers that appear exactly twice and XOR them together. Best approach uses a hash set to track seen numbers and XOR duplicates in a single pass. Time: O(n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Count Occurrences	O(n²)	O(1)	For each number, count how many times it appears, then XOR all numbers that appear twice
Optimized Bit Manipulation	O(n)	O(n)	Use XOR properties to efficiently find numbers that appear twice in a single pass
Frequency Count with Hash Map	O(n)	O(n)	Count frequency of each number using hash map, then XOR all numbers with frequency 2

Brute Force - Count Occurrences — Algorithm Steps

Iterate through each number in the array
Count how many times this number appears in the entire array
If count is 2, XOR it with the result

Visualization

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Step-by-Step Walkthrough

Check Each Number

For each position, count how many times that number appears

Count Occurrences

Scan entire array to count occurrences of current number

XOR Duplicates

If count equals 2, XOR the number with result

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize) {
    int result = 0;
    
    for (int i = 0; i < numsSize; i++) {
        int count = 0;
        for (int j = 0; j < numsSize; j++) {
            if (nums[j] == nums[i]) {
                count++;
            }
        }
        
        if (count == 2) {
            result ^= nums[i];
        }
    }
    
    return result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array format [1,2,3]
    int nums[1000];
    int size = 0;
    
    char* ptr = strchr(line, '[');
    if (ptr) {
        ptr++;
        char* token = strtok(ptr, ",]");
        while (token != NULL) {
            nums[size++] = atoi(token);
            token = strtok(NULL, ",]");
        }
    }
    
    printf("%d\n", solution(nums, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops - for each element, scan entire array to count occurrences

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for counting and result

✓ Linear Space

12.0K Views

Medium Frequency

~10 min Avg. Time

450 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Count Occurrences — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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