Find the XOR of Numbers Which Appear Twice

Find the XOR of Numbers Which Appear Twice - Problem

You're given an array nums where each number appears either once or twice. Your task is to find all numbers that appear exactly twice and return their bitwise XOR.

If no number appears twice, return 0.

Key Points:

Numbers can appear 1 or 2 times only
XOR all the duplicates together
XOR has useful properties: a ⊕ a = 0 and a ⊕ 0 = a

Example: For array [1, 2, 1, 3], only 1 appears twice, so return 1.

Input & Output

example_1.py — Basic case with one duplicate

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

› Output: 1

💡 Note: Only the number 1 appears twice in the array. So we return 1 ⊕ 0 = 1.

example_2.py — Multiple duplicates

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

› Output: 0

💡 Note: Numbers 1, 2, and 3 all appear twice. The XOR is 1 ⊕ 2 ⊕ 3 = 0.

example_3.py — No duplicates

$ Input: nums = [1, 2, 3, 4]

› Output: 0

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

Constraints

1 ≤ nums.length ≤ 50
1 ≤ nums[i] ≤ 50
Each number appears either once or twice

Visualization

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Understanding the Visualization

Guest Arrives

First time seeing a number - add to set

Guest Returns

Second time seeing a number - XOR with result

Guest Leaves

Remove from set since we've processed the duplicate

Key Takeaway

🎯 Key Insight: Use XOR's self-canceling property (a ⊕ a = 0) combined with hash set for O(n) duplicate detection.

Asked in

G Google 15 a Amazon 12 f Meta 8 ⊞ Microsoft 6

The optimal solution uses a hash set to track seen numbers in a single pass. When we encounter a number for the first time, we add it to the set. When we see it again, we XOR it with our result and remove it from the set. This elegantly handles the constraint that numbers appear at most twice, giving us O(n) time and O(n) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Two-Pass Hash Table	O(n)	O(n)	First pass counts frequencies, second pass XORs duplicates
Two Pointers (Sorted Array)	O(n log n)	O(1)	Sort array and use two pointers to identify consecutive duplicates
Brute Force (Nested Loops)	O(n²)	O(1)	For each number, count occurrences by scanning the entire array
One-Pass Hash Set (Optimal)	O(n)	O(n)	Use set to track seen numbers, XOR when removing duplicates

Two-Pass Hash Table — Algorithm Steps

First pass: Count frequency of each number using a hash map
Second pass: Iterate through the hash map
For each number with frequency 2, XOR it with the result
Return the final XOR result

Visualization

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Step-by-Step Walkthrough

Build Frequency Map

Count how many times each number appears

Find Duplicates

Identify numbers that appear exactly twice

XOR Results

XOR all duplicate numbers together

Code -

solution.c — C

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

#define MAX_SIZE 1000

typedef struct {
    int key;
    int value;
} HashEntry;

int duplicateNumbersXOR(int* nums, int n) {
    HashEntry freq[MAX_SIZE];
    int freqSize = 0;
    
    // First pass: count frequencies
    for (int i = 0; i < n; i++) {
        int found = 0;
        for (int j = 0; j < freqSize; j++) {
            if (freq[j].key == nums[i]) {
                freq[j].value++;
                found = 1;
                break;
            }
        }
        if (!found) {
            freq[freqSize].key = nums[i];
            freq[freqSize].value = 1;
            freqSize++;
        }
    }
    
    // Second pass: XOR numbers that appear twice
    int result = 0;
    for (int i = 0; i < freqSize; i++) {
        if (freq[i].value == 2) {
            result ^= freq[i].key;
        }
    }
    
    return result;
}

int main() {
    int nums[1000];
    int n = 0;
    char c;
    
    scanf("%c", &c); // skip '['
    while (scanf("%d%c", &nums[n], &c)) {
        n++;
        if (c == ']') break;
    }
    
    printf("%d\n", duplicateNumbersXOR(nums, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Two passes through the array, each taking O(n) time

✓ Linear Growth

Space Complexity

O(n)

Hash map can store up to n unique elements in worst case

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

890 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Two-Pass Hash Table — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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