Count Triplets with Even XOR Set Bits I

Count Triplets with Even XOR Set Bits I - Problem

Given three integer arrays a, b, and c, return the number of triplets (a[i], b[j], c[k]) such that the bitwise XOR of the elements of each triplet has an even number of set bits.

A set bit is a bit that is equal to 1. The XOR operation combines three numbers: a[i] ^ b[j] ^ c[k]. Count how many bits are 1 in the result - if this count is even, include the triplet in your answer.

Input & Output

Example 1 — Basic Case

$ Input: a = [2,1], b = [3,4], c = [1,2]

› Output: 4

💡 Note: All 4 triplets work: (2,3,1)→0 (even), (2,4,2)→0 (even), (1,3,2)→0 (even), (1,4,1)→4 (even)

Example 2 — Single Elements

$ Input: a = [1], b = [2], c = [3]

› Output: 1

💡 Note: Only triplet is (1,2,3): 1^2^3=0, which has 0 set bits (even), so result is 1

Example 3 — All Same Parity

$ Input: a = [1,3], b = [1,3], c = [1,3]

› Output: 0

💡 Note: All numbers have odd set bits. odd^odd^odd=odd, so no valid triplets

Constraints

1 ≤ a.length, b.length, c.length ≤ 10⁵
0 ≤ a[i], b[i], c[i] ≤ 10⁹

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is that XOR parity follows mathematical rules: numbers with even set bits XORed in certain combinations always yield even results. The optimized approach groups numbers by parity and uses multiplication to count valid combinations. Time: O(n), Space: O(1).

Common Approaches

✓ Sliding Window

⏱️ Time: N/A Space: N/A

Sort First

⏱️ Time: N/A Space: N/A

Brute Force - Try All Triplets

⏱️ Time: O(n³) Space: O(1)

For each element in array a, combine it with each element in array b and each element in array c. Calculate XOR and count set bits to determine if even.

Optimized - Count by Parity

⏱️ Time: O(n) Space: O(1)

Instead of checking every triplet, classify each number as having even or odd set bits. Use the fact that even XOR even = even, odd XOR odd = even to count valid combinations mathematically.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

int countSetBits(int n) {
    int count = 0;
    while (n) {
        count++;
        n &= (n - 1);
    }
    return count;
}

int* parseArray(char* line, int* size) {
    *size = 0;
    int capacity = 100;
    int* arr = (int*)malloc(capacity * sizeof(int));
    
    char* ptr = line;
    while (*ptr && *ptr != '[') ptr++;
    if (*ptr == '[') ptr++;
    
    while (*ptr && *ptr != ']') {
        if (*ptr == ',' || *ptr == ' ') {
            ptr++;
            continue;
        }
        
        char* endptr;
        int num = strtol(ptr, &endptr, 10);
        if (ptr != endptr) {
            if (*size >= capacity) {
                capacity *= 2;
                arr = (int*)realloc(arr, capacity * sizeof(int));
            }
            arr[(*size)++] = num;
            ptr = endptr;
        } else {
            ptr++;
        }
    }
    
    return arr;
}

int main() {
    char line[1000];
    
    // Read array a
    fgets(line, sizeof(line), stdin);
    int sizeA;
    int* a = parseArray(line, &sizeA);
    
    // Read array b
    fgets(line, sizeof(line), stdin);
    int sizeB;
    int* b = parseArray(line, &sizeB);
    
    // Read array c
    fgets(line, sizeof(line), stdin);
    int sizeC;
    int* c = parseArray(line, &sizeC);
    
    int count = 0;
    
    // Check all triplets
    for (int i = 0; i < sizeA; i++) {
        for (int j = 0; j < sizeB; j++) {
            for (int k = 0; k < sizeC; k++) {
                int xorResult = a[i] ^ b[j] ^ c[k];
                int setBits = countSetBits(xorResult);
                if (setBits % 2 == 0) {
                    count++;
                }
            }
        }
    }
    
    printf("%d\n", count);
    
    free(a);
    free(b);
    free(c);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Smart Actions

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