Find the K-or of an Array

Find the K-or of an Array - Problem

You are given an integer array nums and an integer k.

Let's introduce K-or operation by extending the standard bitwise OR. In K-or, a bit position in the result is set to 1 if at least k numbers in nums have a 1 in that position.

Return the K-or of nums.

Input & Output

Example 1 — Basic Case

$ Input: nums = [7,12,9,8,9,15], k = 4

› Output: 9

💡 Note: Binary: 7=0111, 12=1100, 9=1001, 8=1000, 9=1001, 15=1111. Bit 0 appears in 4 numbers (7,9,9,15), bit 3 appears in 4 numbers (8,9,9,12). Result = 1001₂ = 9

Example 2 — Higher Threshold

$ Input: nums = [2,12,1,3,4,5], k = 3

› Output: 0

💡 Note: No bit position appears in 3 or more numbers, so K-or result is 0

Example 3 — Low Threshold

$ Input: nums = [10,8,5,9,11,6,8], k = 2

› Output: 15

💡 Note: Multiple bit positions have count ≥ 2. Result combines all qualifying bits to get 15

Constraints

1 ≤ nums.length ≤ 50
0 ≤ nums[i] < 2³¹
1 ≤ k ≤ nums.length

Visualization

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Asked in

a Amazon 15 M Microsoft 12

The key insight is that K-or treats each bit position independently - we count how many numbers have each bit set, then include bits that appear at least k times. Best approach uses single pass counting with O(32×n) = O(n) time and O(1) space.

Common Approaches

✓ Memoization

⏱️ Time: N/A Space: N/A

Brute Force Bit Counting

⏱️ Time: O(32 × n) Space: O(1)

For each bit position from 0 to 31, count how many numbers have a 1 in that position. If count >= k, set that bit in the result.

Single Pass Bit Accumulation

⏱️ Time: O(n) Space: O(1)

Use an array to count occurrences of each bit position, then build the result by checking which counts meet the threshold k.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

int solution(int* nums, int numsSize, int k) {
    // Count occurrences of each bit position
    int bitCounts[32] = {0};
    
    // Single pass: count all bits for each number
    for (int i = 0; i < numsSize; i++) {
        int num = nums[i];
        for (int bitPos = 0; bitPos < 32; bitPos++) {
            if (num & (1 << bitPos)) {
                bitCounts[bitPos]++;
            }
        }
    }
    
    // Build result from bit counts
    int result = 0;
    for (int bitPos = 0; bitPos < 32; bitPos++) {
        if (bitCounts[bitPos] >= k) {
            result |= (1 << bitPos);
        }
    }
    
    return result;
}

int main() {
    char buffer[1000];
    fgets(buffer, sizeof(buffer), stdin);
    
    // Parse array
    int nums[1000];
    int count = 0;
    char* ptr = buffer + 1; // skip '['
    while (*ptr != ']' && *ptr != '\0') {
        if (*ptr >= '0' && *ptr <= '9') {
            nums[count++] = atoi(ptr);
            while (*ptr >= '0' && *ptr <= '9') ptr++;
        } else {
            ptr++;
        }
    }
    
    // Read k
    fgets(buffer, sizeof(buffer), stdin);
    int k = atoi(buffer);
    
    int result = solution(nums, count, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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