Find the K-or of an Array - Practice Coding Problems

Find the K-or of an Array - Problem

You are given an integer array nums and an integer k. Your task is to compute the K-or of the array, which is a variation of the standard bitwise OR operation.

In the K-or operation, a bit position in the result is set to 1 if and only if at least k numbers in the array have a 1 in that same bit position. Otherwise, the bit is set to 0.

Example: If nums = [7, 12, 9, 8, 9, 15] and k = 4, we need to find which bit positions have at least 4 numbers with a 1 bit. The binary representations are:

7 → 0111
12 → 1100
9 → 1001
8 → 1000
9 → 1001
15 → 1111

Count the 1s in each position and keep only positions where the count ≥ k.

Input & Output

example_1.py — Basic K-or Operation

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

› Output: 9

💡 Note: Binary representations: 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 (12,9,8,15). Result: 1001₂ = 9

example_2.py — High Threshold

$ Input: nums = [2, 12, 1, 3, 4], k = 6

› Output: 0

💡 Note: No bit position appears in 6 or more numbers since we only have 5 numbers total. Result is 0.

example_3.py — All Bits Set

$ Input: nums = [15, 15, 15], k = 2

› Output: 15

💡 Note: 15 in binary is 1111. All bit positions (0,1,2,3) appear in at least 2 numbers. Result: 1111₂ = 15

Constraints

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

Visualization

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Understanding the Visualization

Cast Votes

Each number votes 'yes' on proposals corresponding to its set bits

Count Votes

Tally votes for each of the 32 proposals

Determine Winners

Proposals with ≥ k votes pass and get set in the result

Key Takeaway

🎯 Key Insight: K-or extends regular OR by requiring a minimum vote threshold, making it perfect for finding commonly set bits across multiple numbers.

Asked in

f Meta 15 a Amazon 12 ⊞ Microsoft 8 G Google 5

The optimal solution uses single-pass bit counting with O(n) time complexity. We count occurrences of each bit position across all numbers, then construct the result by setting bits where count ≥ k. This is more efficient than checking each bit position separately.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Check Each Bit Position)	O(32 * n) = O(n)	O(1)	For each bit position, iterate through all numbers to count 1s
One-Pass Bit Counting (Optimal)	O(n)	O(1)	Count all bit positions in a single pass through the array

Brute Force (Check Each Bit Position) — Algorithm Steps

Initialize result to 0
For each bit position i from 0 to 31:
Count how many numbers have bit i set to 1
If count >= k, set bit i in result
Return the result

Visualization

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Step-by-Step Walkthrough

Check Bit 0

Count 1s in the rightmost bit position across all numbers

Check Bit 1

Count 1s in the second bit position

Continue for all 32 bits

Repeat process for each bit position

Build Result

Set bits in result where count >= k

Code -

solution.c — C

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

int findKOr(int* nums, int numsSize, int k) {
    int result = 0;
    
    // Check each bit position (0 to 31 for 32-bit integers)
    for (int bitPos = 0; bitPos < 32; bitPos++) {
        int count = 0;
        
        // Count how many numbers have this bit set
        for (int i = 0; i < numsSize; i++) {
            if (nums[i] & (1 << bitPos)) {
                count++;
            }
        }
        
        // If at least k numbers have this bit set, set it in result
        if (count >= k) {
            result |= (1 << bitPos);
        }
    }
    
    return result;
}

int main() {
    int nums[1000];
    int numsSize = 0;
    int k;
    
    // Simple input parsing for [nums], k format
    char c;
    while (scanf("%d%c", &nums[numsSize], &c)) {
        numsSize++;
        if (c == ']') break;
    }
    scanf(", %d", &k);
    
    printf("%d\n", findKOr(nums, numsSize, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(32 * n) = O(n)

We check 32 bit positions, and for each position we iterate through all n numbers

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for counting and result

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Check Each Bit Position) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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