LeetCode #2917 — EASY

Find the K-or of an Array

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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.

Example 1:

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

Output: 9

Explanation:

Represent numbers in binary:

Number	Bit 3	Bit 2	Bit 1	Bit 0
7	0	1	1	1
12	1	1	0	0
9	1	0	0	1
8	1	0	0	0
9	1	0	0	1
15	1	1	1	1
Result = 9	1	0	0	1

Bit 0 is set in 7, 9, 9, and 15. Bit 3 is set in 12, 9, 8, 9, and 15.
Only bits 0 and 3 qualify. The result is (1001)₂ = 9.

Example 2:

Input: nums = [2,12,1,11,4,5], k = 6

Output: 0

Explanation: No bit appears as 1 in all six array numbers, as required for K-or with k = 6. Thus, the result is 0.

Example 3:

Input: nums = [10,8,5,9,11,6,8], k = 1

Output: 15

Explanation: Since k == 1, the 1-or of the array is equal to the bitwise OR of all its elements. Hence, the answer is 10 OR 8 OR 5 OR 9 OR 11 OR 6 OR 8 = 15.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Bit Manipulation

Example 1

[7,12,9,8,9,15]
4

Core Insight

What unlocks the optimal approach

Fix a <code>bit</code> from the range <code>[0, 31]</code>, then count the number of elements of <code>nums</code> that have <code>bit</code> set in them.
<code>bit</code> is set in integer <code>x</code> if and only if <code>2<sup>bit</sup> AND x == 2<sup>bit</sup></code>, where <code>AND</code> is the bitwise <code>AND</code> operation.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2917: Find the K-or of an Array
class Solution {
    public int findKOr(int[] nums, int k) {
        int ans = 0;
        for (int i = 0; i < 32; ++i) {
            int cnt = 0;
            for (int x : nums) {
                cnt += (x >> i & 1);
            }
            if (cnt >= k) {
                ans |= 1 << i;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2917: Find the K-or of an Array
func findKOr(nums []int, k int) (ans int) {
	for i := 0; i < 32; i++ {
		cnt := 0
		for _, x := range nums {
			cnt += (x >> i & 1)
		}
		if cnt >= k {
			ans |= 1 << i
		}
	}
	return
}

# Accepted solution for LeetCode #2917: Find the K-or of an Array
class Solution:
    def findKOr(self, nums: List[int], k: int) -> int:
        ans = 0
        for i in range(32):
            cnt = sum(x >> i & 1 for x in nums)
            if cnt >= k:
                ans |= 1 << i
        return ans

// Accepted solution for LeetCode #2917: Find the K-or of an Array
fn find_k_or(nums: Vec<i32>, k: i32) -> i32 {
    let mut ret = 0;
    for i in 0..32 {
        let bit = 1 << i;
        let count = nums.iter().filter(|&num| num & bit != 0).count();
        if count >= k as usize {
            ret += bit;
        }
    }

    ret
}

fn main() {
    let nums = vec![7, 12, 9, 8, 9, 15];
    let ret = find_k_or(nums, 4);
    println!("ret={ret}");
}

#[test]
fn test_find_k_or() {
    {
        let nums = vec![7, 12, 9, 8, 9, 15];
        let ret = find_k_or(nums, 4);
        assert_eq!(ret, 9);
    }
    {
        let nums = vec![2, 12, 1, 11, 4, 5];
        let ret = find_k_or(nums, 6);
        assert_eq!(ret, 0);
    }
    {
        let nums = vec![10, 8, 5, 9, 11, 6, 8];
        let ret = find_k_or(nums, 1);
        assert_eq!(ret, 15);
    }
}

// Accepted solution for LeetCode #2917: Find the K-or of an Array
function findKOr(nums: number[], k: number): number {
    let ans = 0;
    for (let i = 0; i < 32; ++i) {
        let cnt = 0;
        for (const x of nums) {
            cnt += (x >> i) & 1;
        }
        if (cnt >= k) {
            ans |= 1 << i;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.