LeetCode #3171 — HARD

Find Subarray With Bitwise OR Closest to K

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an array nums and an integer k. You need to find a subarray of nums such that the absolute difference between k and the bitwise OR of the subarray elements is as small as possible. In other words, select a subarray nums[l..r] such that |k - (nums[l] OR nums[l + 1] ... OR nums[r])| is minimum.

Return the minimum possible value of the absolute difference.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

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

Output: 0

Explanation:

The subarray nums[0..1] has OR value 3, which gives the minimum absolute difference |3 - 3| = 0.

Example 2:

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

Output: 1

Explanation:

The subarray nums[1..1] has OR value 3, which gives the minimum absolute difference |3 - 2| = 1.

Example 3:

Input: nums = [1], k = 10

Output: 9

Explanation:

There is a single subarray with OR value 1, which gives the minimum absolute difference |10 - 1| = 9.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
1 <= k <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an array nums and an integer k. You need to find a subarray of nums such that the absolute difference between k and the bitwise OR of the subarray elements is as small as possible. In other words, select a subarray nums[l..r] such that |k - (nums[l] OR nums[l + 1] ... OR nums[r])| is minimum. Return the minimum possible value of the absolute difference. A subarray is a contiguous non-empty sequence of elements within an array.

Baseline thinking

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

Pattern signal: Array · Binary Search · Bit Manipulation · Segment Tree

Example 1

[1,2,4,5]
3

Example 2

[1,3,1,3]
2

Example 3

[1]
10

Core Insight

What unlocks the optimal approach

Let <code>dp[i]</code> be the set of all the bitwise <code>OR</code> of all the subarrays ending at index <code>i</code>.
We start from <code>nums[i]</code>, taking the bitwise <code>OR</code> result by including elements one by one from <code>i</code> towards left. Notice that only unset bits can become set on adding an element, and set bits never become unset again.
Hence <code>dp[i]</code> can contain at most 30 elements.

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

Largest constraint values

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 #3171: Find Subarray With Bitwise OR Closest to K
class Solution {
    public int minimumDifference(int[] nums, int k) {
        int mx = 0;
        for (int x : nums) {
            mx = Math.max(mx, x);
        }
        int m = 32 - Integer.numberOfLeadingZeros(mx);
        int[] cnt = new int[m];
        int n = nums.length;
        int ans = Integer.MAX_VALUE;
        for (int i = 0, j = 0, s = 0; j < n; ++j) {
            s |= nums[j];
            ans = Math.min(ans, Math.abs(s - k));
            for (int h = 0; h < m; ++h) {
                if ((nums[j] >> h & 1) == 1) {
                    ++cnt[h];
                }
            }
            while (i < j && s > k) {
                for (int h = 0; h < m; ++h) {
                    if ((nums[i] >> h & 1) == 1 && --cnt[h] == 0) {
                        s ^= 1 << h;
                    }
                }
                ++i;
                ans = Math.min(ans, Math.abs(s - k));
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3171: Find Subarray With Bitwise OR Closest to K
func minimumDifference(nums []int, k int) int {
	m := bits.Len(uint(slices.Max(nums)))
	cnt := make([]int, m)
	ans := math.MaxInt32
	s, i := 0, 0
	for j, x := range nums {
		s |= x
		ans = min(ans, abs(s-k))
		for h := 0; h < m; h++ {
			if x>>h&1 == 1 {
				cnt[h]++
			}
		}
		for i < j && s > k {
			y := nums[i]
			for h := 0; h < m; h++ {
				if y>>h&1 == 1 {
					cnt[h]--
					if cnt[h] == 0 {
						s ^= 1 << h
					}
				}
			}
			ans = min(ans, abs(s-k))
			i++
		}
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #3171: Find Subarray With Bitwise OR Closest to K
class Solution:
    def minimumDifference(self, nums: List[int], k: int) -> int:
        m = max(nums).bit_length()
        cnt = [0] * m
        s = i = 0
        ans = inf
        for j, x in enumerate(nums):
            s |= x
            ans = min(ans, abs(s - k))
            for h in range(m):
                if x >> h & 1:
                    cnt[h] += 1
            while i < j and s > k:
                y = nums[i]
                for h in range(m):
                    if y >> h & 1:
                        cnt[h] -= 1
                        if cnt[h] == 0:
                            s ^= 1 << h
                i += 1
                ans = min(ans, abs(s - k))
        return ans

// Accepted solution for LeetCode #3171: Find Subarray With Bitwise OR Closest to K
/**
 * [3171] Find Subarray With Bitwise OR Closest to K
 */
pub struct Solution {}

// submission codes start here
use std::cmp::Ordering;

impl Solution {
    pub fn minimum_difference(nums: Vec<i32>, k: i32) -> i32 {
        let n = nums.len();
        let mut bits_max_pos = vec![None; 31];
        let mut pos_to_bit = Vec::with_capacity(31);
        let mut result = i32::MAX;

        for i in 0..n {
            for j in 0..=30 {
                if nums[i] >> j & 1 == 1 {
                    bits_max_pos[j] = Some(i)
                }
            }

            pos_to_bit.clear();
            for j in 0..=30 {
                if let Some(pos) = bits_max_pos[j] {
                    pos_to_bit.push((pos, j));
                }
            }

            pos_to_bit.sort_by(|a, b| match b.0.cmp(&a.0) {
                Ordering::Less => Ordering::Less,
                Ordering::Equal => b.1.cmp(&a.1),
                Ordering::Greater => Ordering::Greater,
            });

            let mut value = 0;
            let (mut left, mut right) = (0, 0);

            while right < pos_to_bit.len() {
                while right < pos_to_bit.len() && pos_to_bit[right].0 == pos_to_bit[left].0 {
                    value |= 1 << pos_to_bit[right].1;
                    right += 1;
                }

                result = result.min((value - k).abs());
                left = right;
            }
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3171() {
        assert_eq!(0, Solution::minimum_difference(vec![1, 2, 4, 5], 3));
        assert_eq!(1, Solution::minimum_difference(vec![1, 3, 1, 3], 2));
        assert_eq!(9, Solution::minimum_difference(vec![1], 10));
    }
}

// Accepted solution for LeetCode #3171: Find Subarray With Bitwise OR Closest to K
function minimumDifference(nums: number[], k: number): number {
    const m = Math.max(...nums).toString(2).length;
    const n = nums.length;
    const cnt: number[] = Array(m).fill(0);
    let ans = Infinity;
    for (let i = 0, j = 0, s = 0; j < n; ++j) {
        s |= nums[j];
        ans = Math.min(ans, Math.abs(s - k));
        for (let h = 0; h < m; ++h) {
            if ((nums[j] >> h) & 1) {
                ++cnt[h];
            }
        }
        while (i < j && s > k) {
            for (let h = 0; h < m; ++h) {
                if ((nums[i] >> h) & 1 && --cnt[h] === 0) {
                    s ^= 1 << h;
                }
            }
            ans = Math.min(ans, Math.abs(s - k));
            ++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 × log M)

Space

O(log M)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Find Subarray With Bitwise OR Closest to K

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries

Boundary update without `+1` / `-1`