LeetCode #2411 — MEDIUM

Smallest Subarrays With Maximum Bitwise OR

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a 0-indexed array nums of length n, consisting of non-negative integers. For each index i from 0 to n - 1, you must determine the size of the minimum sized non-empty subarray of nums starting at i (inclusive) that has the maximum possible bitwise OR.

In other words, let B_ij be the bitwise OR of the subarray nums[i...j]. You need to find the smallest subarray starting at i, such that bitwise OR of this subarray is equal to max(B_ik) where i <= k <= n - 1.

The bitwise OR of an array is the bitwise OR of all the numbers in it.

Return an integer array answer of size n where answer[i] is the length of the minimum sized subarray starting at i with maximum bitwise OR.

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

Example 1:

Input: nums = [1,0,2,1,3]
Output: [3,3,2,2,1]
Explanation:
The maximum possible bitwise OR starting at any index is 3. 
- Starting at index 0, the shortest subarray that yields it is [1,0,2].
- Starting at index 1, the shortest subarray that yields the maximum bitwise OR is [0,2,1].
- Starting at index 2, the shortest subarray that yields the maximum bitwise OR is [2,1].
- Starting at index 3, the shortest subarray that yields the maximum bitwise OR is [1,3].
- Starting at index 4, the shortest subarray that yields the maximum bitwise OR is [3].
Therefore, we return [3,3,2,2,1].

Example 2:

Input: nums = [1,2]
Output: [2,1]
Explanation:
Starting at index 0, the shortest subarray that yields the maximum bitwise OR is of length 2.
Starting at index 1, the shortest subarray that yields the maximum bitwise OR is of length 1.
Therefore, we return [2,1].

Constraints:

n == nums.length
1 <= n <= 10⁵
0 <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array nums of length n, consisting of non-negative integers. For each index i from 0 to n - 1, you must determine the size of the minimum sized non-empty subarray of nums starting at i (inclusive) that has the maximum possible bitwise OR. In other words, let Bij be the bitwise OR of the subarray nums[i...j]. You need to find the smallest subarray starting at i, such that bitwise OR of this subarray is equal to max(Bik) where i <= k <= n - 1. The bitwise OR of an array is the bitwise OR of all the numbers in it. Return an integer array answer of size n where answer[i] is the length of the minimum sized subarray starting at i with maximum bitwise OR. 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 · Sliding Window

Example 1

[1,0,2,1,3]

Example 2

[1,2]

Core Insight

What unlocks the optimal approach

Consider trying to solve the problem for each bit position separately.
For each bit position, find the position of the next number that has a 1 in that position, if any.
Take the maximum distance to such a number, including the current number.
Iterate backwards to achieve a linear complexity.

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 #2411: Smallest Subarrays With Maximum Bitwise OR
class Solution {
    public int[] smallestSubarrays(int[] nums) {
        int n = nums.length;
        int[] ans = new int[n];
        int[] f = new int[32];
        Arrays.fill(f, -1);
        for (int i = n - 1; i >= 0; --i) {
            int t = 1;
            for (int j = 0; j < 32; ++j) {
                if (((nums[i] >> j) & 1) == 1) {
                    f[j] = i;
                } else if (f[j] != -1) {
                    t = Math.max(t, f[j] - i + 1);
                }
            }
            ans[i] = t;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2411: Smallest Subarrays With Maximum Bitwise OR
func smallestSubarrays(nums []int) []int {
	n := len(nums)
	f := make([]int, 32)
	for i := range f {
		f[i] = -1
	}
	ans := make([]int, n)
	for i := n - 1; i >= 0; i-- {
		t := 1
		for j := 0; j < 32; j++ {
			if ((nums[i] >> j) & 1) == 1 {
				f[j] = i
			} else if f[j] != -1 {
				t = max(t, f[j]-i+1)
			}
		}
		ans[i] = t
	}
	return ans
}

# Accepted solution for LeetCode #2411: Smallest Subarrays With Maximum Bitwise OR
class Solution:
    def smallestSubarrays(self, nums: List[int]) -> List[int]:
        n = len(nums)
        ans = [1] * n
        f = [-1] * 32
        for i in range(n - 1, -1, -1):
            t = 1
            for j in range(32):
                if (nums[i] >> j) & 1:
                    f[j] = i
                elif f[j] != -1:
                    t = max(t, f[j] - i + 1)
            ans[i] = t
        return ans

// Accepted solution for LeetCode #2411: Smallest Subarrays With Maximum Bitwise OR
impl Solution {
    pub fn smallest_subarrays(nums: Vec<i32>) -> Vec<i32> {
        let n = nums.len();
        let mut ans = vec![1; n];
        let mut f = vec![-1; 32];

        for i in (0..n).rev() {
            let mut t = 1;
            for j in 0..32 {
                if (nums[i] >> j) & 1 != 0 {
                    f[j] = i as i32;
                } else if f[j] != -1 {
                    t = t.max(f[j] - i as i32 + 1);
                }
            }
            ans[i] = t;
        }

        ans
    }
}

// Accepted solution for LeetCode #2411: Smallest Subarrays With Maximum Bitwise OR
function smallestSubarrays(nums: number[]): number[] {
    const n = nums.length;
    const ans: number[] = Array(n).fill(1);
    const f: number[] = Array(32).fill(-1);

    for (let i = n - 1; i >= 0; i--) {
        let t = 1;
        for (let j = 0; j < 32; j++) {
            if ((nums[i] >> j) & 1) {
                f[j] = i;
            } else if (f[j] !== -1) {
                t = Math.max(t, f[j] - i + 1);
            }
        }
        ans[i] = t;
    }

    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(log n)

Space

O(1)

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.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.