LeetCode #2419 — MEDIUM

Longest Subarray With Maximum Bitwise AND

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

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array nums of size n.

Consider a non-empty subarray from nums that has the maximum possible bitwise AND.

In other words, let k be the maximum value of the bitwise AND of any subarray of nums. Then, only subarrays with a bitwise AND equal to k should be considered.

Return the length of the longest such subarray.

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

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

Example 1:

Input: nums = [1,2,3,3,2,2]
Output: 2
Explanation:
The maximum possible bitwise AND of a subarray is 3.
The longest subarray with that value is [3,3], so we return 2.

Example 2:

Input: nums = [1,2,3,4]
Output: 1
Explanation:
The maximum possible bitwise AND of a subarray is 4.
The longest subarray with that value is [4], so we return 1.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums of size n. Consider a non-empty subarray from nums that has the maximum possible bitwise AND. In other words, let k be the maximum value of the bitwise AND of any subarray of nums. Then, only subarrays with a bitwise AND equal to k should be considered. Return the length of the longest such subarray. The bitwise AND of an array is the bitwise AND of all the numbers in it. A subarray is a contiguous 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 · Bit Manipulation

Example 1

[1,2,3,3,2,2]

Example 2

[1,2,3,4]

Core Insight

What unlocks the optimal approach

Notice that the bitwise AND of two different numbers will always be strictly less than the maximum of those two numbers.
What does that tell us about the nature of the subarray that we should choose?

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 #2419: Longest Subarray With Maximum Bitwise AND
class Solution {
    public int longestSubarray(int[] nums) {
        int mx = Arrays.stream(nums).max().getAsInt();
        int ans = 0, cnt = 0;
        for (int x : nums) {
            if (x == mx) {
                ans = Math.max(ans, ++cnt);
            } else {
                cnt = 0;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2419: Longest Subarray With Maximum Bitwise AND
func longestSubarray(nums []int) (ans int) {
	mx := slices.Max(nums)
	cnt := 0
	for _, x := range nums {
		if x == mx {
			cnt++
			ans = max(ans, cnt)
		} else {
			cnt = 0
		}
	}
	return
}

# Accepted solution for LeetCode #2419: Longest Subarray With Maximum Bitwise AND
class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        mx = max(nums)
        ans = cnt = 0
        for x in nums:
            if x == mx:
                cnt += 1
                ans = max(ans, cnt)
            else:
                cnt = 0
        return ans

// Accepted solution for LeetCode #2419: Longest Subarray With Maximum Bitwise AND
impl Solution {
    pub fn longest_subarray(nums: Vec<i32>) -> i32 {
        let mx = *nums.iter().max().unwrap();
        let mut ans = 0;
        let mut cnt = 0;

        for &x in nums.iter() {
            if x == mx {
                cnt += 1;
                ans = ans.max(cnt);
            } else {
                cnt = 0;
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #2419: Longest Subarray With Maximum Bitwise AND
function longestSubarray(nums: number[]): number {
    const mx = Math.max(...nums);
    let [ans, cnt] = [0, 0];
    for (const x of nums) {
        if (x === mx) {
            ans = Math.max(ans, ++cnt);
        } else {
            cnt = 0;
        }
    }
    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.