LeetCode #1004 — MEDIUM

Max Consecutive Ones III

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

The Problem

Problem Statement

Given a binary array nums and an integer k, return the maximum number of consecutive 1's in the array if you can flip at most k 0's.

Example 1:

Input: nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2
Output: 6
Explanation: [1,1,1,0,0,1,1,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

Example 2:

Input: nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], k = 3
Output: 10
Explanation: [0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

Constraints:

1 <= nums.length <= 10⁵
nums[i] is either 0 or 1.
0 <= k <= nums.length

Step 01

Brute Force Baseline

Problem summary: Given a binary array nums and an integer k, return the maximum number of consecutive 1's in the array if you can flip at most k 0's.

Baseline thinking

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

Pattern signal: Array · Binary Search · Sliding Window

Example 1

[1,1,1,0,0,0,1,1,1,1,0]
2

Example 2

[0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1]
3

Core Insight

What unlocks the optimal approach

One thing's for sure, we will only flip a zero if it extends an existing window of 1s. Otherwise, there's no point in doing it, right? Think Sliding Window!
Since we know this problem can be solved using the sliding window construct, we might as well focus in that direction for hints. Basically, in a given window, we can never have > K zeros, right?
We don't have a fixed size window in this case. The window size can grow and shrink depending upon the number of zeros we have (we don't actually have to flip the zeros here!).
The way to shrink or expand a window would be based on the number of zeros that can still be flipped and so on.

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 #1004: Max Consecutive Ones III
class Solution {
    public int longestOnes(int[] nums, int k) {
        int l = 0, cnt = 0;
        for (int x : nums) {
            cnt += x ^ 1;
            if (cnt > k) {
                cnt -= nums[l++] ^ 1;
            }
        }
        return nums.length - l;
    }
}

// Accepted solution for LeetCode #1004: Max Consecutive Ones III
func longestOnes(nums []int, k int) int {
	l, cnt := 0, 0
	for _, x := range nums {
		cnt += x ^ 1
		if cnt > k {
			cnt -= nums[l] ^ 1
			l++
		}
	}
	return len(nums) - l
}

# Accepted solution for LeetCode #1004: Max Consecutive Ones III
class Solution:
    def longestOnes(self, nums: List[int], k: int) -> int:
        l = cnt = 0
        for x in nums:
            cnt += x ^ 1
            if cnt > k:
                cnt -= nums[l] ^ 1
                l += 1
        return len(nums) - l

// Accepted solution for LeetCode #1004: Max Consecutive Ones III
struct Solution;

impl Solution {
    fn longest_ones(a: Vec<i32>, k: i32) -> i32 {
        let n = a.len();
        let mut sum = 0;
        let mut res = 0;
        let mut start = 0;
        let mut end = 0;
        while end < n {
            if sum <= k {
                if a[end] == 0 {
                    sum += 1;
                }
                end += 1;
            } else {
                if a[start] == 0 {
                    sum -= 1;
                }
                start += 1;
            }
            if sum <= k {
                res = res.max(end - start);
            }
        }
        res as i32
    }
}

#[test]
fn test() {
    let a = vec![1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0];
    let k = 2;
    let res = 6;
    assert_eq!(Solution::longest_ones(a, k), res);
    let a = vec![0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1];
    let k = 3;
    let res = 10;
    assert_eq!(Solution::longest_ones(a, k), res);
    let a = vec![0, 0, 0, 1];
    let k = 4;
    let res = 4;
    assert_eq!(Solution::longest_ones(a, k), res);
}

// Accepted solution for LeetCode #1004: Max Consecutive Ones III
function longestOnes(nums: number[], k: number): number {
    let [l, cnt] = [0, 0];
    for (const x of nums) {
        cnt += x ^ 1;
        if (cnt > k) {
            cnt -= nums[l++] ^ 1;
        }
    }
    return nums.length - l;
}

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.