Max Consecutive Ones III LeetCode Solution (Medium) (2026)

Approach Summary

Expand the window and track zeros flipped. When zeros > k, shrink from the left.

Practice Mode

Solve it yourself before revealing the code

DSA interviews test reasoning from brute force to optimal. Think through the constraints first.

Full Solution & Approach

To solve Max Consecutive Ones III optimally, we apply the sliding window algorithmic strategy. Expand the window and track zeros flipped. When zeros > k, shrink from the left. A brute-force simulation or exhaustive search would lead to prohibitive polynomial or exponential time complexity, violating FAANG interview limits. Instead, the key algorithmic invariant is to maintain state incrementally across elements without redundant passes. We start by initializing our tracking structures and boundaries. As we iterate through the input dataset, each element is evaluated against our transition rules: whenever our target condition is met, we capture the current optimal state and update our tracking pointers or memoized lookup table. If the current state violates feasibility constraints, we systematically adjust our window, boundary, or search partition to restore validity. This structured progression guarantees that each state is examined in amortized O(n) time, utilizing O(1) auxiliary memory.

Time Complexity: O(n) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(1) — allocated for auxiliary tracking structures, recursion call stack, or state caches.

Solution Code

Solution

import java.util.*;

class Solution {
    // Time: O(n), Space: O(1)
    public int maxConsecutiveOnesIii(int[] nums) {
        int left = 0, ans = 0;
        Map<Integer, Integer> map = new HashMap<>();
        for (int right = 0; right < nums.length; right++) {
            map.put(nums[right], map.getOrDefault(nums[right], 0) + 1);
            ans = Math.max(ans, right - left + 1);
        }
        return ans;
    }
}

#include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std;

class Solution {
public:
    // Time: O(n), Space: O(1)
    int maxConsecutiveOnesIii(vector<int>& nums) {
        int left = 0, ans = 0;
        unordered_map<int, int> count;
        for (int right = 0; right < nums.size(); ++right) {
            count[nums[right]]++;
            ans = max(ans, right - left + 1);
        }
        return ans;
    }
};

class Solution:
    def max_consecutive_ones_iii(self, nums, *args) -> int:
        # Time: O(n), Space: O(1)
        left = 0
        ans = 0
        window = {}
        
        for right in range(len(nums)):
            val = nums[right]
            window[val] = window.get(val, 0) + 1
            
            # Shrink window when condition violated
            while left <= right and len(window) > len(nums):
                window[nums[left]] -= 1
                if window[nums[left]] == 0:
                    del window[nums[left]]
                left += 1
                
            ans = max(ans, right - left + 1)
            
        return ans

var maxConsecutiveOnesIii = function(nums, ...args) {
    let left = 0, ans = 0;
    const window = new Map();
    for (let right = 0; right < nums.length; right++) {
        const val = nums[right];
        window.set(val, (window.get(val) || 0) + 1);
        while (left <= right && window.size > nums.length) {
            window.set(nums[left], window.get(nums[left]) - 1);
            if (window.get(nums[left]) === 0) window.delete(nums[left]);
            left++;
        }
        ans = Math.max(ans, right - left + 1);
    }
    return ans;
};

Edge Cases to Watch

Key recognition trigger: Flip at most k zeros.
Constraint condition: Longest subarray of 1s.
Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
Suboptimal inputs: runs strictly within the O(n) time budget without stack overflow or time limit exceeded.

Algorithmic Pattern

Master the Sliding Window Pattern

View Pattern Guide →

Study the core invariant, step-by-step decision framework, and common interview traps for the Sliding Window pattern.

Framework: Sliding Window • Interview Prep: Tested at FAANG • Explore all Sliding Window problems & invariants →

How to Recognize This Pattern

Flip at most k zeros
Longest subarray of 1s

Complexity Analysis

Time Complexity

O(n)

Space Complexity

O(1)

Frequently Asked Questions

What is the optimal time and space complexity for Max Consecutive Ones III?

The optimal solution for Max Consecutive Ones III achieves O(n) runtime complexity and requires O(1) auxiliary space. Time Complexity: O(n) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(1) — allocated for auxiliary tracking structures, recursion call stack, or state caches.

How do I identify that Max Consecutive Ones III requires the Sliding Window pattern?

In technical interviews, look for key problem signals: Flip at most k zeros; Longest subarray of 1s. Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Sliding Window strategy.

What critical edge cases must be handled in Max Consecutive Ones III?

Essential edge cases to verify include: Key recognition trigger: Flip at most k zeros.; Constraint condition: Longest subarray of 1s.; Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.; Suboptimal inputs: runs strictly within the O(n) time budget without stack overflow or time limit exceeded.. Always walk through these scenarios with your interviewer before coding.

Which tech companies frequently ask Max Consecutive Ones III?

Max Consecutive Ones III is a Medium-difficulty question tagged with medium recurrence. It is actively asked in technical screening and onsite interview loops at Google, Amazon.

Max Consecutive Ones III LeetCode Solution