LeetCode #1869 — EASY

Longer Contiguous Segments of Ones than Zeros

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

Given a binary string s, return true if the longest contiguous segment of 1's is strictly longer than the longest contiguous segment of 0's in s, or return false otherwise.

For example, in s = "110100010" the longest continuous segment of 1s has length 2, and the longest continuous segment of 0s has length 3.

Note that if there are no 0's, then the longest continuous segment of 0's is considered to have a length 0. The same applies if there is no 1's.

Example 1:

Input: s = "1101"
Output: true
Explanation:
The longest contiguous segment of 1s has length 2: "1101"
The longest contiguous segment of 0s has length 1: "1101"
The segment of 1s is longer, so return true.

Example 2:

Input: s = "111000"
Output: false
Explanation:
The longest contiguous segment of 1s has length 3: "111000"
The longest contiguous segment of 0s has length 3: "111000"
The segment of 1s is not longer, so return false.

Example 3:

Input: s = "110100010"
Output: false
Explanation:
The longest contiguous segment of 1s has length 2: "110100010"
The longest contiguous segment of 0s has length 3: "110100010"
The segment of 1s is not longer, so return false.

Constraints:

1 <= s.length <= 100
s[i] is either '0' or '1'.

Step 01

Brute Force Baseline

Problem summary: Given a binary string s, return true if the longest contiguous segment of 1's is strictly longer than the longest contiguous segment of 0's in s, or return false otherwise. For example, in s = "110100010" the longest continuous segment of 1s has length 2, and the longest continuous segment of 0s has length 3. Note that if there are no 0's, then the longest continuous segment of 0's is considered to have a length 0. The same applies if there is no 1's.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"1101"

Example 2

"111000"

Example 3

"110100010"

Core Insight

What unlocks the optimal approach

Check every possible segment of 0s and 1s.
Is there a way to iterate through the string to keep track of the current character and its count?

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 #1869: Longer Contiguous Segments of Ones than Zeros
class Solution {
    public boolean checkZeroOnes(String s) {
        return f(s, '1') > f(s, '0');
    }

    private int f(String s, char x) {
        int cnt = 0, mx = 0;
        for (int i = 0; i < s.length(); ++i) {
            if (s.charAt(i) == x) {
                mx = Math.max(mx, ++cnt);
            } else {
                cnt = 0;
            }
        }
        return mx;
    }
}

// Accepted solution for LeetCode #1869: Longer Contiguous Segments of Ones than Zeros
func checkZeroOnes(s string) bool {
	f := func(x rune) int {
		cnt, mx := 0, 0
		for _, c := range s {
			if c == x {
				cnt++
				mx = max(mx, cnt)
			} else {
				cnt = 0
			}
		}
		return mx
	}
	return f('1') > f('0')
}

# Accepted solution for LeetCode #1869: Longer Contiguous Segments of Ones than Zeros
class Solution:
    def checkZeroOnes(self, s: str) -> bool:
        def f(x: str) -> int:
            cnt = mx = 0
            for c in s:
                if c == x:
                    cnt += 1
                    mx = max(mx, cnt)
                else:
                    cnt = 0
            return mx

        return f("1") > f("0")

// Accepted solution for LeetCode #1869: Longer Contiguous Segments of Ones than Zeros
struct Solution;

impl Solution {
    fn check_zero_ones(s: String) -> bool {
        let s: Vec<char> = s.chars().collect();
        let n = s.len();
        let mut state = '2';
        let mut length = 0;
        let mut zero = 0;
        let mut one = 0;
        for i in 0..n {
            if s[i] == state {
                length += 1;
            } else {
                state = s[i];
                length = 1;
            }
            if s[i] == '0' {
                zero = zero.max(length);
            } else {
                one = one.max(length);
            }
        }
        one > zero
    }
}

#[test]
fn test() {
    let s = "1101".to_string();
    let res = true;
    assert_eq!(Solution::check_zero_ones(s), res);
    let s = "111000".to_string();
    let res = false;
    assert_eq!(Solution::check_zero_ones(s), res);
    let s = "110100010".to_string();
    let res = false;
    assert_eq!(Solution::check_zero_ones(s), res);
}

// Accepted solution for LeetCode #1869: Longer Contiguous Segments of Ones than Zeros
function checkZeroOnes(s: string): boolean {
    const f = (x: string): number => {
        let [mx, cnt] = [0, 0];
        for (const c of s) {
            if (c === x) {
                mx = Math.max(mx, ++cnt);
            } else {
                cnt = 0;
            }
        }
        return mx;
    };
    return f('1') > f('0');
}

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

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.