Leetcode #1869: Longer Contiguous Segments of Ones than Zeros
In this guide, we solve Leetcode #1869 Longer Contiguous Segments of Ones than Zeros in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: String
Intuition
We need to scan characters while tracking positions or counts.
A simple state machine keeps the logic precise.
Approach
Iterate through the string once and update the state for each character.
Use a map or array if you need fast lookups.
Steps:
- Iterate through characters.
- Maintain necessary state.
- Build or validate the output.
Example
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.
Python Solution
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")
Complexity
The time complexity is , where is the length of the string . The space complexity is $O(1)`.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.