## Description

https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/

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`

. Return `false`

*otherwise*.

- For example, in
`s = "`

the longest contiguous segment of__11__01__000__10"`1`

s has length`2`

, and the longest contiguous segment of`0`

s has length`3`

.

Note that if there are no `0`

s, then the longest contiguous segment of `0`

s is considered to have length `0`

. The same applies if there are no `1`

s.

**Example 1:**

Input:s = "1101"Output:trueExplanation: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:falseExplanation: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:falseExplanation: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'`

.

## Explanation

Compares the longest consecutive ones and zeros counts.

## Python Solution

```
class Solution:
def checkZeroOnes(self, s: str) -> bool:
longest_ones = 0
longest_zeros = 0
ones = 0
zeros = 0
for c in s:
if c == '1':
ones += 1
longest_zeros = max(longest_zeros, zeros)
zeros = 0
else:
zeros += 1
longest_ones = max(longest_ones, ones)
ones = 0
longest_zeros = max(longest_zeros, zeros)
longest_ones = max(longest_ones, ones)
return longest_ones > longest_zeros
```

- Time Complexity: O(N).
- Space Complexity: O(1).