# 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`, or return `false` otherwise.

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

SOLUTION:

``````class Solution:
def checkZeroOnes(self, s: str) -> bool:
s += "x"
longest = [0, 0]
curr = [0, 0]
for c in s:
if c == "x":
longest = max(longest, curr)
longest = max(longest, curr)
elif c == "1":
longest = max(longest, curr)
curr = 0
curr += 1
elif c == "0":
longest = max(longest, curr)
curr = 0
curr += 1
return longest > longest
``````