# 1869. Longer Contiguous Segments of Ones than Zeros

## Description

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'.

## Solutions

Solution 1: Two Passes

We design a function $f(x)$, which represents the length of the longest consecutive substring in string $s$ composed of $x$. If $f(1) > f(0)$, then return true, otherwise return false.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is \$O(1).

• 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;
}
}

• class Solution {
public:
bool checkZeroOnes(string s) {
auto f = [&](char x) {
int cnt = 0, mx = 0;
for (char& c : s) {
if (c == x) {
mx = max(mx, ++cnt);
} else {
cnt = 0;
}
}
return mx;
};
return f('1') > f('0');
}
};

• 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")


• 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')
}

• 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');
}


• /**
* @param {string} s
* @return {boolean}
*/
var checkZeroOnes = function (s) {
const f = x => {
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');
};

`