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Formatted question description: https://leetcode.ca/all/1869.html

# 1869. Longer Contiguous Segments of Ones than Zeros

Easy

## Description

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

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

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

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

Loop over s and calculate the lengths of the longest contiguous segment of 0’s and the longest contiguous segment of 1’s. Then compare the two lengths and return true if the longest contiguous segment of 0’s is strictly greater, or false otherwise.

• class Solution {
public boolean checkZeroOnes(String s) {
int length = s.length();
int maxZeros = 0, maxOnes = 0;
int zeros = 0, ones = 0;
char prev = '2';
for (int i = 0; i < length; i++) {
char curr = s.charAt(i);
if (curr == '0') {
if (curr == prev) {
zeros++;
maxZeros = Math.max(maxZeros, zeros);
} else {
zeros = 1;
maxZeros = Math.max(maxZeros, zeros);
}
} else {
if (curr == prev) {
ones++;
maxOnes = Math.max(maxOnes, ones);
} else {
ones = 1;
maxOnes = Math.max(maxOnes, ones);
}
}
prev = curr;
}
return maxOnes > maxZeros;
}
}

• // OJ: https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/
// Time: O(N)
// Space: O(1)
class Solution {
int longest(string s, char c) {
int ans = 0;
for (int i = 0; i < s.size(); ) {
if (s[i] != c) ++i;
else {
int start = i;
while (i < s.size() && s[i] == c) ++i;
ans = max(ans, i - start);
}
}
return ans;
}
public:
bool checkZeroOnes(string s) {
return longest(s, '0') < longest(s, '1');
}
};

• class Solution:
def checkZeroOnes(self, s: str) -> bool:
n0 = n1 = 0
t0 = t1 = 0
for c in s:
if c == '0':
t0 += 1
t1 = 0
else:
t0 = 0
t1 += 1
n0 = max(n0, t0)
n1 = max(n1, t1)
return n1 > n0

############

# 1869. Longer Contiguous Segments of Ones than Zeros
# https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/

class Solution:
def checkZeroOnes(self, s: str) -> bool:
ones = zeroes = 0

for k, g in itertools.groupby(s):
if k == "1":
ones = max(ones, len(list(g)))
else:
zeroes = max(zeroes, len(list(g)))

return ones > zeroes