Solution Explanation: Longer Contiguous Segments of Ones than Zeros

This problem asks to determine if the longest contiguous segment of '1's in a binary string is strictly longer than the longest contiguous segment of '0's.

Approach:

The most efficient approach involves two linear scans of the input string. We can create a helper function f(x) that takes a character ('0' or '1') and returns the length of the longest contiguous segment of that character in the input string. Then, we simply compare f('1') and f('0'). If f('1') is greater, we return true; otherwise, false.

Algorithm:

1. Helper Function f(x):

   • Initialize cnt (current count) and mx (maximum count) to 0.
   • Iterate through the string:
     • If the current character is equal to x, increment cnt and update mx to be the maximum of mx and cnt.
     • Otherwise, reset cnt to 0.
   • Return mx.

2. Main Function:

   • Call f('1') to get the length of the longest contiguous segment of '1's.
   • Call f('0') to get the length of the longest contiguous segment of '0's.
   • Return true if f('1') > f('0'), otherwise false.

Time and Space Complexity:

• Time Complexity: O(n), where n is the length of the string. This is because we iterate through the string twice (once for '1's and once for '0's) in the helper function.
• Space Complexity: O(1). We only use a constant amount of extra space to store variables like cnt and mx.

Code Implementation (Python):

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")

Code Implementation (Java):

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;
    }
}

(Other languages like C++, Go, TypeScript, and JavaScript implementations are similar and follow the same logic. Refer to the provided code snippets for those implementations.)

This solution is efficient and straightforward, making it a good approach for solving this problem. The use of a helper function improves readability and code organization.