In the competitive programming world, Leetcode is a popular platform for honing problem-solving skills. One intriguing problem you might come across is “Longer Contiguous Segments of Ones than Zeros”. This problem tests your understanding of string manipulation, iteration, and elementary algorithmic skills. This comprehensive article will guide you through various approaches to solve this problem, starting with a naive approach and moving on to more optimized solutions.

## Problem Statement

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. So, in this case, you should return `False`

.

Constraints:

`1 <= s.length <= 100`

`s[i]`

is either ‘0’ or ‘1’.

## Initial Thoughts and Brute-force Approach

The most straightforward approach to solving this problem involves iterating through the string and keeping track of the length of contiguous segments of 1s and 0s.

Python Code:

```
def checkZeroOnes(s: str) -> bool:
longest_ones = 0
longest_zeros = 0
current_ones = 0
current_zeros = 0
for ch in s:
if ch == '1':
current_ones += 1
current_zeros = 0
else:
current_zeros += 1
current_ones = 0
longest_ones = max(longest_ones, current_ones)
longest_zeros = max(longest_zeros, current_zeros)
return longest_ones > longest_zeros
```

## Optimized Approaches

### Using Counters

Given that the problem constraints are quite lenient (string length up to 100), the brute-force approach is sufficient for this problem. However, you can still optimize a little using counters and breaking early when possible.

Python Code:

```
def checkZeroOnes(s: str) -> bool:
longest_ones = 0
longest_zeros = 0
current_ones = 0
current_zeros = 0
for ch in s:
if ch == '1':
current_ones += 1
current_zeros = 0
else:
current_zeros += 1
current_ones = 0
longest_ones = max(longest_ones, current_ones)
longest_zeros = max(longest_zeros, current_zeros)
if longest_ones > s.length() // 2:
return True
if longest_zeros > s.length() // 2:
return False
return longest_ones > longest_zeros
```

### State Machine Approach

The problem can also be thought of as a simple state machine where you’re either in the “counting 1s” state or the “counting 0s” state.

Python Code:

```
def checkZeroOnes(s: str) -> bool:
longest = { '1': 0, '0': 0 }
current = { '1': 0, '0': 0 }
last_ch = ''
for ch in s:
if ch != last_ch:
current[ch] = 0
current[ch] += 1
longest[ch] = max(longest[ch], current[ch])
last_ch = ch
return longest['1'] > longest['0']
```

## Performance Analysis

All the above approaches are linear with a time complexity of O(n), where n is the length of the string. Given the problem’s constraints, all these approaches will run in a reasonable time.

## Conclusion

The “Longer Contiguous Segments of Ones than Zeros” problem on Leetcode is a straightforward but interesting string manipulation problem that can be solved using various approaches. The problem tests fundamental programming skills, including iteration and condition checking. Although it can be solved using a brute-force approach, understanding the underlying logic can help you apply similar techniques to more complex problems.