LeetCode 1784 – Check if Binary String Has at Most One Segment of Ones (Day-06)

A very interesting string observation problem from LeetCode that can be solved in one line using find() in C++.

Problem

You are given a binary string s without leading zeros.

Return true if the string contains at most one contiguous segment of '1's, otherwise return false.

Example

Input:  "1001"
Output: false

Explanation:

1 000 1

There are two segments of ones, so the answer is false.

Another example:

Input: "11100"
Output: true

There is only one segment of ones.

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Key Observation

If a binary string has more than one segment of ones, there must be a pattern:

01

Why?

Because after the first segment of 1s ends, we see:

1 → 0

And when another 1 starts again, the string must contain:

01

Example:

1001
   ↑
  "01"

So the trick is simple:

If "01" exists in the string, then there are multiple segments of ones.

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3 Elegant One-Line Solution

class Solution {
public:
    bool checkOnesSegment(string s) {
        return s.find("01") == string::npos;
    }
};

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Explanation

string::npos means pattern not found.

So:

s.find("01") == string::npos

means

"01" does NOT exist in the string

Therefore the string has at most one contiguous segment of ones.

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Complexity

Time Complexity

O(n)

Space Complexity

O(1)

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Interview Insight

Instead of counting segments manually, notice the pattern transition:

1 → 0 → 1

The substring "01" reveals the start of another segment of ones.

Recognizing such pattern-based shortcuts is a powerful trick in coding interviews.