Check if Binary String Has at Most One Segment of Ones - Problem

String Easy

Given a binary string s without leading zeros, return true if s contains at most one contiguous segment of ones. Otherwise, return false.

A contiguous segment of ones is a substring that contains only the character '1' and is surrounded by '0' characters or string boundaries.

Input & Output

Example 1 — Single Segment

$ Input: s = "1001"

› Output: false

💡 Note: There are two segments of ones: "1" at position 0, and "1" at position 3. Since there are more than one segment, return false.

Example 2 — Contiguous Segment

$ Input: s = "110"

› Output: true

💡 Note: There is only one contiguous segment of ones: "11" at positions 0-1. Return true.

Example 3 — All Ones

$ Input: s = "1"

› Output: true

💡 Note: The string contains only one character '1', which forms exactly one segment. Return true.

Constraints

1 ≤ s.length ≤ 100
s[i] is either '0' or '1'
s[0] == '1' (no leading zeros)

Visualization

Tap to expand

Asked in

a Amazon 15 M Microsoft 8

The key insight is that if a binary string has multiple segments of 1s, there must be a '0' followed by a '1' somewhere (the pattern '01'). Since the string has no leading zeros, finding this pattern guarantees multiple segments. Best approach scans for '01' pattern in O(n) time, O(1) space.

Common Approaches

✓ Brute Force - Count All Segments

⏱️ Time: O(n) Space: O(1)

Iterate through the string and identify each contiguous segment of 1s by tracking when we enter and exit segments. Count the total number of segments found.

Optimized - Pattern Detection

⏱️ Time: O(n) Space: O(1)

Since we know the string has no leading zeros, if there are multiple segments of 1s, we must see a '0' followed by a '1' somewhere. The key insight is that the pattern '01' can only exist if there are at least 2 segments.

Brute Force - Count All Segments — Algorithm Steps

Iterate through each character in the string
When transitioning from 0 to 1, start a new segment
When transitioning from 1 to 0, end current segment
Count total segments and check if ≤ 1

Visualization

Tap to expand

Step-by-Step Walkthrough

Initialize

Start with segments=0, inSegment=false

Process

For each character, update state and count segments

Result

Check if segments ≤ 1

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

bool solution(char* s) {
    int segments = 0;
    bool inSegment = false;
    
    for (int i = 0; s[i] != '\0'; i++) {
        if (s[i] == '1' && !inSegment) {
            segments++;
            inSegment = true;
        } else if (s[i] == '0') {
            inSegment = false;
        }
    }
    
    return segments <= 1;
}

int main() {
    char s[100001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    bool result = solution(s);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string of length n

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to track state

✓ Linear Space

23.4K Views

Medium Frequency

~10 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Check if Binary String Has at Most One Segment of Ones - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Count All Segments — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler