Number of Segments in a String

Number of Segments in a String - Problem

String Easy

Given a string s, return the number of segments in the string.

A segment is defined to be a contiguous sequence of non-space characters.

Note that the string may contain leading/trailing spaces or multiple spaces between segments.

Input & Output

Example 1 — Basic Case

$ Input: s = "Hello, my name is John"

› Output: 5

💡 Note: The five segments are: "Hello,", "my", "name", "is", "John"

Example 2 — Multiple Spaces

$ Input: s = "Hello"

› Output: 1

💡 Note: Only one segment: "Hello"

Example 3 — Leading/Trailing Spaces

$ Input: s = " Hello world "

› Output: 2

💡 Note: Two segments: "Hello" and "world", ignoring leading/trailing spaces

Constraints

0 ≤ s.length ≤ 300
s consists of lowercase and uppercase English letters, digits, or one of the following characters: ' ', '.', ',', '?', '!', '"', '/', '\', '(', ')'

Visualization

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Asked in

G Google 15 a Amazon 12

The key insight is to identify contiguous sequences of non-space characters. Best approach is to split by spaces and count non-empty segments. Time: O(n), Space: O(k) where k is number of segments.

Common Approaches

✓ Character by Character Traversal

⏱️ Time: O(n) Space: O(1)

Go through each character in the string. When we find a non-space character after a space (or at the beginning), we've found the start of a new segment. Count each segment as we encounter it.

Split and Filter Approach

⏱️ Time: O(n) Space: O(k)

Split the string by spaces to get all potential segments, then filter out empty strings that result from consecutive spaces. Count the remaining non-empty segments.

Character by Character Traversal — Algorithm Steps

Iterate through each character
Track if we're currently in a segment
When transitioning from space to non-space, increment segment count

Visualization

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Step-by-Step Walkthrough

Initialize

Start with segment count = 0, not in segment

Scan Characters

For each non-space after space, increment count

Count Result

Return total segments found

Code -

solution.c — C

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

int solution(char* s) {
    if (s == NULL || strlen(s) == 0) {
        return 0;
    }
    
    int segments = 0;
    int in_segment = 0;
    
    for (int i = 0; s[i] != '\0'; i++) {
        char ch = s[i];
        if (ch != ' ') {
            if (!in_segment) {
                segments += 1;
                in_segment = 1;
            }
        } else {
            in_segment = 0;
        }
    }
    
    return segments;
}

int main() {
    char s[1000];
    if (fgets(s, sizeof(s), stdin) != NULL) {
        s[strcspn(s, "\n")] = 0;
    } else {
        s[0] = '\0';
    }
    int result = solution(s);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string of length n

✓ Linear Growth

Space Complexity

O(1)

Only uses a few variables for counting and state tracking

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Character by Character Traversal — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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