Number of Segments in a String - Practice Coding Problems

Number of Segments in a String - Problem

String Easy

Imagine you're a text processing software that needs to count meaningful chunks of text. Given a string s, your task is to determine how many segments it contains.

A segment is defined as a contiguous sequence of non-space characters. Think of it as counting words, but more precisely - any group of characters that aren't spaces, separated by one or more spaces.

Examples:

"Hello World" → 2 segments: "Hello" and "World"
" a b " → 2 segments: "a" and "b" (leading/trailing spaces don't matter)
"" → 0 segments (empty string)

Your goal is to return the total count of these segments as an integer.

Input & Output

example_1.py — Basic Case

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

› Output: 5

💡 Note: The string contains 5 segments: 'Hello,', 'my', 'name', 'is', and 'John'. Each segment is separated by a single space.

example_2.py — Multiple Spaces

$ Input: s = "Hello"

› Output: 1

💡 Note: The string contains only one segment: 'Hello' with no spaces.

example_3.py — Edge Case

$ Input: s = ""

› Output: 0

💡 Note: An empty string contains no segments, so the result is 0.

Constraints

0 ≤ s.length ≤ 300
s consists of lowercase and uppercase English letters, digits, or spaces ' '
Note: The string may contain leading or trailing spaces

Visualization

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Understanding the Visualization

Fly Over the Map

Start scanning the string from left to right like flying over water

Spot Land

When you see a non-space character after spaces, you've found a new island

Count Islands

Each transition from 'water' to 'land' represents a new segment to count

Continue Flying

Keep scanning until you've covered the entire string

Key Takeaway

🎯 Key Insight: Just like counting islands from above, we only need to count the transitions from spaces to non-space characters. Each transition marks the beginning of a new segment!

Asked in

G Google 25 ⊞ Microsoft 18 a Amazon 15 Apple 12

The optimal solution uses built-in string splitting methods like split() or Fields() to divide the string by spaces and count non-empty segments. This approach is clean, readable, and efficient with O(n) time complexity. For learning purposes, the manual character-by-character approach helps understand the underlying logic of segment detection.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Character by Character)	O(n)	O(1)	Check every character and manually track segments
Built-in Split Method (Optimal)	O(n)	O(n)	Use language built-in string splitting functionality

Brute Force (Character by Character) — Algorithm Steps

Initialize segment counter and a flag to track if we're in a segment
Iterate through each character in the string
If current character is not a space and we're not in a segment, increment counter and mark as in segment
If current character is a space, mark as not in segment
Return the segment counter

Visualization

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Step-by-Step Walkthrough

Initialize

Start with count=0, inSegment=false

Scan Characters

Process each character left to right

Detect Transitions

When we hit non-space after space, increment count

Update State

Track whether we're currently in a segment

Code -

solution.c — C

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

int countSegments(char* s) {
    int count = 0;
    bool inSegment = false;
    
    for (int i = 0; s[i] != '\0'; i++) {
        if (s[i] != ' ') {
            if (!inSegment) {
                count++;
                inSegment = true;
            }
        } else {
            inSegment = false;
        }
    }
    
    return count;
}

int main() {
    char s[1000];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    if (s[0] == '"') {
        memmove(s, s+1, strlen(s));
        s[strlen(s)-1] = '\0';
    }
    printf("%d\n", countSegments(s));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We visit each character exactly once

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables regardless of input size

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Character by Character) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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