Length of Last Word - Practice Coding Problems

Length of Last Word - Problem

String Easy

Given a string s consisting of words and spaces, return the length of the last word in the string.

A word is a maximal substring consisting of non-space characters only.

Input & Output

Example 1 — Basic Case

$ Input: s = "Hello World"

› Output: 5

💡 Note: The last word is "World" which has length 5

Example 2 — Trailing Spaces

$ Input: s = " fly me to the moon "

› Output: 4

💡 Note: The last word is "moon" which has length 4, ignoring trailing spaces

Example 3 — Single Word

$ Input: s = "luffy"

› Output: 5

💡 Note: Only one word "luffy" with length 5

Constraints

1 ≤ s.length ≤ 10⁴
s consists of only English letters and spaces ' '

Visualization

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Understanding the Visualization

Input String

String with words separated by spaces

Identify Last Word

Find the rightmost word ignoring trailing spaces

Count Length

Return the character count of last word

Key Takeaway

🎯 Key Insight: Work backwards from the end to avoid processing the entire string unnecessarily

Asked in

Apple 12 M Microsoft 8

The key insight is to scan backwards from the end, skipping trailing spaces and counting characters until hitting the next space. The reverse iteration approach is optimal with Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Split and Count	O(n)	O(n)	Split the string into words and return the length of the last word
Reverse Iteration	O(n)	O(1)	Scan backwards from end, skip trailing spaces, count last word length

Split and Count — Algorithm Steps

Split string by spaces to get array of words
Return length of last word in array

Visualization

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Step-by-Step Walkthrough

Split String

Divide string into array of words

Get Last Word

Access the final word in array

Count Length

Return length of last word

Code -

solution.c — C

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

int solution(char* s) {
    char* words[1000];
    int wordCount = 0;
    char* token = strtok(s, " ");
    
    while (token != NULL) {
        words[wordCount++] = token;
        token = strtok(NULL, " ");
    }
    
    return strlen(words[wordCount - 1]);
}

int main() {
    char s[10000];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    int result = solution(s);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Must scan entire string to split into words

✓ Linear Growth

Space Complexity

O(n)

Split creates array storing all words from the string

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Split and Count — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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