Length of Last Word

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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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

✓ Reverse Iteration

⏱️ Time: O(n) Space: O(1)

Start from the end of the string and work backwards. First skip any trailing spaces, then count characters until we hit another space or reach the beginning. This avoids processing the entire string.

Split and Count

⏱️ Time: O(n) Space: O(n)

Split the string by spaces to get all words, then return the length of the last word in the array. This processes the entire string even though we only need the last word.

Reverse Iteration — Algorithm Steps

Start from end of string
Skip trailing spaces
Count characters of last word going backwards

Visualization

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Step-by-Step Walkthrough

Start from End

Begin at last character

Skip Spaces

Move backwards past trailing spaces

Count Letters

Count characters until next space

Code -

solution.c — C

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

int solution(char* s) {
    int i = strlen(s) - 1;
    
    // Skip trailing spaces
    while (i >= 0 && s[i] == ' ') {
        i--;
    }
    
    // Count length of last word
    int length = 0;
    while (i >= 0 && s[i] != ' ') {
        length++;
        i--;
    }
    
    return length;
}

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)

In worst case, might need to scan entire string if it's one word

✓ Linear Growth

Space Complexity

O(1)

Only uses a few variables, no extra data structures

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Reverse Iteration — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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