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 defined as a maximal substring consisting of non-space characters only. This means spaces act as delimiters between words, and you need to find the final word before any trailing spaces.

Example: In the string "Hello World", the last word is "World" with length 5.

Note: The input string may contain leading and/or trailing spaces, but you should focus only on the actual words separated by spaces.

Input & Output

example_1.py — Basic Case

$ Input: s = "Hello World"

› Output: 5

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

example_2.py — Trailing Spaces

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

› Output: 4

💡 Note: The last word is "moon" (length 4). The trailing spaces are ignored.

example_3.py — Single Word

$ Input: s = "luffy"

› Output: 5

💡 Note: There is only one word "luffy" in the string, so it's also the last word with length 5.

Constraints

1 ≤ s.length ≤ 10⁴
s consists of only English letters and spaces ' '
The string contains at least one word

Visualization

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Understanding the Visualization

Start from the Right

Begin at the end of the string, like starting from the right side of a book spine

Skip Extra Spaces

Move left past any trailing spaces, like ignoring empty space on the spine

Count the Last Word

Count each character of the last word until you hit a space

Return the Count

You now have the length of the last word without reading the entire spine

Key Takeaway

🎯 Key Insight: Starting from the end eliminates unnecessary processing and naturally handles edge cases like trailing spaces

Asked in

G Google 15 a Amazon 12 ⊞ Microsoft 8 Apple 5

The optimal solution uses a two-pointer approach starting from the end of the string. We first skip any trailing spaces, then count backwards until we hit another space or reach the beginning. This approach is time-efficient (often faster than O(n)) and uses constant space, making it ideal for this problem.

Common Approaches

Approach	Time	Space	Notes
✓ Two Pointers (Optimal)	O(n)	O(1)	Start from the end, skip trailing spaces, then count the last word
Brute Force (Split and Index)	O(n)	O(n)	Split the string into words and return the length of the last word

Two Pointers (Optimal) — Algorithm Steps

Start from the end of the string (right pointer)
Skip any trailing spaces by moving left
Count non-space characters from right to left
Stop when we hit a space or reach the beginning
Return the count of characters in the last word

Visualization

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Step-by-Step Walkthrough

Start from End

Position pointer at the last character

Skip Trailing Spaces

Move left past any trailing spaces

Count Last Word

Count characters until we hit a space

Return Count

Return the length of the last word

Code -

solution.c — C

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

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

int main() {
    char s[10001];
    fgets(s, sizeof(s), stdin);
    // Remove newline and quotes
    s[strcspn(s, "\n")] = 0;
    if (s[0] == '"' && s[strlen(s)-1] == '"') {
        s[strlen(s)-1] = 0;
        memmove(s, s+1, strlen(s));
    }
    printf("%d\n", lengthOfLastWord(s));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

In worst case, we might scan the entire string, but often much faster as we start from the end

✓ Linear Growth

Space Complexity

O(1)

We only use a few integer variables to track position and count

✓ Linear Space

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Medium Frequency

~8 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Two Pointers (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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