Truncate Sentence

Truncate Sentence - Problem

Array String Easy

A sentence is a list of words that are separated by a single space with no leading or trailing spaces. Each of the words consists of only uppercase and lowercase English letters (no punctuation).

For example, "Hello World", "HELLO", and "hello world hello world" are all sentences.

You are given a sentence s and an integer k. You want to truncate s such that it contains only the first k words. Return s after truncating it.

Input & Output

Example 1 — Basic Case

$ Input: s = "My dog ate his homework", k = 3

› Output: "My dog ate"

💡 Note: Take the first 3 words: "My", "dog", "ate". Join them with spaces to get "My dog ate".

Example 2 — Single Word

$ Input: s = "chomp chomp chomp", k = 1

› Output: "chomp"

💡 Note: Take only the first word "chomp" since k=1.

Example 3 — All Words

$ Input: s = "Hello world", k = 3

› Output: "Hello world"

💡 Note: k=3 but sentence only has 2 words, so return entire sentence "Hello world".

Constraints

1 ≤ s.length ≤ 500
k is a positive integer
1 ≤ k ≤ number of words in s
s consists of only lowercase and uppercase English letters and spaces

Visualization

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Asked in

a Amazon 25 M Microsoft 18

The key insight is to identify word boundaries and stop processing once we have k complete words. Best approach is split and join for simplicity: split by spaces, take first k elements, join back. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force - Character by Character

⏱️ Time: O(n) Space: O(1)

Walk through the string character by character, counting spaces to determine word boundaries. Build result string until we've seen k words.

Split and Join Approach

⏱️ Time: O(n) Space: O(n)

Use built-in string split function to convert sentence into array of words, take the first k elements, then join them back into a string.

Brute Force - Character by Character — Algorithm Steps

Iterate through each character in the string
Count spaces to track word completion
Build result string character by character
Stop when k words are found

Visualization

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Step-by-Step Walkthrough

Start Processing

Begin reading characters one by one

Count Words

Count spaces to track word boundaries

Stop at Target

Stop when k words are found

Code -

solution.c — C

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

char* solution(char* s, int k) {
    int len = strlen(s);
    char* result = malloc(len + 1);
    strcpy(result, s);
    
    int wordCount = 0;
    
    for (int i = 0; i < len; i++) {
        if (result[i] == ' ') {
            wordCount++;
            if (wordCount == k) {
                result[i] = '\0';
                break;
            }
        }
    }
    
    return result;
}

int main() {
    char s[1000];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    int k;
    scanf("%d", &k);
    
    char* result = solution(s, k);
    printf("%s\n", result);
    free(result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through characters until k words found

✓ Linear Growth

Space Complexity

O(1)

Only using variables to track position and count

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Character by Character — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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