Solution Explanation for Truncate Sentence

The problem asks to truncate a given sentence s to contain only the first k words. The solution involves iterating through the sentence and counting the spaces to determine the end of the k-th word.

Approach

The most efficient approach is to leverage the built-in string manipulation functions in each programming language. The core idea is to:

1. Split the sentence into words: This is done using the split() method (or equivalent) which splits the string at each space character, resulting in a list or array of words.

2. Take the first k words: We slice the array of words to keep only the first k elements. This efficiently extracts the desired portion of the sentence.

3. Join the words back together: The sliced array of words is joined back into a string using the join() method (or equivalent), with a space character as the separator to reconstruct the truncated sentence.

Time and Space Complexity Analysis

The time complexity of this approach is dominated by the split() and join() operations. These operations typically have a linear time complexity, O(n), where n is the length of the string s. Therefore, the overall time complexity is O(n).

The space complexity depends on the size of the intermediate array of words. In the worst-case scenario (where all characters are letters except for spaces between words), the array size is proportional to the number of words. The space required for the final truncated sentence is also proportional to the number of words (or the length of the string). Therefore, the overall space complexity is O(n) in the worst case. In many cases where k is smaller than the total words, the space complexity is better than O(n).

Code Examples in Multiple Languages

The following code snippets demonstrate the solution using different programming languages:

Python

class Solution:
    def truncateSentence(self, s: str, k: int) -> str:
        words = s.split()
        return " ".join(words[:k])
 

Java

class Solution {
    public String truncateSentence(String s, int k) {
        String[] words = s.split(" ");
        StringBuilder result = new StringBuilder();
        for (int i = 0; i < k; i++) {
            result.append(words[i]);
            if (i < k - 1) {
                result.append(" ");
            }
        }
        return result.toString();
    }
}

C++

#include <string>
#include <sstream>
#include <vector>
 
class Solution {
public:
    string truncateSentence(string s, int k) {
        stringstream ss(s);
        string word;
        vector<string> words;
        while (ss >> word) {
            words.push_back(word);
        }
        string result = "";
        for (int i = 0; i < k; ++i) {
            result += words[i];
            if (i < k - 1) {
                result += " ";
            }
        }
        return result;
    }
};

JavaScript

/**
 * @param {string} s
 * @param {number} k
 * @return {string}
 */
var truncateSentence = function(s, k) {
    const words = s.split(' ');
    return words.slice(0, k).join(' ');
};

Go

func truncateSentence(s string, k int) string {
    words := strings.Split(s, " ")
    return strings.Join(words[:k], " ")
}

These code examples all follow the same basic approach: split, slice, and join. They demonstrate the efficiency and readability of this solution. Note that the choice of using StringBuilder in Java or string concatenation in other languages is a matter of optimization. For smaller strings, it might not have a big impact. For very large strings, StringBuilder is generally more efficient in Java because it avoids repeated string creation.