# 1859. Sorting the Sentence

## Description

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

A sentence can be shuffled by appending the 1-indexed word position to each word then rearranging the words in the sentence.

• For example, the sentence "This is a sentence" can be shuffled as "sentence4 a3 is2 This1" or "is2 sentence4 This1 a3".

Given a shuffled sentence s containing no more than 9 words, reconstruct and return the original sentence.

Example 1:

Input: s = "is2 sentence4 This1 a3"
Output: "This is a sentence"
Explanation: Sort the words in s to their original positions "This1 is2 a3 sentence4", then remove the numbers.


Example 2:

Input: s = "Myself2 Me1 I4 and3"
Output: "Me Myself and I"
Explanation: Sort the words in s to their original positions "Me1 Myself2 and3 I4", then remove the numbers.


Constraints:

• 2 <= s.length <= 200
• s consists of lowercase and uppercase English letters, spaces, and digits from 1 to 9.
• The number of words in s is between 1 and 9.
• The words in s are separated by a single space.
• s contains no leading or trailing spaces.

## Solutions

Solution 1: String Splitting

 First, we split the string $s$ by spaces to get the string array $words$. Then, we create a string array $ans$ of length $words$ to store the answer.
 Next, we iterate over each string $w$ in the string array $words$, find the position $i$ represented by the last character of $w$, then take the first $w -1$ characters of $w$ as the new string $w’$, and place $w’$ in the $i$th position of the array $ans$.

Finally, we join the array $ans$ into a string by spaces, which is the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string $s$.

• class Solution {
public String sortSentence(String s) {
String[] ws = s.split(" ");
int n = ws.length;
String[] ans = new String[n];
for (int i = 0; i < n; ++i) {
String w = ws[i];
ans[w.charAt(w.length() - 1) - '1'] = w.substring(0, w.length() - 1);
}
return String.join(" ", ans);
}
}

• class Solution {
public:
string sortSentence(string s) {
istringstream iss(s);
string w;
vector<string> ws;
while (iss >> w) {
ws.push_back(w);
}
vector<string> ss(ws.size());
for (auto& w : ws) {
ss[w.back() - '1'] = w.substr(0, w.size() - 1);
}
string ans;
for (auto& w : ss) {
ans += w + " ";
}
ans.pop_back();
return ans;
}
};

• class Solution:
def sortSentence(self, s: str) -> str:
ws = [(w[:-1], int(w[-1])) for w in s.split()]
ws.sort(key=lambda x: x[1])
return ' '.join(w for w, _ in ws)


• func sortSentence(s string) string {
ws := strings.Split(s, " ")
ans := make([]string, len(ws))
for _, w := range ws {
ans[w[len(w)-1]-'1'] = w[:len(w)-1]
}
return strings.Join(ans, " ")
}

• function sortSentence(s: string): string {
const ws = s.split(' ');
const ans = Array(ws.length);
for (const w of ws) {
ans[w.charCodeAt(w.length - 1) - '1'.charCodeAt(0)] = w.slice(0, -1);
}
return ans.join(' ');
}


• /**
* @param {string} s
* @return {string}
*/
var sortSentence = function (s) {
const ws = s.split(' ');
const ans = Array(ws.length);
for (const w of ws) {
ans[w.charCodeAt(w.length - 1) - '1'.charCodeAt(0)] = w.slice(0, -1);
}
return ans.join(' ');
};