LeetCode #1859 — EASY

Sorting the Sentence

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"is2 sentence4 This1 a3"

Example 2

"Myself2 Me1 I4 and3"

Core Insight

What unlocks the optimal approach

Divide the string into the words as an array of strings
Sort the words by removing the last character from each word and sorting according to it

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1859: Sorting the Sentence
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);
    }
}

// Accepted solution for LeetCode #1859: Sorting the Sentence
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, " ")
}

# Accepted solution for LeetCode #1859: Sorting the Sentence
class Solution:
    def sortSentence(self, s: str) -> str:
        ws = s.split()
        ans = [None] * len(ws)
        for w in ws:
            ans[int(w[-1]) - 1] = w[:-1]
        return " ".join(ans)

// Accepted solution for LeetCode #1859: Sorting the Sentence
struct Solution;

impl Solution {
    fn sort_sentence(s: String) -> String {
        let mut pairs: Vec<(String, String)> = s
            .split_whitespace()
            .map(|s| {
                let n = s.len();
                let p = s[n - 1..].to_string();
                let word = s[0..n - 1].to_string();
                (p, word)
            })
            .collect();
        pairs.sort_unstable();
        let words: Vec<String> = pairs.into_iter().map(|(_, word)| word).collect();
        words.join(" ")
    }
}

#[test]
fn test() {
    let s = "is2 sentence4 This1 a3".to_string();
    let res = "This is a sentence".to_string();
    assert_eq!(Solution::sort_sentence(s), res);
    let s = "Myself2 Me1 I4 and3".to_string();
    let res = "Me Myself and I".to_string();
    assert_eq!(Solution::sort_sentence(s), res);
}

// Accepted solution for LeetCode #1859: Sorting the Sentence
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(' ');
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.