LeetCode #1451 — MEDIUM

Rearrange Words in a Sentence

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

Given a sentence text (A sentence is a string of space-separated words) in the following format:

First letter is in upper case.
Each word in text are separated by a single space.

Your task is to rearrange the words in text such that all words are rearranged in an increasing order of their lengths. If two words have the same length, arrange them in their original order.

Return the new text following the format shown above.

Example 1:

Input: text = "Leetcode is cool"
Output: "Is cool leetcode"
Explanation: There are 3 words, "Leetcode" of length 8, "is" of length 2 and "cool" of length 4.
Output is ordered by length and the new first word starts with capital letter.

Example 2:

Input: text = "Keep calm and code on"
Output: "On and keep calm code"
Explanation: Output is ordered as follows:
"On" 2 letters.
"and" 3 letters.
"keep" 4 letters in case of tie order by position in original text.
"calm" 4 letters.
"code" 4 letters.

Example 3:

Input: text = "To be or not to be"
Output: "To be or to be not"

Constraints:

text begins with a capital letter and then contains lowercase letters and single space between words.
1 <= text.length <= 10^5

Step 01

Brute Force Baseline

Problem summary: Given a sentence text (A sentence is a string of space-separated words) in the following format: First letter is in upper case. Each word in text are separated by a single space. Your task is to rearrange the words in text such that all words are rearranged in an increasing order of their lengths. If two words have the same length, arrange them in their original order. Return the new text following the format shown above.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"Leetcode is cool"

Example 2

"Keep calm and code on"

Example 3

"To be or not to be"

Step 02

Core Insight

What unlocks the optimal approach

Store each word and their relative position. Then, sort them by length of words in case of tie by their original order.

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 #1451: Rearrange Words in a Sentence
class Solution {
    public String arrangeWords(String text) {
        String[] words = text.split(" ");
        words[0] = words[0].toLowerCase();
        Arrays.sort(words, Comparator.comparingInt(String::length));
        words[0] = words[0].substring(0, 1).toUpperCase() + words[0].substring(1);
        return String.join(" ", words);
    }
}

// Accepted solution for LeetCode #1451: Rearrange Words in a Sentence
func arrangeWords(text string) string {
	words := strings.Split(text, " ")
	words[0] = strings.ToLower(words[0])
	sort.SliceStable(words, func(i, j int) bool { return len(words[i]) < len(words[j]) })
	words[0] = strings.Title(words[0])
	return strings.Join(words, " ")
}

# Accepted solution for LeetCode #1451: Rearrange Words in a Sentence
class Solution:
    def arrangeWords(self, text: str) -> str:
        words = text.split()
        words[0] = words[0].lower()
        words.sort(key=len)
        words[0] = words[0].title()
        return " ".join(words)

// Accepted solution for LeetCode #1451: Rearrange Words in a Sentence
struct Solution;

impl Solution {
    fn arrange_words(text: String) -> String {
        let mut words: Vec<(usize, usize, String)> = text
            .split_whitespace()
            .enumerate()
            .map(|(i, s)| (s.len(), i, s.to_lowercase()))
            .collect();
        words.sort();
        let words: Vec<String> = words.into_iter().map(|(_, _, s)| s).collect();
        let mut res = words.join(" ");
        res[0..1].make_ascii_uppercase();
        res
    }
}

#[test]
fn test() {
    let text = "Leetcode is cool".to_string();
    let res = "Is cool leetcode".to_string();
    assert_eq!(Solution::arrange_words(text), res);
    let text = "Keep calm and code on".to_string();
    let res = "On and keep calm code".to_string();
    assert_eq!(Solution::arrange_words(text), res);
    let text = "To be or not to be".to_string();
    let res = "To be or to be not".to_string();
    assert_eq!(Solution::arrange_words(text), res);
}

// Accepted solution for LeetCode #1451: Rearrange Words in a Sentence
function arrangeWords(text: string): string {
    let words: string[] = text.split(' ');
    words[0] = words[0].toLowerCase();
    words.sort((a, b) => a.length - b.length);
    words[0] = words[0].charAt(0).toUpperCase() + words[0].slice(1);
    return words.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(1)

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.