LeetCode #1813 — MEDIUM

Sentence Similarity III

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given two strings sentence1 and sentence2, each representing a sentence composed of words. A sentence is a list of words that are separated by a single space with no leading or trailing spaces. Each word consists of only uppercase and lowercase English characters.

Two sentences s1 and s2 are considered similar if it is possible to insert an arbitrary sentence (possibly empty) inside one of these sentences such that the two sentences become equal. Note that the inserted sentence must be separated from existing words by spaces.

For example,

s1 = "Hello Jane" and s2 = "Hello my name is Jane" can be made equal by inserting "my name is" between "Hello" and "Jane" in s1.
s1 = "Frog cool" and s2 = "Frogs are cool" are not similar, since although there is a sentence "s are" inserted into s1, it is not separated from "Frog" by a space.

Given two sentences sentence1 and sentence2, return true if sentence1 and sentence2 are similar. Otherwise, return false.

Example 1:

Input: sentence1 = "My name is Haley", sentence2 = "My Haley"

Output: true

Explanation:

sentence2 can be turned to sentence1 by inserting "name is" between "My" and "Haley".

Example 2:

Input: sentence1 = "of", sentence2 = "A lot of words"

Output: false

Explanation:

No single sentence can be inserted inside one of the sentences to make it equal to the other.

Example 3:

Input: sentence1 = "Eating right now", sentence2 = "Eating"

Output: true

Explanation:

sentence2 can be turned to sentence1 by inserting "right now" at the end of the sentence.

Constraints:

1 <= sentence1.length, sentence2.length <= 100
sentence1 and sentence2 consist of lowercase and uppercase English letters and spaces.
The words in sentence1 and sentence2 are separated by a single space.

Step 01

Brute Force Baseline

Problem summary: You are given two strings sentence1 and sentence2, each representing a sentence composed of words. A sentence is a list of words that are separated by a single space with no leading or trailing spaces. Each word consists of only uppercase and lowercase English characters. Two sentences s1 and s2 are considered similar if it is possible to insert an arbitrary sentence (possibly empty) inside one of these sentences such that the two sentences become equal. Note that the inserted sentence must be separated from existing words by spaces. For example, s1 = "Hello Jane" and s2 = "Hello my name is Jane" can be made equal by inserting "my name is" between "Hello" and "Jane" in s1. s1 = "Frog cool" and s2 = "Frogs are cool" are not similar, since although there is a sentence "s are" inserted into s1, it is not separated from "Frog" by a space. Given two sentences sentence1 and sentence2, return

Baseline thinking

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

Pattern signal: Array · Two Pointers

Example 1

"My name is Haley"
"My Haley"

Example 2

"of"
"A lot of words"

Example 3

"Eating right now"
"Eating"

Step 02

Core Insight

What unlocks the optimal approach

One way to look at it is to find one sentence as a concatenation of a prefix and suffix from the other sentence.
Get the longest common prefix between them and the longest common suffix.

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 #1813: Sentence Similarity III
class Solution {
    public boolean areSentencesSimilar(String sentence1, String sentence2) {
        var words1 = sentence1.split(" ");
        var words2 = sentence2.split(" ");
        if (words1.length < words2.length) {
            var t = words1;
            words1 = words2;
            words2 = t;
        }
        int m = words1.length, n = words2.length;
        int i = 0, j = 0;
        while (i < n && words1[i].equals(words2[i])) {
            ++i;
        }
        while (j < n && words1[m - 1 - j].equals(words2[n - 1 - j])) {
            ++j;
        }
        return i + j >= n;
    }
}

// Accepted solution for LeetCode #1813: Sentence Similarity III
func areSentencesSimilar(sentence1 string, sentence2 string) bool {
	words1, words2 := strings.Fields(sentence1), strings.Fields(sentence2)
	if len(words1) < len(words2) {
		words1, words2 = words2, words1
	}
	m, n := len(words1), len(words2)
	i, j := 0, 0
	for i < n && words1[i] == words2[i] {
		i++
	}
	for j < n && words1[m-1-j] == words2[n-1-j] {
		j++
	}
	return i+j >= n
}

# Accepted solution for LeetCode #1813: Sentence Similarity III
class Solution:
    def areSentencesSimilar(self, sentence1: str, sentence2: str) -> bool:
        words1, words2 = sentence1.split(), sentence2.split()
        m, n = len(words1), len(words2)
        if m < n:
            words1, words2 = words2, words1
            m, n = n, m
        i = j = 0
        while i < n and words1[i] == words2[i]:
            i += 1
        while j < n and words1[m - 1 - j] == words2[n - 1 - j]:
            j += 1
        return i + j >= n

// Accepted solution for LeetCode #1813: Sentence Similarity III
struct Solution;

use std::collections::VecDeque;

impl Solution {
    fn are_sentences_similar(sentence1: String, sentence2: String) -> bool {
        let words1: Vec<String> = sentence1
            .split_whitespace()
            .map(|s| s.to_string())
            .collect();
        let words2: Vec<String> = sentence2
            .split_whitespace()
            .map(|s| s.to_string())
            .collect();
        let mut q1 = VecDeque::from(words1);
        let mut q2 = VecDeque::from(words2);
        while let (Some(w1), Some(w2)) = (q1.front(), q2.front()) {
            if w1 == w2 {
                q1.pop_front();
                q2.pop_front();
            } else {
                break;
            }
        }
        while let (Some(w1), Some(w2)) = (q1.back(), q2.back()) {
            if w1 == w2 {
                q1.pop_back();
                q2.pop_back();
            } else {
                break;
            }
        }
        q1.is_empty() || q2.is_empty()
    }
}

#[test]
fn test() {
    let sentence1 = "My name is Haley".to_string();
    let sentence2 = "My Haley".to_string();
    let res = true;
    assert_eq!(Solution::are_sentences_similar(sentence1, sentence2), res);
    let sentence1 = "of".to_string();
    let sentence2 = "A lot of words".to_string();
    let res = false;
    assert_eq!(Solution::are_sentences_similar(sentence1, sentence2), res);
    let sentence1 = "Eating right now".to_string();
    let sentence2 = "Eating".to_string();
    let res = true;
    assert_eq!(Solution::are_sentences_similar(sentence1, sentence2), res);
    let sentence1 = "Luky".to_string();
    let sentence2 = "Lucccky".to_string();
    let res = false;
    assert_eq!(Solution::are_sentences_similar(sentence1, sentence2), res);
}

// Accepted solution for LeetCode #1813: Sentence Similarity III
function areSentencesSimilar(sentence1: string, sentence2: string): boolean {
    const [words1, words2] = [sentence1.split(' '), sentence2.split(' ')];
    const [m, n] = [words1.length, words2.length];

    if (m > n) return areSentencesSimilar(sentence2, sentence1);

    let [l, r] = [0, 0];
    for (let i = 0; i < n; i++) {
        if (l === i && words1[i] === words2[i]) l++;
        if (r === i && words2[n - i - 1] === words1[m - r - 1]) r++;
    }

    return l + r >= m;
}

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(L)

Space

O(L)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

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.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.