LeetCode #1662 — EASY

Check If Two String Arrays are Equivalent

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

Given two string arrays word1 and word2, return true if the two arrays represent the same string, and false otherwise.

A string is represented by an array if the array elements concatenated in order forms the string.

Example 1:

Input: word1 = ["ab", "c"], word2 = ["a", "bc"]
Output: true
Explanation:
word1 represents string "ab" + "c" -> "abc"
word2 represents string "a" + "bc" -> "abc"
The strings are the same, so return true.

Example 2:

Input: word1 = ["a", "cb"], word2 = ["ab", "c"]
Output: false

Example 3:

Input: word1  = ["abc", "d", "defg"], word2 = ["abcddefg"]
Output: true

Constraints:

1 <= word1.length, word2.length <= 10³
1 <= word1[i].length, word2[i].length <= 10³
1 <= sum(word1[i].length), sum(word2[i].length) <= 10³
word1[i] and word2[i] consist of lowercase letters.

Step 01

Brute Force Baseline

Problem summary: Given two string arrays word1 and word2, return true if the two arrays represent the same string, and false otherwise. A string is represented by an array if the array elements concatenated in order forms the string.

Baseline thinking

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

Pattern signal: Array

Example 1

["ab", "c"]
["a", "bc"]

Example 2

["a", "cb"]
["ab", "c"]

Example 3

["abc", "d", "defg"]
["abcddefg"]

Core Insight

What unlocks the optimal approach

Concatenate all strings in the first array into a single string in the given order, the same for the second array.
Both arrays represent the same string if and only if the generated strings are the same.

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 #1662: Check If Two String Arrays are Equivalent
class Solution {
    public boolean arrayStringsAreEqual(String[] word1, String[] word2) {
        return String.join("", word1).equals(String.join("", word2));
    }
}

// Accepted solution for LeetCode #1662: Check If Two String Arrays are Equivalent
func arrayStringsAreEqual(word1 []string, word2 []string) bool {
	return strings.Join(word1, "") == strings.Join(word2, "")
}

# Accepted solution for LeetCode #1662: Check If Two String Arrays are Equivalent
class Solution:
    def arrayStringsAreEqual(self, word1: List[str], word2: List[str]) -> bool:
        return ''.join(word1) == ''.join(word2)

// Accepted solution for LeetCode #1662: Check If Two String Arrays are Equivalent
impl Solution {
    pub fn array_strings_are_equal(word1: Vec<String>, word2: Vec<String>) -> bool {
        word1.join("") == word2.join("")
    }
}

// Accepted solution for LeetCode #1662: Check If Two String Arrays are Equivalent
function arrayStringsAreEqual(word1: string[], word2: string[]): boolean {
    return word1.join('') === word2.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(m)

Space

O(m)

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.