How do you solve Check If Two String Arrays are Equivalent on LeetCode?

Check If Two String Arrays are Equivalent (LeetCode #1662) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Concatenating strings can be inefficient; direct comparison is often better.

What is the brute force solution for Check If Two String Arrays are Equivalent?

The brute force approach for Check If Two String Arrays are Equivalent is Brute Force, with O(n²) time complexity and O(1) space complexity. We can concatenate all strings in both arrays and then compare the two resulting strings. This is straightforward but not the most efficient solution. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check If Two String Arrays are Equivalent?

Check If Two String Arrays are Equivalent uses the following concepts: Array, String. The recommended patterns to study are: Two Pointers, String Manipulation.

What are the interview tips for Check If Two String Arrays are Equivalent?

Always consider edge cases, such as empty strings or arrays. Explain your thought process clearly; this shows your understanding. Practice writing clean and efficient code; readability matters.

What are common mistakes when solving Check If Two String Arrays are Equivalent?

Forgetting to check if both arrays have been fully traversed at the end. Not handling the index correctly when moving to the next string in the array.

#1662

Check If Two String Arrays are Equivalent

Easy

ArrayStringTwo PointersString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of creating two large strings, we can compare the elements of both arrays directly. This avoids unnecessary memory usage and speeds up the process.

⚙️

Algorithm

5 steps

1Step 1: Initialize two pointers, i and j, to 0.
2Step 2: Initialize two variables, index1 and index2, to 0 to track the current character position in each string.
3Step 3: While both pointers are within their respective array bounds, compare characters at index1 of word1 and index2 of word2.
4Step 4: If characters match, move to the next character in both strings. If they don't match, return false.
5Step 5: If we finish comparing all characters and both pointers have reached the end of their respective arrays, return true.

solution.py14 lines

1def arrayStringsAreEqual(word1, word2):
2    i, j, index1, index2 = 0, 0, 0, 0
3    while i < len(word1) and j < len(word2):
4        if word1[i][index1] != word2[j][index2]:
5            return False
6        index1 += 1
7        if index1 == len(word1[i]):
8            i += 1
9            index1 = 0
10        index2 += 1
11        if index2 == len(word2[j]):
12            j += 1
13            index2 = 0
14    return i == len(word1) and j == len(word2)

ℹ

Complexity note: The time complexity is O(n) because we traverse through each character of both arrays only once. The space complexity is O(1) since we are using a constant amount of extra space.

1Concatenating strings can be inefficient; direct comparison is often better.
2Understanding how to traverse and compare elements in arrays is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.