1662. Check If Two String Arrays are Equivalent - Explanation

1. Concatenate Strings

class Solution:
    def arrayStringsAreEqual(self, word1: List[str], word2: List[str]) -> bool:
        return "".join(word1) == "".join(word2)

Time & Space Complexity

Time complexity: $O(n + m)$
Space complexity: $O(n + m)$

Where $n$ and $m$ are the total number of characters in both the arrays $word1$ and $word2$ , respectively.

2. Concatenate Strings Of One Array

class Solution:
    def arrayStringsAreEqual(self, word1: List[str], word2: List[str]) -> bool:
        s1 = "".join(word1)
        i = 0
        for w in word2:
            for c in w:
                if i == len(s1) or s1[i] != c:
                    return False
                i += 1
        return i == len(s1)

Time & Space Complexity

Time complexity: $O(n + m)$
Space complexity: $O(n)$

Where $n$ and $m$ are the total number of characters in both the arrays $word1$ and $word2$ , respectively.

3. Two Pointers

class Solution:
    def arrayStringsAreEqual(self, word1: List[str], word2: List[str]) -> bool:
        w1 = w2 = 0  # Index of word
        i = j = 0    # Index of character

        while w1 < len(word1) and w2 < len(word2):
            if word1[w1][i] != word2[w2][j]:
                return False

            i, j = i + 1, j + 1

            if i == len(word1[w1]):
                w1 += 1
                i = 0
            if j == len(word2[w2]):
                w2 += 1
                j = 0

        return w1 == len(word1) and w2 == len(word2)

Time & Space Complexity

Time complexity: $O(n + m)$
Space complexity: $O(1)$ extra space.

Where $n$ and $m$ are the total number of characters in both the arrays $word1$ and $word2$ , respectively.