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Check If Two String Arrays Are Equivalent  Leetcode Solution
LeetCode: Check If Two String Arrays Are Equivalent Leetcode Solution
Difficulty: Easy
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.
Solution:
To solve this problem, we can concatenate the strings in both arrays and compare them using the == operator. If they are equal, we can return true, else we can return false. Let's build the code step by step.

Initialize an empty string for both word1 and word2.

Traverse through the array word1 and concatenate the strings in it to form a single string.

Traverse through the array word2 and concatenate the strings in it to form a single string.

Compare the two strings using the == operator.

If they are equal, return true, else return false.
Here is the Python code for the solution:
class Solution: def arrayStringsAreEqual(self, word1: List[str], word2: List[str]) > bool:
# Step 1: Initialize an empty string for both word1 and word2.
w1 = ""
w2 = ""
# Step 2: Traverse through the array word1 and concatenate the strings it to form a single string.
for s in word1:
w1 += s
# Step 3: Traverse through the array word2 and concatenate the strings it to form a single string.
for s in word2:
w2 += s
# Step 4: Compare the two strings using the == operator.
# Step 5: If they are equal, return true, else return false.
return w1 == w2
Time Complexity:
Traversing through the arrays word1 and word2 takes O(n) time where n is the length of the array. Concatenating the strings takes O(m) time where m is the length of the string. Therefore, the time complexity of the solution is O(n*m).
Space Complexity:
The space required to store the two strings w1 and w2 is O(m) where m is the length of the string. Therefore, the space complexity of the solution is O(m).
Check If Two String Arrays Are Equivalent Solution Code
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