Check If Two String Arrays are Equivalent

Problem Statement

The task involves comparing two string arrays, word1 and word2, and determining whether they both represent the same string when their individual elements are concatenated in their respective orders. A string is said to be represented by an array when the concatenation of its elements forms the string. The function should return true if the concatenated results of word1 and word2 are identical, and false otherwise.

Examples

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.

Approach and Intuition

To solve this problem, you can approach it with the following intuitive steps:

1. Concatenate all the strings in word1 to form a single string.

2. Concatenate all the strings in word2 to form another single string.

3. Compare the two resulting strings from word1 and word2.

   • If they are the same, return true.
   • Otherwise, return false.

Examples for better understanding:

• For arrays word1 = ["ab", "c"] and word2 = ["a", "bc"], the concatenated result for word1 is "abc", and for word2 is also "abc". Since both strings are identical, the result is true.
• For arrays word1 = ["a", "cb"] and word2 = ["ab", "c"], the concatenated result for word1 is "acb", whereas for word2 it is "abc". These strings are different, hence the result is false.

Considering the constraints where the total length of concatenated strings can go up to 103, this approach is efficient. The operations involved are straightforward string concatenations followed by a single equality check, ensuring that the solution is optimal within the provided limits.

Solutions

class Solution {
public:
    bool compareStringArrays(vector<string>& arr1, vector<string>& arr2) {
        string concatenatedSecond = accumulate(arr2.begin(), arr2.end(), string());
        int pos = 0;
        for (const auto& part : arr1) {
            for (char ch : part) {
                if (pos >= concatenatedSecond.length() || ch != concatenatedSecond[pos]) {
                    return false;
                }
                pos++;
            }
        }
        return pos == concatenatedSecond.length();
    }
};