1662

1662. Check If Two String Arrays are Equivalent (Easy)

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.

Companies: Apple, Amazon, Facebook

Related Topics:
Array, String

Similar Questions:

• Check if an Original String Exists Given Two Encoded Strings (Hard)

Hints:

• 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.

Solution 1. Brute Force

// OJ: https://leetcode.com/problems/check-if-two-string-arrays-are-equivalent/
// Author: github.com/lzl124631x
// Time: O(A + B)
// Space: O(A + B)
class Solution {
public:
    bool arrayStringsAreEqual(vector<string>& A, vector<string>& B) {
        string a, b;
        for (auto &x : A) a += x;
        for (auto &x : B) b += x;
        return a == b;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/check-if-two-string-arrays-are-equivalent/
// Author: github.com/lzl124631x
// Time: O(A + B)
// Space: O(1)
class Solution {
public:
    bool arrayStringsAreEqual(vector<string>& A, vector<string>& B) {
        int i = 0, a = 0, j = 0, b = 0;
        while (i < A.size() && j < B.size()) {
            if (A[i][a] != B[j][b]) return false;
            if (++a >= A[i].size()) ++i, a = 0;
            if (++b >= B[j].size()) ++j, b = 0;
        }
        return i == A.size() && j == B.size();
    }
};