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This approach involves concatenating all the elements in each string array to form two complete strings, and then comparing these strings for equality. Since we need to ensure that the concatenated results of the two arrays are the same, this approach is both intuitive and straightforward.
The time complexity is O(n), where n is the sum of the lengths of both arrays' strings, as each character is traversed once during concatenation. The space complexity is O(n) due to the storage of the concatenated strings.
1#include <iostream>
2#include <vector>
3
4using namespace std;
5
6bool arrayStringsAreEqual(vector<string>& word1, vector<string>& word2) {
7 string s1 = "", s2 = "";
8 for (const string &w : word1) s1 += w;
9 for (const string &w : word2) s2 += w;
10 return s1 == s2;
11}
12
13int main() {
14 vector<string> word1 = {"abc", "d", "defg"};
15 vector<string> word2 = {"abcddefg"};
16 cout << (arrayStringsAreEqual(word1, word2) ? "true" : "false") << endl;
17 return 0;
18}This C++ solution utilizes the powerful string handling capabilities in C++. It concatenates each vector's contents into two strings and then returns a boolean indicating whether the resultant strings are equal.
Instead of explicitly concatenating the arrays into strings, this approach uses two pointers to iterate through the characters of the arrays simultaneously. It allows checking one character at a time across both arrays, ensuring they match without creating additional strings.
Time complexity is O(n), where n is the total number of characters in both arrays. Space complexity is minimized to O(1) as no additional storage is required beyond basic variables.
This C solution uses multiple iterators to navigate through the arrays without creating concatenated strings. Characters are compared directly during iteration until all characters in both arrays have been evaluated.