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This approach uses a recursive backtracking strategy, where we build each possible concatenated string by considering elements one by one. We use a set to track characters for uniqueness and maximize the length only if all characters are unique.
Time Complexity: O(2^n), where n is the number of strings.
Space Complexity: O(n), for the recursive stack and current string.
1#include <iostream>
2#include <vector>
3#include <string>
4#include <unordered_set>
5
6using namespace std;
7
8bool isUnique(string &s) {
9 vector<int> chars(26, 0);
10 for (char c : s) {
11 if (chars[c - 'a']++ > 0) return false;
12 }
13 return true;
14}
15
16void backtrack(vector<string> &arr, string current, int index, int &max_length) {
17 if (!isUnique(current)) return;
18 if (current.size() > max_length) {
19 max_length = current.size();
20 }
21 for (int i = index; i < arr.size(); i++) {
22 backtrack(arr, current + arr[i], i + 1, max_length);
23 }
24}
25
26int maxLength(vector<string> &arr) {
27 int max_length = 0;
28 backtrack(arr, "", 0, max_length);
29 return max_length;
30}
31
32int main() {
33 vector<string> arr = {"un", "iq", "ue"};
34 cout << maxLength(arr) << endl;
35 return 0;
36}This C++ solution leverages backtracking to explore combinations of strings while checking for uniqueness using a vector to track characters.
This approach utilizes bitmasking to efficiently determine if characters are unique when combining strings. Each character is represented by a distinct position in a 32-bit integer, allowing for quick checks and updates.
Time Complexity: O(2^n), similar to previous approaches for evaluating combinations.
Space Complexity: O(n), due to the recursive stack with depth dependent on input size.
Java code utilizes bitmasking to handle character collisions efficiently. It performs bit manipulations to identify and ignore recurring characters across concats.