2022-10-24 Daily Challenge
Today I have done leetcode's October LeetCoding Challenge with cpp.
October LeetCoding Challenge 24
Description
Maximum Length of a Concatenated String with Unique Characters
You are given an array of strings arr. A string s is formed by the concatenation of a subsequence of arr that has unique characters.
Return the maximum possible length of s.
A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.
Example 1:
Input: arr = ["un","iq","ue"]
Output: 4
Explanation: All the valid concatenations are:
- ""
- "un"
- "iq"
- "ue"
- "uniq" ("un" + "iq")
- "ique" ("iq" + "ue")
Maximum length is 4.
Example 2:
Input: arr = ["cha","r","act","ers"]
Output: 6
Explanation: Possible longest valid concatenations are "chaers" ("cha" + "ers") and "acters" ("act" + "ers").
Example 3:
Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26 Explanation: The only string in arr has all 26 characters.
Constraints:
1 <= arr.length <= 161 <= arr[i].length <= 26arr[i]contains only lowercase English letters.
Solution
class Solution {
int answer = 0;
int len;
vector<int> masks;
vector<int> lens;
void init(vector<string>& arr) {
len = arr.size();
masks.resize(len);
lens.resize(len);
for(int i = 0; i < len; ++i) {
lens[i] = arr[i].length();
for(auto c : arr[i]) {
int m = (1 << (c - 'a'));
if(m & masks[i]) {
masks[i] = 0;
lens[i] = 0;
break;
}
masks[i] |= m;
}
}
}
void solve(int index = 0, int mask = 0, int length = 0) {
if(index == len) {
if(length > answer) answer = length;
return;
}
solve(index + 1, mask, length);
if(lens[index] && (masks[index] & mask) == 0) {
solve(index + 1, mask | masks[index], length + lens[index]);
}
}
public:
int maxLength(vector<string>& arr) {
init(arr);
solve();
return answer;
}
};
// Accepted
// 89/89 cases passed (4 ms)
// Your runtime beats 95.93 % of cpp submissions
// Your memory usage beats 92.84 % of cpp submissions (8.1 MB)