2024-05-22 Daily Challenge
Today I have done leetcode's May LeetCoding Challenge with cpp
.
May LeetCoding Challenge 22
Description
Palindrome Partitioning
Given a string s
, partition s
such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s
.
Example 1:
Input: s = "aab" Output: [["a","a","b"],["aa","b"]]
Example 2:
Input: s = "a" Output: [["a"]]
Constraints:
1 <= s.length <= 16
s
contains only lowercase English letters.
Solution
class Solution {
int len;
vector<vector<bool>> isPalindrome;
void init(string &s) {
len = s.length();
isPalindrome.resize(len+1, vector<bool>(len+1));
for(int i = 0; i < len; ++i) {
isPalindrome[i][i] = true;
isPalindrome[i][i+1] = true;
}
for(int i = 2; i <= len; ++i) {
for(int j = 0; j+i <= len; ++j) {
isPalindrome[j][j+i] = isPalindrome[j+1][j+i-1] && (s[j] == s[j+i-1]);
}
}
}
void dfs(vector<vector<string>> &answer, vector<string> &container, string &s, int index) {
if(index == len) {
answer.push_back(container);
}
for(int i = 1; index+i <= len; ++i) {
if(isPalindrome[index][index+i]) {
container.push_back(s.substr(index, i));
dfs(answer, container, s, index+i);
container.pop_back();
}
}
}
public:
vector<vector<string>> partition(string s) {
init(s);
vector<vector<string>> answer;
vector<string> container;
dfs(answer, container, s, 0);
return answer;
}
};
// Accepted
// 32/32 cases passed (125 ms)
// Your runtime beats 51.96 % of cpp submissions
// Your memory usage beats 89.38 % of cpp submissions (49.2 MB)