LeetCode 1745: Palindrome Partitioning IV
LeetCode 1745 Solution Explanation
Explanation
To solve this problem, we can iterate through all possible partitions of the string s and check if each partition results in three palindromic substrings. We can use dynamic programming to check if a given substring is a palindrome quickly. By checking all possible partitions efficiently, we can determine if it is possible to split the string into three palindromic substrings.
Algorithm:
- Use dynamic programming to pre-calculate if each substring is a palindrome.
- Iterate through all possible partitions of the string
s. - Check if each partition results in three palindromic substrings.
- Return true if a valid partition is found, otherwise return false.
Time Complexity: O(n^2), where n is the length of the input string s.
Space Complexity: O(n^2), for storing the palindrome information.
LeetCode 1745 Solutions in Java, C++, Python
class Solution {
public boolean checkPartitioning(String s) {
int n = s.length();
boolean[][] isPalindrome = new boolean[n][n];
for (int i = n - 1; i >= 0; i--) {
for (int j = i; j < n; j++) {
if (s.charAt(i) == s.charAt(j) && (j - i <= 2 || isPalindrome[i + 1][j - 1])) {
isPalindrome[i][j] = true;
}
}
}
for (int i = 1; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
if (isPalindrome[0][i - 1] && isPalindrome[i][j - 1] && isPalindrome[j][n - 1]) {
return true;
}
}
}
return false;
}
}Interactive Code Editor for LeetCode 1745
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