In this post, we are going to solve the Palindrome Partitioning II Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.
Problem
Given a string s, partition s such that every substring of the partition is a palindrome.
Return the minimum cuts needed for a palindrome partitioning of s.
Example 1:
Input: s = "aab" Output: 1 Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.
Example 2:
Input: s = "a" Output: 0
Example 3:
Input: s = "ab" Output: 1
Constraints:
1 <= s.length <= 2000sconsists of lowercase English letters only.
Now, let’s see the leetcode solution of Palindrome Partitioning II Leetcode Solution.
Palindrome Partitioning II Leetcode Solution in Python
class Solution:
def minCut(self, s: str) -> int:
n = len(s)
cut = [0] * n
dp = [[False] * n for _ in range(n)]
for i in range(n):
mini = i
for j in range(i + 1):
if s[j] == s[i] and (j + 1 > i - 1 or dp[j + 1][i - 1]):
dp[j][i] = True
mini = 0 if j == 0 else min(mini, cut[j - 1] + 1)
cut[i] = mini
return cut[n - 1]Palindrome Partitioning II Leetcode Solution in CPP
class Solution {
public:
int minCut(string s) {
const int n = s.length();
// isPalindrome[i][j] := true if s[i..j] is a palindrome
vector<vector<bool>> isPalindrome(n, vector<bool>(n, true));
// dp[i] := min cuts needed for a palindrome partitioning of s[0..i]
vector<int> dp(n, n);
for (int l = 2; l <= n; ++l)
for (int i = 0, j = l - 1; j < n; ++i, ++j)
isPalindrome[i][j] = s[i] == s[j] && isPalindrome[i + 1][j - 1];
for (int i = 0; i < n; ++i) {
if (isPalindrome[0][i]) {
dp[i] = 0;
continue;
}
// Try all possible partitions
for (int j = 0; j < i; ++j)
if (isPalindrome[j + 1][i])
dp[i] = min(dp[i], dp[j] + 1);
}
return dp.back();
}
};Palindrome Partitioning II Leetcode Solution in Java
class Solution {
public int minCut(String s) {
final int n = s.length();
// isPalindrome[i][j] := true if s[i..j] is a palindrome
boolean[][] isPalindrome = new boolean[n][n];
for (boolean[] row : isPalindrome)
Arrays.fill(row, true);
// dp[i] := min cuts needed for a palindrome partitioning of s[0..i]
int[] dp = new int[n];
Arrays.fill(dp, n);
for (int l = 2; l <= n; ++l)
for (int i = 0, j = l - 1; j < n; ++i, ++j)
isPalindrome[i][j] = s.charAt(i) == s.charAt(j) && isPalindrome[i + 1][j - 1];
for (int i = 0; i < n; ++i) {
if (isPalindrome[0][i]) {
dp[i] = 0;
continue;
}
// Try all possible partitions
for (int j = 0; j < i; ++j)
if (isPalindrome[j + 1][i])
dp[i] = Math.min(dp[i], dp[j] + 1);
}
return dp[n - 1];
}
}Note: This problem Palindrome Partitioning II is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purpose.
