Palindrome Partitioning II LeetCode Solution

Problem – Palindrome Partitioning II LeetCode Solution

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 <= 2000
  • s consists of lowercase English letters only.

Palindrome Partitioning II LeetCode Solution in Java

public int minCut(String s) {
    char[] c = s.toCharArray();
    int n = c.length;
    int[] cut = new int[n];
    boolean[][] pal = new boolean[n][n];
    
    for(int i = 0; i < n; i++) {
        int min = i;
        for(int j = 0; j <= i; j++) {
            if(c[j] == c[i] && (j + 1 > i - 1 || pal[j + 1][i - 1])) {
                pal[j][i] = true;  
                min = j == 0 ? 0 : Math.min(min, cut[j - 1] + 1);
            }
        }
        cut[i] = min;
    }
    return cut[n - 1];
}

Palindrome Partitioning II LeetCode Solution in C++

    class Solution {
    public:
        int minCut(string s) {
            const int N = s.size();
            if(N<=1) return 0;
            int i,j;
            bool isPalin[N][N];
            fill_n(&isPalin[0][0], N*N, false);
            int minCuts[N+1];
            for(i=0; i<=N; ++i) minCuts[i] = i-1;
            
            for(j=1; j<N; ++j)
            {
                for(i=j; i>=0; --i)
                {
                    if( (s[i] == s[j]) && ( ( j-i < 2 ) || isPalin[i+1][j-1] ) )
                    {
                        isPalin[i][j] = true;
                        minCuts[j+1] = min(minCuts[j+1], 1 + minCuts[i]);    
                    }
                }
            }
            return minCuts[N];
            
        }
    };

Palindrome Partitioning II LeetCode Solution in Python

class Solution:
    def minCut(self, s: str) -> int:
        n = len(s)
        
        @lru_cache(None)
        def isPalindrome(l, r):  # l, r inclusive
            if l >= r: return True
            if s[l] != s[r]: return False
            return isPalindrome(l+1, r-1)
        
        @lru_cache(None)
        def dp(i):  # s[i..n-1]
            if i == n:
                return 0
            ans = math.inf
            for j in range(i, n):
                if (isPalindrome(i, j)):
                    ans = min(ans, dp(j+1) + 1)
            return ans
        
        return dp(0) - 1
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