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

Palindrome Partitioning II LeetCode Solution Review:

In our experience, we suggest you solve this Palindrome Partitioning II LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the Palindrome Partitioning II LeetCode Solution

Find on Leetcode

Conclusion:

I hope this Palindrome Partitioning II LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

Keep Learning!

More Coding Solutions >>

LeetCode Solutions

Hacker Rank Solutions

CodeChef Solutions