132. Palindrome Partitioning II

Description

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.

Constraints

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public int minCut(String s) {
        if(s == null || s.length() <= 1) return 0;
        
        int n = s.length();
        boolean[][] isPalindrome = new boolean[n][n];
        int[] cut = new int[n];
        
        for(int i = 1; i < n; i++) {
            int min = i;
            for(int j = 0; j <= i; j++) {
                if(s.charAt(i) == s.charAt(j) && 
                   (i <= j+1 || isPalindrome[i-1][j+1])) {
                    isPalindrome[i][j] = true;
                    if(j == 0) {
                        min = Math.min(min, 0);
                    } else {
                        min = Math.min(min, 1+cut[j-1]);
                    }
                }
            }
            cut[i] = min;
        }
        
        return cut[n-1];
    }
}

Solution 2

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public int minCut(String s) {
        if(s == null || s.length() <= 1) return 0;
        
        int n = s.length();
        int[] dp = new int[n];
        char[] sChars = s.toCharArray();
        Arrays.fill(dp, n-1);
        
        for(int i = 0; i < n; i++) {
            isPalindrome(sChars, i, i, dp);
            isPalindrome(sChars, i, i+1, dp);
        }
        
        return dp[n-1];
    }
    
    private void isPalindrome(char[] sChars, int left, int right, int[] dp) {
        while(left >= 0 && 
              right < sChars.length && 
              sChars[left] == sChars[right]) {
            dp[right] = Math.min(dp[right], (left == 0)? 0: 1+dp[left-1]);
            left--;
            right++;
        }
    }
}

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