Q23 Palindrome Partitioning II - LeetCode 132

PHOTO

Fri Mar 24 2023 14:29:54 GMT+0000 (Coordinated Universal Time)

Saved by @Ayush_dabas07

class Solution {
    public int minCut(String s) {
        /* dp playlist video 28 , 2nd approach O(n^2)
        */

        //make boolean 2d array
        int n = s.length();
        boolean[][] arr = new boolean[n][n];

        
        //traversing diagonally up
        for(int gap = 0 ; gap < n ; gap++){

            for(int i = 0 , j = gap ; j < n ; i++ , j++){

                if(gap == 0)
                arr[i][j] = true;

                else if(gap == 1){
                    if(s.charAt(i) == s.charAt(j)) 
                    arr[i][j] = true;
                }
                

                else{
                    if(s.charAt(i) == s.charAt(j) && arr[i+1][j-1]==true)
                    arr[i][j] = true;
                }
                
            }
        }

        //making 1d dp array 
        int dp[] = new int [n];
        dp[0] = 0;  //0 letters need no cuts
        
        //now checking every suffix from 
        for(int i = 1 ; i < n ; i++){
            
            int min = Integer.MAX_VALUE;
            //making suffix and checking from boolean table
            for(int j = i ; j >= 0 ;j-- ){
                //if suffix is already a pallindrome
                if(j==0 && arr[j][i])
                min = 0;

                else if(arr[j][i])
                min = Math.min(min , dp[j-1] + 1);
            
            }
            dp[i] = min;

        }

        return dp[n-1];
    }
}

https://leetcode.com/problems/palindrome-partitioning-ii/

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DP level 2

@Ayush_dabas07

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