#### Palindromic Partitioning

S
A
```package com.thealgorithms.dynamicprogramming;

import java.util.Scanner;

/**
* @file @brief Implements [Palindrome
* Partitioning](https://leetcode.com/problems/palindrome-partitioning-ii/)
* algorithm, giving you the minimum number of partitions you need to make
*
* @details palindrome partitioning uses dynamic programming and goes to all the
* possible partitions to find the minimum you are given a string and you need
* to give minimum number of partitions needed to divide it into a number of
* palindromes [Palindrome Partitioning]
* (https://www.geeksforgeeks.org/palindrome-partitioning-dp-17/) overall time
* complexity O(n^2) For example: example 1:- String : "nitik" Output : 2 => "n
* | iti | k" For example: example 2:- String : "ababbbabbababa" Output : 3 =>
* "aba | b | bbabb | ababa"
* @author [Syed] (https://github.com/roeticvampire)
*/
public class PalindromicPartitioning {

public static int minimalpartitions(String word) {
int len = word.length();
/* We Make two arrays to create a bottom-up solution.
minCuts[i] = Minimum number of cuts needed for palindrome partitioning of substring word[0..i]
isPalindrome[i][j] = true if substring str[i..j] is palindrome
Base Condition: C[i] is 0 if P[i]= true
*/
int[] minCuts = new int[len];
boolean[][] isPalindrome = new boolean[len][len];

int i, j, L; // different looping variables

// Every substring of length 1 is a palindrome
for (i = 0; i < len; i++) {
isPalindrome[i][i] = true;
}

/* L is substring length. Build the solution in bottom up manner by considering all substrings of length starting from 2 to n. */
for (L = 2; L <= len; L++) {
// For substring of length L, set different possible starting indexes
for (i = 0; i < len - L + 1; i++) {
j = i + L - 1; // Ending index
// If L is 2, then we just need to
// compare two characters. Else need to
// check two corner characters and value
// of P[i+1][j-1]
if (L == 2) {
isPalindrome[i][j] = (word.charAt(i) == word.charAt(j));
} else {
if ((word.charAt(i) == word.charAt(j)) && isPalindrome[i + 1][j - 1]) {
isPalindrome[i][j] = true;
} else {
isPalindrome[i][j] = false;
}

}
}
}

//We find the minimum for each index
for (i = 0; i < len; i++) {
if (isPalindrome[i] == true) {
minCuts[i] = 0;
} else {
minCuts[i] = Integer.MAX_VALUE;
for (j = 0; j < i; j++) {
if (isPalindrome[j + 1][i] == true && 1 + minCuts[j] < minCuts[i]) {
minCuts[i] = 1 + minCuts[j];
}
}
}
}

// Return the min cut value for complete
// string. i.e., str[0..n-1]
return minCuts[len - 1];
}

public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String word;
System.out.println("Enter the First String");
word = input.nextLine();
// ans stores the final minimal cut count needed for partitioning
int ans = minimalpartitions(word);
System.out.println(
"The minimum cuts needed to partition \"" + word + "\" into palindromes is " + ans);
input.close();
}
}
```  