Add Palindromic Paritioning (#2386)

DynamicProgramming/PalindromicPartitioning.java

@@ -0,0 +1,94 @@
+package DynamicProgramming;
+/**
+ * @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)
+ */
+
+import java.util.Scanner;
+
+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[0][i]= true
+        */
+           int[] minCuts = new int[len];
+           boolean[][] isPalindrome = new boolean[len][len];
+    
+           int i, j, k, 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[0][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();
+  }
+}