Create MinimumCutsForPalindromicPartition.java

Author: 17tanya

DynamicProgramming/MinimumCutsForPalindromicPartition.java

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+/*
+https://www.techiedelight.com/find-minimum-cuts-needed-palindromic-partition-string/
+https://youtu.be/WPr1jDh3bUQ
+*/
+
+//Bottom Up Dynamic Programming
+
+//Pre Computation - Bottom Up DP to find if substring is a palindrome or not
+public void findAllPlindromes(String s, boolean isPalin[][]){
+    //We only use the values above the principal diagonal
+    //isPalin[i][j] stores if s[i..j] is a palindrome or not
+    
+    //we start from the end because we require the value of (i+1).  
+    for(int i = s.length() - 1 ; i >= 0 ; i--){
+        for(int j = i ; j < s.length() ; j++){
+            
+            //single character is always a palindrome
+            if(i == j) isPalin[i][j] = true;
+            
+            //if the border characters match --> check the inner string
+            else if(s.charAt(i) == s.charAt(j)){
+                //if the length of the substring being considered is 2 --> directly put true
+                //else check if the inner string [i+1 ... j-1] is a palindrome or not
+                isPalin[i][j] = (j-i == 1) ? true : isPalin[i+1][j-1];
+            }
+            
+            //when the border characters don't match --> not possible to be a palindrome
+            else isPlain[i][j] = false;
+        }
+    }
+    
+}
+
+public int minCutsPlainPartition(String s, boolean isPlain[][]){
+    int n = s.length();
+    
+    //cuts[i] indicates the minimum cuts required in s[0...i] to form palindrome partition
+    int cuts[] = new int[n];
+    
+    for(int i=0;i<n;i++){
+        //check if the current substring is partition --> if it is a palindrome we need 0 cuts
+        if(isPlain[0][i]) cuts[i] = 0;
+        else{
+            int minCuts = Integer.MAX_VALUE;
+            
+            //we partition the substring s[0...i] at j into two substrings --> s[0...j] and s[j+1...i]
+            //since we need to make palindromic partitions, we check if the 2nd substring s[j+1...i] is a palindrome or not
+            //we are basically CHOOSING A PARTITION(j) WHICH KEEPS THE 2nd SUBSTRING AS IT IS(bacuse it is a palindrome) AND 
+            //DECOMPOSES THE 1ST SUBSTRING INTO SUBPROBLEMS TO FIND MINIMUM CUTS
+            for(int j=0;j<i;j++){ //j creates the partition
+            
+                //1. Check if 2nd substring is a palindrome
+                //2. solve the subproblem --> we are using the bottom up appraoch, subproblem is already solved
+                //   subproblem is s[0...j] --> we solve it by considering cuts[j]
+                //3. Consider all valid partitions and chose the cheapest
+                if(isPalin[j+1][i] && minCuts > cuts[j]+1){
+                    minCuts = cuts[j] + 1;
+                }
+            }
+            cuts[i] = minCuts;
+        }
+    }
+    
+    //result for minimum cuts of s[0...n-1]
+    return cuts[n-1];
+}
+
+/*
+Time Complexity - O(n^2)
+Space Complexity - O(n^2) [n^2 to store isPlain and n to store cuts --> isPlain is more]
+*/
+