From 2438624f7358604ac0452ce19f92ea6948ef152d Mon Sep 17 00:00:00 2001
From: Jethendra-17 <146614641+Jethendra-17@users.noreply.github.com>
Date: Wed, 25 Oct 2023 22:32:06 +0530
Subject: [PATCH] Create Palindromic Partitioning.cpp
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Given a string str, a partitioning of the string is a palindrome partitioning if every sub-string of the partition is a palindrome. Determine the fewest cuts needed for palindrome partitioning of the given string.


Example 1:

Input: str = "ababbbabbababa"
Output: 3
Explaination: After 3 partitioning substrings
are "a", "babbbab", "b", "ababa".
Example 2:

Input: str = "aaabba"
Output: 1
Explaination: The substrings after 1
partitioning are "aa" and "abba".

Your Task:
You do not need to read input or print anything, Your task is to complete the function palindromicPartition() which takes the string str as the input parameter and returns the minimum number of partitions required.


Expected Time Complexity: O(n*n) [n is the length of the string str]
Expected Auxiliary Space: O(n*n)


Constraints:
1 ≤ length of str ≤ 500
---
 Palindromic Partitioning.cpp | 33 +++++++++++++++++++++++++++++++++
 1 file changed, 33 insertions(+)
 create mode 100644 Palindromic Partitioning.cpp

diff --git a/Palindromic Partitioning.cpp b/Palindromic Partitioning.cpp
new file mode 100644
index 0000000..de61f55
--- /dev/null
+++ b/Palindromic Partitioning.cpp	
@@ -0,0 +1,33 @@
+// User function Template for C++
+
+class Solution{
+public:
+  bool isPalindrome(int i, int j, string& str) {
+        while(i<j) {
+            if(str[i]!=str[j]) return false;
+            i++,j--;
+        }
+        return true;
+    }
+    int solve(int i,int n,string &str,vector<int>&dp) {
+        if(i==n) {
+            return 0;
+        }
+        int mini = INT_MAX;
+        if(dp[i]!=-1) return dp[i];
+        
+        for(int j = i; j<n; j++) {
+            if(isPalindrome(i,j,str)) {
+                int cost = 1 + solve(j+1, n, str,dp);
+                mini = min(mini,cost);
+            }
+               }
+        return dp[i]=mini;
+    }
+    int palindromicPartition(string str)
+    {
+        int n = str.size();
+        vector<int>dp(n,-1);
+        return solve(0,n,str,dp)-1;
+    }
+};