Beginner friendly code and Easy to understand 😊😊
🧠 IntuitionWe want to:
Split the string s into exactly k parts.
For each part, calculate the minimum number of changes required to make that part a palindrome.
# 🧠 Intuition
We want to:

* Split the string s into exactly k parts.
* For each part, calculate the minimum number of changes required to make that part a palindrome.
* Return the minimum total across all partitions.

So, this becomes a Dynamic Programming + Recursion problem where:

* At each index, we try all possible partitions,
* Recursively compute the cost for the rest of the string,
* Memoize to avoid recomputation.

# 🛠️ Approach
1. Recursive Function: lessChangeReq(s, ind, k, dp)
* ind: Current index in the string.

* k: Number of partitions remaining.

* dp[ind][k]: Memoization table.

Base Cases:
* If k == 0 and ind == s.length(): Done partitioning and used all characters → return 0.

* If k <= 0: Invalid state → return large number (1e6 here as infinity).

* If already computed (dp[ind][k] != -1) → use memoized result.

Transition:
* Loop from current index ind to end of string.

* For each i, take the substring from ind to i.

* Count the minimum changes needed to make it a palindrome.

* Recurse on the rest of the string from i+1 with k-1 partitions.

* Take the minimum over all possible splits.

# 🔁 Helper Function: palindrome(st)
* Calculates minimum changes needed to make the string st a palindrome.

* Compares characters from both ends and increments a counter whenever there's a mismatch.

# 🧩 Why Substring Works?
Even though substring creates a new string every time, it's acceptable here since:

* Input size is limited (s.length <= 100).

* We're focusing on correctness and clarity.

# ✅ Time Complexity
* Roughly O(n^2 * k) where:
--> n is length of string.
--> Each state (i, k) may consider up to n splits.
--> palindrome() takes O(n) time per call.

# 📝 Final Notes
This is a classic top-down DP with memoization problem where:
* The state is defined by (index, partitions left).
* You try all valid splits and store the result for reuse.

# Code
```java []
class Solution {
    public int palindromePartition(String s, int k) {
        int[][] dp = new int[s.length()+1][k+1];
        for(int[] ele: dp){
            Arrays.fill(ele, -1);
        }
        return lessChangeReq(s, 0, k, dp);
    }
    public int lessChangeReq(String st, int ind, int k,int[][] dp){
        if(k == 0 && ind == st.length()){
            return 0;
        }
        if(k <= 0){
            return 1000000;
        }
        if(dp[ind][k] != -1){
            return dp[ind][k];
        }
        int ans = 1000000;
        for(int i = ind;i < st.length();i++){
            int reqChanges = palindrome(st.substring(ind, i+1));
            ans = Math.min(ans, lessChangeReq(st, i+1, k-1, dp)+reqChanges);
        }
        return dp[ind][k] = ans;
    }
    public int palindrome(String st){
        int i = 0;
        int j = st.length()-1;
        int cnt = 0;
        while(i < j){
            if(st.charAt(i) != st.charAt(j)){
                cnt++;
            }
            i++;
            j--;
        }
        return cnt;
    }
}
```
['String', 'Dynamic Programming', 'Memoization', 'Java']