EXPLAINED|| BEGINNER FRIENDLY || DP.
Understanding the Code: Minimum Insertions to Make a Palindrome ,But before that ! what is your problem ppl? Why dont you shameless guys upvote my posts , I wri
## Understanding the Code: Minimum Insertions to Make a Palindrome ,But before that ! what is your problem ppl? Why dont you shameless guys upvote my posts , I write such long post puttings so much time and you wont wanna give an upvote ;(\n\n### Problem:\nGiven a string, find the minimum number of insertions required to make it a palindrome.\n\n### Solution Approach:\nThe code uses a dynamic programming approach to solve this problem. It calculates the minimum number of insertions needed for every substring of the given string and stores the results in a 2D array `dp`.\n\n### Code Breakdown:\n\n1. **Initialization:**\n   - A 2D array `dp` is created to store the minimum number of insertions for each substring. Initially, all values are set to `Integer.MAX_VALUE`.\n\n2. **Recursive Function `solve`:**\n   - This function calculates the minimum insertions for a substring defined by indices `i` and `j`.\n   - **Base case:** If `i` is greater than or equal to `j`, it means the substring is empty or has only one character, so no insertions are needed, and 0 is returned.\n   - **Memoization:** If the result for the current substring is already calculated (stored in `dp[i][j]`), it is returned directly.\n   - **Matching Characters:** If the characters at indices `i` and `j` match, the problem reduces to finding the minimum insertions for the substring `i+1` to `j-1`.\n   - **Non-matching Characters:** If the characters don\'t match, the minimum insertions are the minimum of two possibilities:\n     - Inserting a character at index `i` and solving for the substring `i+1` to `j`.\n     - Inserting a character at index `j` and solving for the substring `i` to `j-1`.\n\n3. **Main Function `minInsertions`:**\n   - Creates the `dp` array and initializes it with `Integer.MAX_VALUE`.\n   - Calls the `solve` function with the entire string and stores the result in `dp[0][str.length()-1]`.\n   - Returns the calculated minimum number of insertions.\n\n### Key Points:\n- The code uses a bottom-up approach to populate the `dp` array efficiently.\n- Memoization is used to avoid redundant calculations.\n- The recursive nature of the problem is handled effectively using the `solve` function.\n\n### Time and Space Complexity:\n- Time complexity: O(n^2) due to the nested loops in the `solve` function.\n- Space complexity: O(n^2) for the `dp` array.\n\nThe memoization part is easy but you can make a dp array of half the size , It will work. \n# NOW GO ND UPVOTE.\n\n\n# Code\n```\nclass Solution{\n    public int minInsertions(String str){\n        int [][] dp=new int [str.length()][str.length()];\n        for(int i=0;i<str.length();i++){\n            for(int j=0;j<str.length();j++){\n                dp[i][j]=Integer.MAX_VALUE;\n            }\n        }\n        return solve(str,0,str.length()-1,dp);\n        //return dp[0][str.length()-1];\n    }\n    public int solve(String str, int i, int j,int [][] dp){\n        if(i>=j){\n            return 0;\n        }\n        if(dp[i][j]!=Integer.MAX_VALUE){\n            return dp[i][j];\n        }\n        if(str.charAt(i)==str.charAt(j)){\n            return solve(str,i+1,j-1,dp);\n        }\n        \n        return dp[i][j]=1+Math.min(solve(str,i+1,j,dp),solve(str,i,j-1,dp));\n    }\n}\n\n\n```
['Dynamic Programming', 'Java']