Minimum number of Insertion and deletion needed to convert String A to String B or make both the strings equal (same as lcs)

Problem:
Given two strings str1 and str2. The task is to remove or insert the minimum number of characters from/in str1 so as to transform it into str2. It could be possible that the same character needs to be removed/deleted from one point of str1 and inserted to some another point.

Example 1:

Input: str1 = "heap", str2 = "pea"
Output: 3

Solution:

Intuition : Find the largest common subsequence of the two strings and substract it from 
the length of the two strings and finally add the subtractions. i.e. m+n - 2 *lcs(StringA, StringB)

Bottom up Approach(Memoization) :
Time complexity : O(m*n) where is m and n is the length of two strings a and b
space complexity : o(m*n) for using 2d dp and O(m+n) for auxiliary stack space because in worst case we will make m+n recursive calls.

Coding Ninjas Problem

import java.util.*;
public class Solution {
    public static int canYouMake(String str, String ptr) {
        // Write your code here.
        // we will use bottom up approach for solving this problem.
        int dp[][] =new int[str.length()][ptr.length()];
        for(int row[] : dp){
            Arrays.fill(row,-1);
        }
        return str.length()+ptr.length() - 2 *lcs(str,ptr, str.length()-1, ptr.length()-1,dp);
    }
    public static int lcs(String str, String ptr, int m,int n,int dp[][]){
        if(m<0 || n<0){
            return 0;
        }
        if(dp[m][n]!=-1) return dp[m][n];
        if(str.charAt(m)==ptr.charAt(n)){
            dp[m][n] = 1 + lcs(str,ptr,m-1,n-1,dp);
        }
        else{
            dp[m][n] = 0 +  Integer.max(lcs(str,ptr,m-1,n,dp), lcs(str,ptr,m,n-1,dp));
        }
        return dp[m][n];
    }

}

Top Down Approach(Tabulation) :
Time complexity : O(m*n) where is m and n is the length of two strings a and b
space complexity : o(m*n) for using 2d dp

GeeksForGeeksProblem

class Solution
{
    public int minOperations(String str1, String str2) 
    { 
       // we will use top down approach to solve this 
       int dp[][] =new int[str1.length()+1][str2.length()+1];
       for(int i =0;i<=str1.length();i++){
           dp[i][0] = 0;
       }
       for(int i =0;i<=str2.length();i++){
           dp[0][i] = 0;
       }
       for(int i =1;i<=str1.length();i++){
           for(int j =1;j<=str2.length();j++){
               if(str1.charAt(i-1)==str2.charAt(j-1)){
                   dp[i][j] = 1 + dp[i-1][j-1];
               }
               else dp[i][j] = Integer.max(dp[i-1][j],dp[i][j-1]);
           }
       }
       return str1.length() + str2.length() - 2 * dp[str1.length()][str2.length()];
    }
}