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.
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
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()]; } } 
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