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

Dp Learnings (27 Part Series)

1 House Robber leetcode problem
2 Jump game leetcode problem
… 23 more parts…
3 Jump game II leetcode problem
4 Cherry Pickup II Leetcode Problem
5 Partition Equals Subset Sum Leetcode problem or Subset sum equals to target
6 Maximum Sum Problem GeeksForGeeks
7 Count Palindrome Sub-Strings of a String GeeksForGeeks
8 Length of Longest Common Subsequence (LCS) and Longest common subsequence string leetcode
9 0/1 Knapsack Problem GeeksForGeeks both bounded and Unbounded
10 Shortest distance from source to all other nodes in a Directed Acyclic Graph
11 Coin change Leetcode
12 Coin Change 2 Leetcode
13 Target Sum or Partition with given difference Leetcode
14 Rod cutting Problem (similar to unbounded knapsack)
15 Length of longest common substring
16 Longest Palindromic subsequence (Same as Lcs)
17 Minimum Insertion steps needed to make a string palindrome (Same as LCS)
18 Minimum number of Insertion and deletion needed to convert String A to String B or make both the strings equal (same as lcs)
19 Shortest Common Super-sequence Leetcode (Same as Lcs string)
20 Distinct Subsequence Leetcode
21 Partition with given difference
22 Two best non overlapping events
23 Longest increasing subsequence
24 Edit distance
25 Maximum product subarray
26 Target sum
27 Maximum Multiplication Score

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

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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)

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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];
    }

}

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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()];
    }
}

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Dp Learnings (27 Part Series)

1 House Robber leetcode problem
2 Jump game leetcode problem
… 23 more parts…
3 Jump game II leetcode problem
4 Cherry Pickup II Leetcode Problem
5 Partition Equals Subset Sum Leetcode problem or Subset sum equals to target
6 Maximum Sum Problem GeeksForGeeks
7 Count Palindrome Sub-Strings of a String GeeksForGeeks
8 Length of Longest Common Subsequence (LCS) and Longest common subsequence string leetcode
9 0/1 Knapsack Problem GeeksForGeeks both bounded and Unbounded
10 Shortest distance from source to all other nodes in a Directed Acyclic Graph
11 Coin change Leetcode
12 Coin Change 2 Leetcode
13 Target Sum or Partition with given difference Leetcode
14 Rod cutting Problem (similar to unbounded knapsack)
15 Length of longest common substring
16 Longest Palindromic subsequence (Same as Lcs)
17 Minimum Insertion steps needed to make a string palindrome (Same as LCS)
18 Minimum number of Insertion and deletion needed to convert String A to String B or make both the strings equal (same as lcs)
19 Shortest Common Super-sequence Leetcode (Same as Lcs string)
20 Distinct Subsequence Leetcode
21 Partition with given difference
22 Two best non overlapping events
23 Longest increasing subsequence
24 Edit distance
25 Maximum product subarray
26 Target sum
27 Maximum Multiplication Score

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