Minimize removals to remove another string as a subsequence of a given string

Given two strings str and X of length N and M respectively, the task is to find the minimum characters required to be removed from the string str such that string str doesn’t contain the string X as a subsequence.

Examples:

Input: str = “btagd”, X = “bad”
Output: 1
Explanation:
String “btag” has does not contain “bad” as a subsequence. Therefore, only one removal is required.

Input: str = “bbaaddd”, X = “bad”
Output: 2

Approach: This problem can be solved by Dynamic Programming. Follow the steps below to solve the problem:

• Traverse the string.
• Initialize a 2D array dp[N][M], where N is the length of string str and M is the length of string X.
• dp[i][j] represents the minimum number of character removal required in the substring str[0, i] such that there is no occurrence of substring X[0, j] as a subsequence.
• The two transition states are:
  • If str[i] is equal to X[j],
    • Case 1: Remove the character str[i]
    • Case 2: Maintain the character str[i], by ensuring that there is no occurrence of X[0, j-1] as a subsequence in str[0, i]
    • Update dp[i][j] = min(dp[i − 1][j − 1], dp[i − 1][j] + 1)
  • If str[i] is not equal to X[j], str[i] can be kept as it is.
    • Update dp[i][j] = dp[i – 1][j]
• Print the minimum removals i.e, dp[N – 1][M – 1]

Below is the implementation of the above approach:

C++

// C++ implementation of
// the above approach
#include <bits/stdc++.h>

using namespace std;

// Function to print the minimum number of
// character removals required to remove X
// as a subsequence from the string str
void printMinimumRemovals(string str, string X)
{

    // Length of the string str
    int N = str.size();

    // Length of the string X
    int M = X.size();

    // Stores the dp states
    int dp[N][M] = {};

    // Fill first row of dp[][]
    for (int j = 0; j < M; j++) {

        // If X[j] matches with str[0]
        if (str[0] == X[j]) {

            dp[0][j] = 1;
        }
    }

    for (int i = 1; i < N; i++) {

        for (int j = 0; j < M; j++) {

            // If str[i] is equal to X[j]
            if (str[i] == X[j]) {

                // Update state after removing str[i[
                dp[i][j] = dp[i - 1][j] + 1;

                // Update state after keeping str[i]
                if (j != 0)
                    dp[i][j]
                        = min(dp[i][j], dp[i - 1][j - 1]);
            }

            // If str[i] is not equal to X[j]
            else {

                dp[i][j] = dp[i - 1][j];
            }
        }
    }

    // Print the minimum number
    // of characters removals
    cout << dp[N - 1][M - 1];
}

// Driver Code
int main()
{
    // Input
    string str = "btagd";
    string X = "bad";

    // Function call to get minimum
    // number of character removals
    printMinimumRemovals(str, X);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {

  // Function to print the minimum number of
  // character removals required to remove X
  // as a subsequence from the string str
  static void printMinimumRemovals(String str, String X)
  {

    // Length of the string str
    int N = str.length();

    // Length of the string X
    int M = X.length();

    // Stores the dp states
    int dp[][] = new int[N][M];

    // Fill first row of dp[][]
    for (int j = 0; j < M; j++) {

      // If X[j] matches with str[0]
      if (str.charAt(0) == X.charAt(j)) {

        dp[0][j] = 1;
      }
    }

    for (int i = 1; i < N; i++) {

      for (int j = 0; j < M; j++) {

        // If str[i] is equal to X[j]
        if (str.charAt(i) == X.charAt(j)) {

          // Update state after removing str[i[
          dp[i][j] = dp[i - 1][j] + 1;

          // Update state after keeping str[i]
          if (j != 0)
            dp[i][j] = Math.min(
            dp[i][j], dp[i - 1][j - 1]);
        }

        // If str[i] is not equal to X[j]
        else {

          dp[i][j] = dp[i - 1][j];
        }
      }
    }

    // Print the minimum number
    // of characters removals
    System.out.println(dp[N - 1][M - 1]);
  }

  // Driver code
  public static void main(String[] args)
  {
    // Input
    String str = "btagd";
    String X = "bad";

    // Function call to get minimum
    // number of character removals
    printMinimumRemovals(str, X);
  }
}

// This code is contributed by Kingash.

Python3

# Python 3 implementation of
# the above approach

# Function to print the minimum number of
# character removals required to remove X
# as a subsequence from the string str
def printMinimumRemovals(s, X):

    # Length of the string str
    N = len(s)

    # Length of the string X
    M = len(X)

    # Stores the dp states
    dp = [[0 for x in range(M)] for y in range(N)]

    # Fill first row of dp[][]
    for j in range(M):

        # If X[j] matches with str[0]
        if (s[0] == X[j]):
            dp[0][j] = 1

    for i in range(1, N):
        for j in range(M):

            # If s[i] is equal to X[j]
            if (s[i] == X[j]):

                # Update state after removing str[i[
                dp[i][j] = dp[i - 1][j] + 1

                # Update state after keeping str[i]
                if (j != 0):
                    dp[i][j] = min(dp[i][j], dp[i - 1][j - 1])

            # If str[i] is not equal to X[j]
            else:
                dp[i][j] = dp[i - 1][j]

    # Print the minimum number
    # of characters removals
    print(dp[N - 1][M - 1])

# Driver Code
if __name__ == "__main__":

    # Input
    s = "btagd"
    X = "bad"

    # Function call to get minimum
    # number of character removals
    printMinimumRemovals(s, X)

    # This code is contributed by ukasp.

C#

// C# program for above approach
using System;
public class GFG
{

  // Function to print the minimum number of
  // character removals required to remove X
  // as a subsequence from the string str
  static void printMinimumRemovals(string str, string X)
  {

    // Length of the string str
    int N = str.Length;

    // Length of the string X
    int M = X.Length;

    // Stores the dp states
    int[,] dp = new int[N, M];

    // Fill first row of dp[][]
    for (int j = 0; j < M; j++) {

      // If X[j] matches with str[0]
      if (str[0] == X[j]) {

        dp[0, j] = 1;
      }
    }

    for (int i = 1; i < N; i++) {

      for (int j = 0; j < M; j++) {

        // If str[i] is equal to X[j]
        if (str[i] == X[j]) {

          // Update state after removing str[i[
          dp[i, j] = dp[i - 1, j] + 1;

          // Update state after keeping str[i]
          if (j != 0)
            dp[i, j] = Math.Min(
            dp[i, j], dp[i - 1, j - 1]);
        }

        // If str[i] is not equal to X[j]
        else {

          dp[i, j] = dp[i - 1, j];
        }
      }
    }

    // Print the minimum number
    // of characters removals
    Console.WriteLine(dp[N - 1, M - 1]);
  }

// Driver code
public static void Main(String[] args)
{

    // Input
    string str = "btagd";
    string X = "bad";

    // Function call to get minimum
    // number of character removals
    printMinimumRemovals(str, X);
}
}

// This code is contributed by sanjoy_62.

Javascript

<script>

// Javascript program for the above approach

// Function to print the minimum number of
// character removals required to remove X
// as a subsequence from the string str
function printMinimumRemovals(str,X)
{
    // Length of the string str
    let N = str.length;

    // Length of the string X
    let M = X.length;

    // Stores the dp states
    let dp = new Array(N);
    for(let i=0;i<N;i++)
    {
        dp[i]=new Array(M);
        for(let j=0;j<M;j++)
        {
            dp[i][j]=0;
        }

    }

    // Fill first row of dp[][]
    for (let j = 0; j < M; j++) {

      // If X[j] matches with str[0]
      if (str[0] == X[j]) {

        dp[0][j] = 1;
      }
    }

    for (let i = 1; i < N; i++) {

      for (let j = 0; j < M; j++) {

        // If str[i] is equal to X[j]
        if (str[i] == X[j]) {

          // Update state after removing str[i[
          dp[i][j] = dp[i - 1][j] + 1;

          // Update state after keeping str[i]
          if (j != 0)
            dp[i][j] = Math.min(
            dp[i][j], dp[i - 1][j - 1]);
        }

        // If str[i] is not equal to X[j]
        else {

          dp[i][j] = dp[i - 1][j];
        }
      }
    }

    // Print the minimum number
    // of characters removals
    document.write(dp[N - 1][M - 1]);
}

 // Driver code
let str = "btagd";
let X = "bad";
// Function call to get minimum
// number of character removals
printMinimumRemovals(str, X);


// This code is contributed by avanitrachhadiya2155
</script>

Output:

1

Time Complexity: O(N * M)
Auxiliary Space: O(N * M)