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]
- If str[i] is equal to X[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 strvoid 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 Codeint 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 approachimport 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 strdef 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 Codeif __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 approachusing 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 codepublic 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 strfunction 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 codelet str = "btagd";let X = "bad";// Function call to get minimum// number of character removalsprintMinimumRemovals(str, X);// This code is contributed by avanitrachhadiya2155</script> |
Output:
1
Time Complexity: O(N * M)
Auxiliary Space: O(N * M)
Efficient Approach : using array instead of 2d matrix to optimize space complexity
In previous code we can se that dp[i][j] is dependent upon dp[i+1][j-1] or dp[i][j-1] so we can assume that dp[i+1] is current row and dp[i] is previous row.
Implementations Steps :
- It initializes two vectors curr and prev of length N+1 with all elements as 0.
- Then it iterates over the length of the string str and X using nested loops.
- If str[i] is equal to X[j], it updates the current state after removing str[i] and also the current state after keeping str[i] and removing X[j].
- If str[i] is not equal to X[j], it updates the current state with the previous state value of the same index.
- Finally, it prints the last element of the curr vector as the minimum number of character removals.
Implementation :
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 strvoid 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 rows dp states in 2 vectors vector<int>curr(M+1 ,0); vector<int>prev(M+1 ,0); // Fill first row of dp[][] for (int j = 0; j < M; j++) { // If X[j] matches with str[0] if (str[0] == X[j]) { curr[j] = 1; prev[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[ curr[j] = prev[j] + 1; // Update state after keeping str[i] if (j != 0) curr[j] = min(curr[j], prev[j - 1]); } // If str[i] is not equal to X[j] else { curr[j] = prev[j]; } } prev = curr; } // Print the minimum number // of characters removals cout << curr[M - 1];}// Driver Codeint main(){ // Input string str = "btagd"; string X = "bad"; // Function call to get minimum // number of character removals printMinimumRemovals(str, X); return 0;}// this code is contributed by bhardwajji |
Java
import java.util.Arrays;public class Main { // Function to print the minimum number of // character removals required to remove X // as a subsequence from the string str public 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 rows dp states in 2 arrays int[] curr = new int[M+1]; int[] prev = new int[M+1]; // 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)) { curr[j] = 1; prev[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] curr[j] = prev[j] + 1; // Update state after keeping str[i] if (j != 0) curr[j] = Math.min(curr[j], prev[j - 1]); } // If str[i] is not equal to X[j] else { curr[j] = prev[j]; } } prev = Arrays.copyOf(curr, curr.length); // Update the previous row with current row } // Print the minimum number // of characters removals System.out.println(curr[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); }} |
Python3
def printMinimumRemovals(str, X): # Length of the string str N = len(str) # Length of the string X M = len(X) # Stores the rows dp states in 2 vectors curr = [0]*(M+1) prev = [0]*(M+1) # Fill first row of dp[][] for j in range(M): # If X[j] matches with str[0] if str[0] == X[j]: curr[j] = 1 prev[j] = 1 for i in range(1, N): for j in range(M): # If str[i] is equal to X[j] if str[i] == X[j]: # Update state after removing str[i[ curr[j] = prev[j] + 1 # Update state after keeping str[i] if j != 0: curr[j] = min(curr[j], prev[j - 1]) # If str[i] is not equal to X[j] else: curr[j] = prev[j] prev = curr.copy() # Print the minimum number # of characters removals print(curr[M - 1])# Driver Codestr = "btagd"X = "bad"# Function call to get minimum# number of character removalsprintMinimumRemovals(str, X) |
C#
using System;using System.Collections.Generic;class Program{ // 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 rows dp states in 2 lists List<int> curr = new List<int>(M + 1); List<int> prev = new List<int>(M + 1); for (int i = 0; i < M + 1; i++) { curr.Add(0); prev.Add(0); } // Fill first row of dp[][] for (int j = 0; j < M; j++) { // If X[j] matches with str[0] if (str[0] == X[j]) { curr[j] = 1; prev[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[ curr[j] = prev[j] + 1; // Update state after keeping str[i] if (j != 0) curr[j] = Math.Min(curr[j], prev[j - 1]); } // If str[i] is not equal to X[j] else { curr[j] = prev[j]; } } prev = new List<int>(curr); } // Print the minimum number // of characters removals Console.WriteLine(curr[M - 1]); } // Driver Code static void Main(string[] args) { // Input string str = "btagd"; string X = "bad"; // Function call to get minimum // number of character removals PrintMinimumRemovals(str, X); }} |
Javascript
// JavaScript implementation of the above approachfunction printMinimumRemovals(str, X) { // Length of the string strlet N = str.length;// Length of the string Xlet M = X.length;// Stores the rows dp states in 2 arrays let curr = Array(M + 1).fill(0);let prev = Array(M + 1).fill(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]) { curr[j] = 1; prev[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[ curr[j] = prev[j] + 1; // Update state after keeping str[i] if (j !== 0) curr[j] = Math.min(curr[j], prev[j - 1]); } // If str[i] is not equal to X[j] else { curr[j] = prev[j]; } } prev = [...curr];}// Print the minimum number// of characters removalsconsole.log(curr[M - 1]);}// Driver Codelet str = "btagd";let X = "bad";// Function call to get minimum// number of character removalsprintMinimumRemovals(str, X); |
Output:
1
Time Complexity : O(N*M), where N is the size of the first string and M is the size of the second string
Auxiliary Space: O(M), Space used for storing previous values
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!
