最小化删除以删除另一个字符串作为给定字符串

给定两个长度分别为N和M 的字符串str和X ，任务是找到需要从字符串str 中删除的最少字符，以便字符串str不包含字符串X作为子序列。

例子：

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

方法：这个问题可以通过动态规划解决。请按照以下步骤解决问题：

• 遍历字符串。
• 初始化一个二维数组dp[N][M] ，其中N是字符串str的长度， M是字符串X的长度。
• dp[i][j]表示子串str[0, i] 中所需的最小字符移除次数，使得子串X[0, j]作为子序列不出现。
• 两个过渡状态是：
  • 如果str[i] 等于 X[j] ，
    • 情况 1：删除字符str[i]
    • 情况 2：通过确保没有出现X[0, j-1]作为str[0, i] 中的子序列来维护字符str[ i]
    • 更新dp[i][j] = min(dp[i − 1][j − 1], dp[i − 1][j] + 1)
  • 如果str[i] 不等于 X[j]，则str[i] 可以保持原样。
    •  更新dp[i][j] = dp[i – 1][j]
• 打印最小移除，即dp[N – 1][M – 1]

下面是上述方法的实现：

C++

// C++ implementation of
// the above approach
#include 
 
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

 
// 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

输出：

1

时间复杂度： O(N * M)
辅助空间： O(N * M)

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