将一个字符串转换为另一个字符串的最少删除和插入次数

给定两个分别大小为 m 和 n 的字符串’str1′ 和 ‘str2’。任务是从/在 str1 中移除/删除和插入最少数量的字符，以将其转换为 str2。可能需要从 str1 的一个点删除/删除相同的字符并插入到另一个点。

示例 1：

Input : 
str1 = "heap", str2 = "pea" 
Output : 
Minimum Deletion = 2 and
Minimum Insertion = 1
Explanation:
p and h deleted from heap
Then, p is inserted at the beginning
One thing to note, though p was required yet
it was removed/deleted first from its position and
then it is inserted to some other position.
Thus, p contributes one to the deletion_count
and one to the insertion_count.

示例 2：

Input : 
str1 = "geeksforgeeks", str2 = "geeks"
Output : 
Minimum Deletion = 8
Minimum Insertion = 0       

简单的方法：

一个简单的方法是考虑 str1 的所有子序列，并为每个子序列计算最小的删除和插入，以便将其转换为 str2。一种非常复杂的方法，该解决方案的时间复杂度是指数级的。

有效的方法：

一种有效的方法使用找到给定两个序列的最长公共子序列长度的概念。

算法：

• str1和str2是给定的字符串。
• m和n分别是它们的长度。
• len是str1和str2的最长公共子序列的长度
• 最小删除次数 minDel = m – len
• 最小插入次数 minInsert = n – len

下面是上面代码的实现：

C++

// Dynamic Programming C++ implementation to find
// minimum number of deletions and insertions
#include 
 
using namespace std;
 
// Returns length of length common subsequence
// for str1[0..m-1], str2[0..n-1]
int lcs(string str1, string str2, int m, int n)
{
    int L[m + 1][n + 1];
    int i, j;
 
    // Following steps build L[m+1][n+1] in bottom
    // up fashion. Note that L[i][j] contains
    // length of LCS of str1[0..i-1] and str2[0..j-1]
    for (i = 0; i <= m; i++) {
        for (j = 0; j <= n; j++) {
            if (i == 0 || j == 0)
                L[i][j] = 0;
 
            else if (str1.at(i - 1) == str2.at(j - 1))
                L[i][j] = L[i - 1][j - 1] + 1;
 
            else
                L[i][j] = max(L[i - 1][j], L[i][j - 1]);
        }
    }
 
    // L[m][n] contains length of LCS
    // for X[0..n-1] and Y[0..m-1]
    return L[m][n];
}
 
// function to find minimum number
// of deletions and insertions
void printMinDelAndInsert(string str1, string str2)
{
    int m = str1.size();
    int n = str2.size();
 
    int len = lcs(str1, str2, m, n);
 
    cout << "Minimum number of deletions = " << (m - len)
         << endl;
 
    cout << "Minimum number of insertions = " << (n - len)
         << endl;
}
 
// Driver Code
int main()
{
    string str1 = "heap";
    string str2 = "pea";
   
      // Function Call
    printMinDelAndInsert(str1, str2);
    return 0;
}

Java

// Dynamic Programming Java implementation
// to find minimum number of deletions and
// insertions
import java.io.*;
 
class GFG {
 
    // Returns length of length common
    // subsequence for str1[0..m-1],
    // str2[0..n-1]
    static int lcs(String str1, String str2, int m, int n)
    {
        int L[][] = new int[m + 1][n + 1];
        int i, j;
 
        // Following steps build L[m+1][n+1] in
        // bottom up fashion. Note that L[i][j]
        // contains length of LCS of str1[0..i-1]
        // and str2[0..j-1]
        for (i = 0; i <= m; i++) {
            for (j = 0; j <= n; j++) {
                if (i == 0 || j == 0)
                    L[i][j] = 0;
 
                else if (str1.charAt(i - 1)
                         == str2.charAt(j - 1))
                    L[i][j] = L[i - 1][j - 1] + 1;
 
                else
                    L[i][j] = Math.max(L[i - 1][j],
                                       L[i][j - 1]);
            }
        }
 
        // L[m][n] contains length of LCS
        // for X[0..n-1] and Y[0..m-1]
        return L[m][n];
    }
 
    // function to find minimum number
    // of deletions and insertions
    static void printMinDelAndInsert(String str1,
                                     String str2)
    {
        int m = str1.length();
        int n = str2.length();
 
        int len = lcs(str1, str2, m, n);
 
        System.out.println("Minimum number of "
                           + "deletions = ");
        System.out.println(m - len);
 
        System.out.println("Minimum number of "
                           + "insertions = ");
        System.out.println(n - len);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String str1 = new String("heap");
        String str2 = new String("pea");
       
          // Function Call
        printMinDelAndInsert(str1, str2);
    }
}
// This code is contributed by Prerna Saini

Python3

# Dynamic Programming Python3
# implementation to find minimum
# number of deletions and insertions
 
# Returns length of length
# common subsequence for
# str1[0..m-1], str2[0..n-1]
 
 
def lcs(str1, str2, m, n):
 
    L = [[0 for i in range(n + 1)]
         for i in range(m + 1)]
 
    # Following steps build L[m+1][n+1]
    # in bottom up fashion. Note that
    # L[i][j] contains length of LCS
    # of str1[0..i-1] and str2[0..j-1]
    for i in range(m + 1):
        for j in range(n + 1):
            if (i == 0 or j == 0):
                L[i][j] = 0
            elif(str1[i - 1] == str2[j - 1]):
                L[i][j] = L[i - 1][j - 1] + 1
            else:
                L[i][j] = max(L[i - 1][j],
                              L[i][j - 1])
 
    # L[m][n] contains length of LCS
    # for X[0..n-1] and Y[0..m-1]
    return L[m][n]
 
# function to find minimum number
# of deletions and insertions
 
 
def printMinDelAndInsert(str1, str2):
    m = len(str1)
    n = len(str2)
    leng = lcs(str1, str2, m, n)
    print("Minimum number of deletions = ",
          m - leng, sep=' ')
    print("Minimum number of insertions = ",
          n - leng, sep=' ')
 
 
# Driver Code
str1 = "heap"
str2 = "pea"
 
# Function Call
printMinDelAndInsert(str1, str2)
 
# This code is contributed
# by sahilshelangia

C#

// Dynamic Programming C# implementation
// to find minimum number of deletions and
// insertions
using System;
 
class GFG {
 
    // Returns length of length common
    // subsequence for str1[0..m-1],
    // str2[0..n-1]
    static int lcs(string str1, string str2, int m, int n)
    {
        int[, ] L = new int[m + 1, n + 1];
        int i, j;
 
        // Following steps build L[m+1][n+1] in
        // bottom up fashion. Note that L[i][j]
        // contains length of LCS of str1[0..i-1]
        // and str2[0..j-1]
        for (i = 0; i <= m; i++) {
            for (j = 0; j <= n; j++) {
                if (i == 0 || j == 0)
                    L[i, j] = 0;
 
                else if (str1[i - 1] == str2[j - 1])
                    L[i, j] = L[i - 1, j - 1] + 1;
 
                else
                    L[i, j] = Math.Max(L[i - 1, j],
                                       L[i, j - 1]);
            }
        }
 
        // L[m][n] contains length of LCS
        // for X[0..n-1] and Y[0..m-1]
        return L[m, n];
    }
 
    // function to find minimum number
    // of deletions and insertions
    static void printMinDelAndInsert(string str1,
                                     string str2)
    {
        int m = str1.Length;
        int n = str2.Length;
 
        int len = lcs(str1, str2, m, n);
 
        Console.Write("Minimum number of "
                      + "deletions = ");
        Console.WriteLine(m - len);
 
        Console.Write("Minimum number of "
                      + "insertions = ");
        Console.Write(n - len);
    }
 
    // Driver code
    public static void Main()
    {
        string str1 = new string("heap");
        string str2 = new string("pea");
       
          // Function Call
        printMinDelAndInsert(str1, str2);
    }
}
 
// This code is contributed by nitin mittal.

Javascript

    // Dynamic Programming Javascript implementation
    // to find minimum number of deletions and
    // insertions
 
    // Returns length of length common
    // subsequence for str1[0..m-1],
    // str2[0..n-1]
    function lcs(str1, str2, m, n)
    {
        let L = new Array(m + 1);
             
        let i, j;
         
        for (i = 0; i <= m; i++)
        {
            L[i] = new Array(n + 1);
            for (j = 0; j <= n; j++)
            {
                L[i][j] = 0;
            }
        }
  
        // Following steps build L[m+1][n+1] in
        // bottom up fashion. Note that L[i][j]
        // contains length of LCS of str1[0..i-1]
        // and str2[0..j-1]
        for (i = 0; i <= m; i++) {
            for (j = 0; j <= n; j++) {
                if (i == 0 || j == 0)
                    L[i][j] = 0;
  
                else if (str1[i - 1]
                         == str2[j - 1])
                    L[i][j] = L[i - 1][j - 1] + 1;
  
                else
                    L[i][j] = Math.max(L[i - 1][j],
                                       L[i][j - 1]);
            }
        }
  
        // L[m][n] contains length of LCS
        // for X[0..n-1] and Y[0..m-1]
        return L[m][n];
    }
  
    // function to find minimum number
    // of deletions and insertions
    function printMinDelAndInsert(str1, str2)
    {
        let m = str1.length;
        let n = str2.length;
  
        let len = lcs(str1, str2, m, n);
  
        document.write("Minimum number of "
                           + "deletions = ");
        document.write((m - len) + "</br>");
  
        document.write("Minimum number of "
                           + "insertions = ");
        document.write((n - len) + "</br>");
    }
     
    let str1 = "heap";
    let str2 = "pea";
 
    // Function Call
    printMinDelAndInsert(str1, str2);
     
    // This code is contributed by rameshtravel07.

输出

Minimum number of deletions = 2
Minimum number of insertions = 1

时间复杂度： O(m * n)

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