在给定条件下最大化要从字符中删除的字符串

给定一个字符串s 。任务是最大限度地从s中删除字符。如果至少一个相邻字符是拉丁英文字母表中的前一个字母，则可以删除任何字符。

例子：

Input: s = “bacabcab”
Output: 4
Explanation: Following are the removals that maximize the answer.

During the first move, the first character can be removed as s1 = b because s2= a.
The string updates to s= acabcab.
During the second move, fifth character can be removed as s5= c because s4= b.
The string updates to s= acabab.
During the third move, sixth character can be removed s6=’b’ because s5= a.
The string updates to s= acaba.
During the fourth move, the only character that can be removed is s4= b, because s3= a (or s5= a).
The string updates to s= acaa. Now it is not possible to remove any character.

Therefore, 4 is required answer which is maximum possible.

Input: s = “bcda”
Output: 3

编程需要懂一点英语

方法：这个问题可以通过使用贪心方法来解决。请按照以下步骤解决给定的问题。

选择可以删除的最大可能（按字母顺序）字母，然后将其删除。
如果最大字母p被删除，则可用于删除字符串s中的一些其他字母q 。
很明显，在这种情况下s[p]+1=s[q] 。
如果s[p]和s[q]之间没有其他字母，则s[p]不是最大字母。
现在假设如果s[p]和s[q]之间的所有字母都可以删除。然后首先选择s[q] ，然后再选择 s [p] 。
考虑最后一种情况，在s[p]和s[q]之间至少有一个字母s[k ] 。因为我们无法删除 s[k]，所以只有两种情况： s[k] – 1 > s[q] 或 s[k] < s[p] – 1 。那么我们根本不能使用 s[p] 来删除s[q] 。

下面是上述方法的实现。

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to find maximum characters that can
// be removed with given conditions
int removeMaxCharacters(string s, int n)
{
    // Iterate from the last of english alphabet
    // so that bigger alphabets are processed first
    for (int i = 'z'; i > 'a'; i--) {
 
        // Iterate in the given string
        for (int k = 0; k < s.size(); k++) {
            if (s[k] == i) {
                if (s[k - 1] == i - 1
                    || s[k + 1] == i - 1) {
                    s.erase(k, 1);
                    k = -1;
                }
            }
        }
    }
 
    // Return the n-remaining size of s
    return n - s.size();
}
 
// Driver Code
int main()
{
    string s = "abcde";
 
    // Size of string
    int N = s.size();
 
    // Function Call
    cout << removeMaxCharacters(s, N);
}

Java

// Java program for above approach
import java.util.*;
public class GFG
{
 
  // Function to find maximum characters that can
  // be removed with given conditions
  static int removeMaxCharacters(String s, int n)
  {
    // Iterate from the last of english alphabet
    // so that bigger alphabets are processed first
    for (int i = 'z'; i > 'a'; i--) {
 
      // Iterate in the given string
      for (int k = 0; k < s.length(); k++) {
        if (s.charAt(k) == i) {
          if (s.charAt(k - 1) == i - 1
              || s.charAt(k + 1) == i - 1) {
            s = s.substring(k);
          }
        }
      }
    }
 
    // Return the n-remaining size of s
    return n - s.length();
  }
 
  // Driver Code
  public static void main(String args[])
  {
    String s = "abcde";
 
    // Size of string
    int N = s.length();
 
    // Function Call
    System.out.println(removeMaxCharacters(s, N));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

C#

// C# program for above approach
using System;
 
class GFG
{
 
  // Function to find maximum characters that can
  // be removed with given conditions
  static int removeMaxCharacters(string s, int n)
  {
    // Iterate from the last of english alphabet
    // so that bigger alphabets are processed first
    for (int i = 'z'; i > 'a'; i--) {
 
      // Iterate in the given string
      for (int k = 0; k < s.Length; k++) {
        if (s[k] == i) {
          if (s[k - 1] == i - 1
              || s[k + 1] == i - 1) {
            s = s.Substring(k, 1);
          }
        }
      }
    }
 
    // Return the n-remaining size of s
    return n - s.Length;
  }
 
  // Driver Code
  public static void Main()
  {
    string s = "abcde";
 
    // Size of string
    int N = s.Length;
 
    // Function Call
    Console.WriteLine(removeMaxCharacters(s, N));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

输出

时间复杂度： O(N ² )，其中 N 是字符串的长度。
辅助空间： O(1)