所需的最小删除次数，使得长度为 2 的子序列不会出现超过一次

给定一个由N个小写字符组成的字符串S ，任务是修改给定的字符串，使长度为 2 的子序列在字符串中不重复，方法是删除最少字符数。

例子：

Input: S = “abcaadbcd”
Output: abcd
Explanation: Removing the characters at indices {2, 3, 4, 5, 6, 7} modifies the string to “abcd”, that contains every subsequence of length 2 exactly once.

Input: S = “cadbcc”
Output: cadbc

方法：给定的问题可以通过观察最终字符串只能包含唯一字符来解决，但第一个字符可以在字符串中出现 2 次，一个在开头，另一个在结尾（如果可能） .请按照以下步骤解决问题：

• 初始化一个空字符串，比如说ans来存储最终的字符串。
• 初始化一个大小为26的布尔数组C[]以检查字符是否存在于最终字符串中。
• 初始化一个变量，例如pos为0以存储添加到字符串的最后一个字符的索引ans 。
• 遍历给定的字符串S ，如果当前字符不在ans中，则将其附加到ans ，在数组C[]中将其标记为已访问，并将pos的值更新为i 。
• 使用变量i遍历[pos + 1, N – 1]范围，如果S[i]等于S[0] ，则将其附加到最终字符串ans并跳出循环。
• 完成上述步骤后，打印字符串ans作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to remove the minimum count
// of characters from the string such
// that no subsequence of length 2 repeats
void RemoveCharacters(string s)
{
    // Initialize the final string
    string ans = "";
 
    // Stores if any character occurs
    // in the final string or not
    bool c[26];
 
    for (int i = 0; i < 26; i++)
        c[i] = 0;
 
    // Store the index of the last
    // character added in the string
    int pos = 0;
 
    // Traverse the string
    for (int i = 0; i < s.size(); i++) {
 
        // Add all the unique
        // characters
        if (c[s[i] - 'a'] == 0) {
            c[s[i] - 'a'] = 1;
            pos = i;
            ans += s[i];
        }
    }
 
    // Check if S[0] appears in the
    // range [pos+1, N-1]
    for (int i = pos + 1;
         i < (int)s.size(); i++) {
 
        // If the characters are the
        // same
        if (s[i] == s[0]) {
            ans += s[i];
            break;
        }
    }
 
    // Print the resultant string
    cout << ans;
}
 
// Driver Code
int main()
{
    string S = "abcaadbcd";
    RemoveCharacters(S);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to remove the minimum count
// of characters from the string such
// that no subsequence of length 2 repeats
static void RemoveCharacters(String s)
{
   
    // Initialize the final string
    String ans = "";
 
    // Stores if any character occurs
    // in the final string or not
    int []c = new int[26];
 
    for (int i = 0; i < 26; i++)
        c[i] = 0;
 
    // Store the index of the last
    // character added in the string
    int pos = 0;
 
    // Traverse the string
    for (int i = 0; i < s.length(); i++) {
 
        // Add all the unique
        // characters
        if (c[(int)s.charAt(i) - 97] == 0) {
            c[(int)s.charAt(i) - 97] = 1;
            pos = i;
            ans += s.charAt(i);
        }
    }
 
    // Check if S[0] appears in the
    // range [pos+1, N-1]
    for (int i = pos + 1;
         i < s.length(); i++) {
 
        // If the characters are the
        // same
        if (s.charAt(i) == s.charAt(0)) {
            ans += s.charAt(i);
            break;
        }
    }
 
    // Print the resultant string
    System.out.println(ans);
}
 
// Driver code
public static void main(String[] args)
{
    String S = "abcaadbcd";
    RemoveCharacters(S);
}
}
 
// This code is contributed by code_hunt.

Python3

# Python 3 program for the above approach
 
# Function to remove the minimum count
# of characters from the string such
# that no subsequence of length 2 repeats
def RemoveCharacters(s):
   
    # Initialize the final string
    ans = ""
 
    # Stores if any character occurs
    # in the final string or not
    c = [0 for i in range(26)]
 
    # Store the index of the last
    # character added in the string
    pos = 0
 
    # Traverse the string
    for i in range(len(s)):
       
        # Add all the unique
        # characters
        if (c[ord(s[i]) - 97] == 0):
            c[ord(s[i]) - 97] = 1
            pos = i
            ans += s[i]
 
    # Check if S[0] appears in the
    # range [pos+1, N-1]
    for i in range(pos + 1, len(s), 1):
       
        # If the characters are the
        # same
        if (s[i] == s[0]):
            ans += s[i]
            break
 
    # Print the resultant string
    print(ans)
 
# Driver Code
if __name__ == '__main__':
    S = "abcaadbcd"
    RemoveCharacters(S)
 
    # This code is contributed by ipg2016107.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
  
// Function to remove the minimum count
// of characters from the string such
// that no subsequence of length 2 repeats
static void RemoveCharacters(string s)
{
   
    // Initialize the final string
    string ans = "";
 
    // Stores if any character occurs
    // in the final string or not
    int []c = new int[26];
 
    for (int i = 0; i < 26; i++)
        c[i] = 0;
 
    // Store the index of the last
    // character added in the string
    int pos = 0;
 
    // Traverse the string
    for (int i = 0; i < s.Length; i++) {
 
        // Add all the unique
        // characters
        if (c[(int)s[i] - 97] == 0) {
            c[(int)s[i] - 97] = 1;
            pos = i;
            ans += s[i];
        }
    }
 
    // Check if S[0] appears in the
    // range [pos+1, N-1]
    for (int i = pos + 1;
         i < s.Length; i++) {
 
        // If the characters are the
        // same
        if (s[i] == s[0]) {
            ans += s[i];
            break;
        }
    }
 
    // Print the resultant string
    Console.Write(ans);
}
 
// Driver Code
public static void Main()
{
    string S = "abcaadbcd";
    RemoveCharacters(S);
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript

 
// javascript program for the above approach
 
// Function to remove the minimum count
// of characters from the string such
// that no subsequence of length 2 repeats
function RemoveCharacters(s)
{
    // Initialize the final string
    var ans = "";
 
    // Stores if any character occurs
    // in the final string or not
    var c = new Array(26);
    var i;
    for (i = 0; i < 26; i++)
        c[i] = 0;
 
    // Store the index of the last
    // character added in the string
    var pos = 0;
 
    // Traverse the string
    for (i = 0; i < s.length; i++) {
 
        // Add all the unique
        // characters
        if (c[s[i].charCodeAt() - 97] == 0) {
            c[s[i].charCodeAt() - 97] = 1;
            pos = i;
            ans += s[i];
        }
    }
 
    // Check if S[0] appears in the
    // range [pos+1, N-1]
    for (i = pos + 1;
         i < s.length; i++) {
 
        // If the characters are the
        // same
        if (s[i] == s[0]) {
            ans += s[i];
            break;
        }
    }
 
    // Print the resultant string
    document.write(ans);
}
 
// Driver Code
    var S = "abcaadbcd";
    RemoveCharacters(S);
 
// This code is contributed by bgangwar59.

输出：

abcd

时间复杂度： O(N)
辅助空间： O(N)