检查字符串是否由K个交替字符

给定一个字符串str和一个整数K ，任务是检查它是否由K个交替字符。

例子：

Input: str = “acdeac”, K = 4
Output: Yes

Input: str = “abcdcab”, K = 2
Output: No

方法：为了使字符串由K个交替的字符，它必须满足以下条件：

1. (i + mK)索引处的所有字符必须相同，其中i是当前索引， mK代表K的第m^倍。这意味着在每K个索引之后，必须重复该字符。
2. 相邻字符不能相同。这是因为，如果字符串的类型为“ AAAAA”，并且单个字符重复任意次，则上述条件将被匹配，但是在这种情况下，答案必须为“否”。

   下面的代码是上述方法的实现：

   C++

   // C++ implementation of the approach
     
   #include 
   using namespace std;
     
   // Function to check if a string
   // is made up of k alternating characters
   bool isKAlternating(string s, int k)
   {
       if (s.length() < k)
           return false;
     
       int checker = 0;
     
       // Check if all the characters at
       // indices 0 to K-1 are different
       for (int i = 0; i < k; i++) {
     
           int bitAtIndex = s[i] - 'a';
     
           // If that bit is already set in
           // checker, return false
           if ((checker & (1 << bitAtIndex)) > 0) {
               return false;
           }
     
           // Otherwise update and continue by
           // setting that bit in the checker
           checker = checker | (1 << bitAtIndex);
       }
     
       for (int i = k; i < s.length(); i++)
           if (s[i - k] != s[i])
               return false;
     
       return true;
   }
     
   // Driver code
   int main()
   {
       string str = "acdeac";
       int K = 4;
     
       if (isKAlternating(str, K))
           cout << "Yes" << endl;
       else
           cout << "No" << endl;
     
       return 0;
   }

   Java

   // Java implementation of the approach
   class GFG{
      
   // Function to check if a String
   // is made up of k alternating characters
   static boolean isKAlternating(String s, int k)
   {
       if (s.length() < k)
           return false;
      
       int checker = 0;
      
       // Check if all the characters at
       // indices 0 to K-1 are different
       for (int i = 0; i < k; i++) {
      
           int bitAtIndex = s.charAt(i) - 'a';
      
           // If that bit is already set in
           // checker, return false
           if ((checker & (1 << bitAtIndex)) > 0) {
               return false;
           }
      
           // Otherwise update and continue by
           // setting that bit in the checker
           checker = checker | (1 << bitAtIndex);
       }
      
       for (int i = k; i < s.length(); i++)
           if (s.charAt(i - k) != s.charAt(i) )
               return false;
      
       return true;
   }
      
   // Driver code
   public static void main(String[] args)
   {
       String str = "acdeac";
       int K = 4;
      
       if (isKAlternating(str, K))
           System.out.print("Yes" +"\n");
       else
           System.out.print("No" +"\n"); 
   }
   }
     
   // This code is contributed by sapnasingh4991

   Python3

   # Python 3 implementation of the approach
      
   # Function to check if a string
   # is made up of k alternating characters
   def isKAlternating( s, k):
       if (len(s) < k):
           return False
      
       checker = 0
      
       # Check if all the characters at
       # indices 0 to K-1 are different
       for i in range( k):
      
           bitAtIndex = ord(s[i]) - ord('a')
      
           # If that bit is already set in
           # checker, return false
           if ((checker & (1 << bitAtIndex)) > 0):
               return False
      
           # Otherwise update and continue by
           # setting that bit in the checker
           checker = checker | (1 << bitAtIndex)
      
       for i in range(k,len(s)):
           if (s[i - k] != s[i]):
               return False
      
       return True
      
   # Driver code
   if __name__ =="__main__":
     
       st = "acdeac"
       K = 4
      
       if (isKAlternating(st, K)):
           print ("Yes")
       else:
           print ("No")
      
   # This code is contributed by chitranayal

   C#

   // C# implementation of the approach
   using System;
     
   class GFG{
     
   // Function to check if a String
   // is made up of k alternating characters
   static bool isKAlternating(String s, int k)
   {
       if (s.Length < k)
           return false;
     
       int checker = 0;
     
       // Check if all the characters at
       // indices 0 to K-1 are different
       for (int i = 0; i < k; i++) {
     
           int bitAtIndex = s[i] - 'a';
     
           // If that bit is already set in
           // checker, return false
           if ((checker & (1 << bitAtIndex)) > 0) {
               return false;
           }
     
           // Otherwise update and continue by
           // setting that bit in the checker
           checker = checker | (1 << bitAtIndex);
       }
     
       for (int i = k; i < s.Length; i++)
           if (s[i - k] != s[i] )
               return false;
     
       return true;
   }
     
   // Driver code
   public static void Main() 
   {
       String str = "acdeac";
       int K = 4;
     
       if (isKAlternating(str, K))
           Console.WriteLine("Yes");
       else
           Console.WriteLine("No");
   }
   }
     
   // This article contributed by AbhiThakur

   输出：

   Yes
   

   时间复杂度： O(N)