用于查询给定字符串在恒定时间内的旋转和第 K 个字符的 Python3 程序

给定字符串str ，任务是对给定字符串执行以下类型的查询：

1. (1, K)：将字符串左旋转K个字符。
2. (2, K)：打印字符串的第 K^个字符。

例子：

Input: str = “abcdefgh”, q[][] = {{1, 2}, {2, 2}, {1, 4}, {2, 7}} 
Output: 
d 
e 
Query 1: str = “cdefghab” 
Query 2: 2^nd character is d 
Query 3: str = “ghabcdef” 
Query 4: 7^th character is e
Input: str = “abc”, q[][] = {{1, 2}, {2, 2}} 
Output: 
a 
 

方法：这里的主要观察是字符串不需要在每个查询中旋转，而是我们可以创建一个指针ptr指向字符串的第一个字符，并且可以在每次旋转时更新它ptr = (ptr + K ) %N其中K是字符串需要旋转的整数， N是字符串的长度。现在对于第二种类型的每个查询，可以通过str[(ptr + K – 1) % N]找到第 K^个字符。
下面是上述方法的实现：

Python3

# Python3 implementation of the approach 
size = 2
  
# Function to perform the required 
# queries on the given string 
def performQueries(string, n, queries, q) : 
  
    # Pointer pointing to the current starting 
    # character of the string 
    ptr = 0; 
  
    # For every query 
    for i in range(q) :
  
        # If the query is to rotate the string 
        if (queries[i][0] == 1) : 
  
            # Update the pointer pointing to the 
            # starting character of the string 
            ptr = (ptr + queries[i][1]) % n; 
              
        else :
  
            k = queries[i][1]; 
  
            # Index of the kth character in the 
            # current rotation of the string 
            index = (ptr + k - 1) % n; 
  
            # Print the kth character 
            print(string[index]); 
  
# Driver code 
if __name__ == "__main__" : 
  
    string = "abcdefgh"; 
    n = len(string); 
  
    queries = [[ 1, 2 ], [ 2, 2 ], 
               [ 1, 4 ], [ 2, 7 ]]; 
    q = len(queries); 
  
    performQueries(string, n, queries, q); 
      
# This code is contributed by AnkitRai01

输出：

d
e

有关详细信息，请参阅关于给定字符串的旋转和第 K 个字符的查询的完整文章！