在恒定时间内查询给定字符串的旋转和第 K 个字符

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, K)：将字符串左旋转K个字符。
(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

编程需要懂一点英语

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：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
#define size 2
 
// Function to perform the required
// queries on the given string
void performQueries(string str, int n,
                    int queries[][size], int q)
{
 
    // Pointer pointing to the current starting
    // character of the string
    int ptr = 0;
 
    // For every query
    for (int i = 0; i < q; i++) {
 
        // 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 {
 
            int k = queries[i][1];
 
            // Index of the kth character in the
            // current rotation of the string
            int index = (ptr + k - 1) % n;
 
            // Print the kth character
            cout << str[index] << "\n";
        }
    }
}
 
// Driver code
int main()
{
    string str = "abcdefgh";
    int n = str.length();
 
    int queries[][size] = { { 1, 2 }, { 2, 2 },
                            { 1, 4 }, { 2, 7 } };
    int q = sizeof(queries) / sizeof(queries[0]);
 
    performQueries(str, n, queries, q);
 
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;
class GFG
{
static int size = 2;
 
// Function to perform the required
// queries on the given string
static void performQueries(String str, int n,
                           int queries[][], int q)
{
 
    // Pointer pointing to the current
    // starting character of the string
    int ptr = 0;
 
    // For every query
    for (int i = 0; i < q; i++)
    {
 
        // 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
        {
            int k = queries[i][1];
 
            // Index of the kth character in the
            // current rotation of the string
            int index = (ptr + k - 1) % n;
 
            // Print the kth character
            System.out.println(str.charAt(index));
        }
    }
}
 
// Driver code
public static void main(String[] args)
{
    String str = "abcdefgh";
    int n = str.length();
 
    int queries[][] = { { 1, 2 }, { 2, 2 },
                        { 1, 4 }, { 2, 7 } };
    int q = queries.length;
 
    performQueries(str, n, queries, q);
}
}
 
// This code is contributed by 29AjayKumar

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

C#

// C# implementation of the approach
using System;
     
class GFG
{
static int size = 2;
 
// Function to perform the required
// queries on the given string
static void performQueries(String str, int n,
                           int [,]queries, int q)
{
 
    // Pointer pointing to the current
    // starting character of the string
    int ptr = 0;
 
    // For every query
    for (int i = 0; i < q; i++)
    {
 
        // 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
        {
            int k = queries[i, 1];
 
            // Index of the kth character in the
            // current rotation of the string
            int index = (ptr + k - 1) % n;
 
            // Print the kth character
            Console.WriteLine(str[index]);
        }
    }
}
 
// Driver code
public static void Main(String[] args)
{
    String str = "abcdefgh";
    int n = str.Length;
 
    int [,]queries = { { 1, 2 }, { 2, 2 },
                       { 1, 4 }, { 2, 7 } };
    int q = queries.GetLength(0);
 
    performQueries(str, n, queries, q);
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

输出：

d
e