从具有两个字符的字符串的排列中找出第 K 个最大的字符串

鉴于两个整数N和K，任务是找到字典序^第K^个最大的大小为N从仅包含两个字符“X”和“Y”一组字符串，其中字符“X”是存在的字符串中的字符串(N – 2) 次并且字符’y’ 只出现 2 次。
例子：

Input: N = 4, K = 3 
Output: yxxy 
Explanation: 
All the strings of size 4 – 
{ xxyy, xyxy, xyyx, yxxy, yxyx, yyxx } 
The 3^rd smallest string will be – yxxy
Input: N = 3, K = 2 
Output: yxy 
Explanation: 
All the strings of size 3 – 
{ xyy, yxy, yyx } 
 

方法：
这个想法是观察字典序最大的字符串在开始时将有 2 个“y”，然后是所有的“x”。字典序第二大的字符串会将第二个 ‘y’ 向前移动一个索引，即 ‘yxyxxxxx……’ 等等。
这个问题的关键观察是字符’y’在字典序最大的字符串的前面出现两次，然后在每一步中，第二个字符’y’向前移动一步，直到它到达字符串的末尾以生成下一个最小的字符串。一旦第二个 ‘y’ 到达字符串 的末尾，下一个最小的字符串将在索引 1 和 2 处有两个 ‘y’，然后该过程继续。
因此，我们的想法是找到字符’y’ 的第一个和第二个位置，然后用 ‘y’字符打印这些位置，并用 ‘x’字符填充所有其他位置。
下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// Function to print the
// kth largest string
void kthString(int n, int k)
{
    int total = 0;
    int i = 1;
 
    // loop to iterate through
    // series
    while (total < k) {
        // total takes the position
        // of second y
        total = total + n - i;
 
        // i takes the position of
        // first y
        i++;
    }
 
    // calculating first y postion
    int first_y_position = i - 1;
 
    // calculating second y position
    // from first y
    int second_y_position = k - (total - n + first_y_position);
 
    // print all x before first y
    for (int j = 1; j < first_y_position; j++)
        cout << "x";
 
    // print first y
    cout << "y";
 
    int j = first_y_position + 1;
 
    // print all x between first y
    // and second y
    while (second_y_position > 1) {
        cout << "x";
        second_y_position--;
        j++;
    }
 
    // print second y
    cout << "y";
 
    // print x which occur
    // after second y
    while (j < n) {
        cout << "x";
        j++;
    }
}
 
// Driver code
int main()
{
    int n = 5;
 
    int k = 7;
 
    kthString(n, k);
 
    return 0;
}

Java

// Java implementation of above approach
class GFG{
  
// Function to print the
// kth largest String
static void kthString(int n, int k)
{
    int total = 0;
    int i = 1;
  
    // loop to iterate through
    // series
    while (total < k) {
        // total takes the position
        // of second y
        total = total + n - i;
  
        // i takes the position of
        // first y
        i++;
    }
  
    // calculating first y postion
    int first_y_position = i - 1;
  
    // calculating second y position
    // from first y
    int second_y_position = k - (total - n + first_y_position);
  
    // print all x before first y
    for (int j = 1; j < first_y_position; j++)
        System.out.print("x");
  
    // print first y
    System.out.print("y");
  
    int j = first_y_position + 1;
  
    // print all x between first y
    // and second y
    while (second_y_position > 1) {
        System.out.print("x");
        second_y_position--;
        j++;
    }
  
    // print second y
    System.out.print("y");
  
    // print x which occur
    // after second y
    while (j < n) {
        System.out.print("x");
        j++;
    }
}
  
// Driver code
public static void main(String[] args)
{
    int n = 5;
  
    int k = 7;
  
    kthString(n, k);
}
}
 
// This code is contributed by sapnasingh4991

Python 3

# Python 3 implementation of above approach
 
# Function to print the
# kth largest string
def kthString(n,k):
    total = 0
    i = 1
 
    # loop to iterate through
    # series
    while (total < k):
        # total takes the position
        # of second y
        total = total + n - i
 
        # i takes the position of
        # first y
        i += 1
 
    # calculating first y postion
    first_y_position = i - 1
 
    # calculating second y position
    # from first y
    second_y_position = k - (total - n + first_y_position)
 
    # print all x before first y
    for j in range(1,first_y_position,1):
        print("x",end = "")
 
    # print first y
    print("y",end = "")
 
    j = first_y_position + 1
 
    # print all x between first y
    # and second y
    while (second_y_position > 1):
        print("x",end = "")
        second_y_position -= 1
        j += 1
 
    # print second y
    print("y",end = "")
 
    # print x which occur
    # after second y
    while (j < n):
        print("x")
        j += 1
 
# Driver code
if __name__ == '__main__':
    n = 5
    k = 7
    kthString(n, k)
 
# This code is contributed by Surendra_Gangwar

C#

// C# implementation of above approach
using System;
 
class GFG{
 
    // Function to print the
    // kth largest string
    static void kthString(int n, int k)
    {
        int total = 0;
        int i = 1;
     
        // loop to iterate through
        // series
        while (total < k) {
            // total takes the position
            // of second y
            total = total + n - i;
     
            // i takes the position of
            // first y
            i++;
        }
     
        // calculating first y postion
        int first_y_position = i - 1;
     
        // calculating second y position
        // from first y
        int second_y_position = k - (total - n + first_y_position);
         
        int j;
         
        // print all x before first y
        for (j = 1; j < first_y_position; j++)
            Console.Write("x");
     
        // print first y
        Console.Write("y");
     
        j = first_y_position + 1;
     
        // print all x between first y
        // and second y
        while (second_y_position > 1) {
            Console.Write("x");
            second_y_position--;
            j++;
        }
     
        // print second y
        Console.Write("y");
     
        // print x which occur
        // after second y
        while (j < n) {
            Console.Write("x");
            j++;
        }
    }
     
    // Driver code
    static public void Main ()
    {
        int n = 5;
 
        int k = 7;
     
        kthString(n, k);
    }
}
 
// This code is contributed by shubhamsingh10

Javascript

 
// Javascript implementation of above approach
 
// Function to print the
// kth largest string
function kthString(n, k)
{
    var total = 0;
    var i = 1;
 
    // loop to iterate through
    // series
    while (total < k) {
        // total takes the position
        // of second y
        total = total + n - i;
 
        // i takes the position of
        // first y
        i++;
    }
 
    // calculating first y postion
    var first_y_position = i - 1;
 
    // calculating second y position
    // from first y
    var second_y_position = k - (total - n + first_y_position);
 
    // print all x before first y
    for (var j = 1; j < first_y_position; j++)
        document.write( "x");
 
    // print first y
    document.write( "y");
 
    var j = first_y_position + 1;
 
    // print all x between first y
    // and second y
    while (second_y_position > 1) {
        document.write( "x");
        second_y_position--;
        j++;
    }
 
    // print second y
    document.write( "y");
 
    // print x which occur
    // after second y
    while (j < n) {
        document.write( "x");
        j++;
    }
}
 
// Driver code
var n = 5;
var k = 7;
kthString(n, k);
 
// This code is contributed by rutvik_56.

输出：

xyxxy

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