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Find Kth largest string from the permutations of the string with two characters
• Last Updated : 09 Apr, 2020

Given two integers N and K, the task is find the lexicographically Kth largest string of size N from the set of strings containing only two characters ‘x’ and ‘y’, where character ‘x’ is present in the string (N – 2) times and the character ‘y’ is present only 2 times.

Examples:

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

Input: N = 3, K = 2
Output: yxy
Explanation:
All the strings of size 3 –
{ xyy, yxy, yyx }

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
The idea is to observe that the lexicographically largest string will have 2 ‘y’ at start followed by all ‘x’. The lexicographically second largest string will have the second ‘y’ moved one index ahead that is, ‘yxyxxxxx……’ and so on.

The key observation in this problem is that character ‘y’ is present two times at the front in the lexicographically largest string and then in each step, the second character ‘y’ moves one step ahead until it reaches the end of the string to generate next smallest string. Once, the second ‘y’ reaches end of string, the next smallest string will have two ‘y’ at index 1 and 2 and then the process continues.

Therefore, the idea is to find the first and second positions for the character ‘y’ and then print these positions with ‘y’ character and all other positions filled with ‘x’ character.

Below is the implementation of the above approach:

## 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`
Output:
```xyxxy
```
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