# Find the k-th string in lexicographical order consisting of n-2 X’s and 2 Y’s

Last Updated : 31 Jul, 2022

Given two numbers N and K, the task is to find the Kth string in lexicographical order if the starting string contains (N-2) x’s first and then 2 Y’s
Note:

1 ? K ? N*(N-1)/2, N*(N-1)/2 are the number of possible permutations

Examples:

Input : N = 5, K = 7
Output : YXXXY
The possible strings in lexicographical order
1. XXXYY
2. XXYXY
3. XXYYX
4. XYXXY
5. XYXYX
6. XYYXX
7. YXXXY
8. YXXYX
9. YXYXX
10. YYXXX
Input : N = 8, K = 20
Output : XYXYXXXX

Approach
Inorder to find the kth position string, we have to follow the following steps-

1. we have to find the position of the leftmost occurrence of ‘Y’ by iterating over all positions from n-2 to 0.
2. Now while iterating, if k<=n-i-1 then this is the required position of the leftmost occurrence of ‘Y’ and the position of rightmost occurrence is n-k so we can print the answer.
3. Otherwise, let’s decrease k by n-i-1, i.e, remove all strings which have the leftmost ‘Y’ at the current position and proceed to the next position. In this way we consider all possible strings in lexicographic order.

Below is the implementation of the above approach.

## C++

 `// C++ program find the Kth string in` `// lexicographical order consisting` `// of N-2 xâ€™s and 2 yâ€™s` `#include ` `using` `namespace` `std;`   `// Function to find the Kth string` `// in lexicographical order which` `// consists of N-2 xâ€™s and 2 yâ€™s` `void` `kth_string(``int` `n, ``int` `k)` `{` `    ``// Iterate for all possible positions of` `    ``// Left most Y` `    ``for` `(``int` `i = n - 2; i >= 0; i--) {` `        ``// If i is the left most position of Y` `        ``if` `(k <= (n - i - 1)) {` `            ``// Put Y in their positions` `            ``for` `(``int` `j = 0; j < n; j++) {` `                ``if` `(j == i or j == n - k)` `                    ``cout << ``'Y'``;` `                ``else` `                    ``cout << ``'X'``;` `            ``}`   `            ``break``;` `        ``}` `        ``k -= (n - i - 1);` `    ``}` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 5, k = 7;`   `    ``// Function call` `    ``kth_string(n, k);`   `    ``return` `0;` `}`

## Java

 `// Java program find the Kth String in` `// lexicographical order consisting` `// of N-2 xâ€™s and 2 yâ€™s` `class` `GFG{`   `// Function to find the Kth String` `// in lexicographical order which` `// consists of N-2 xâ€™s and 2 yâ€™s` `static` `void` `kth_String(``int` `n, ``int` `k)` `{` `    `  `    ``// Iterate for all possible ` `    ``// positions of eft most Y` `    ``for``(``int` `i = n - ``2``; i >= ``0``; i--) ` `    ``{` `       ``// If i is the left ` `       ``// most position of Y` `       ``if` `(k <= (n - i - ``1``))` `       ``{` `           ``// Put Y in their positions` `           ``for``(``int` `j = ``0``; j < n; j++)` `           ``{` `              ``if` `(j == i || j == n - k)` `                  ``System.out.print(``'Y'``);` `              ``else` `                  ``System.out.print(``'X'``);` `           ``}`   `            ``break``;` `        ``}` `        `  `        ``k -= (n - i - ``1``);` `    ``}` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `n = ``5``, k = ``7``;`   `    ``// Function call` `    ``kth_String(n, k);` `}` `}`   `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 program find the Kth string in ` `# lexicographical order consisting ` `# of N-2 xâ€™s and 2 yâ€™s `   `# Function to find the Kth string ` `# in lexicographical order which ` `# consists of N-2 xâ€™s and 2 yâ€™s ` `def` `kth_string(n, k): `   `    ``# Iterate for all possible positions of ` `    ``# left most Y ` `    ``for` `i ``in` `range``(n ``-` `2``, ``-``1``, ``-``1``):` `        `  `        ``# If i is the left most position of Y ` `        ``if` `k <``=` `(n ``-` `i ``-` `1``): ` `            `  `            ``# Put Y in their positions ` `            ``for` `j ``in` `range``(n): ` `                ``if` `(j ``=``=` `i ``or` `j ``=``=` `n ``-` `k): ` `                    ``print``(``'Y'``, end ``=` `"") ` `                ``else``:` `                    ``print``(``'X'``, end ``=` `"")` `            ``break`   `        ``k ``-``=` `(n ``-` `i ``-` `1``) `   `# Driver code ` `n ``=` `5` `k ``=` `7`   `# Function call ` `kth_string(n, k)`   `# This code is contributed by divyamohan123`

## C#

 `// C# program find the Kth String in` `// lexicographical order consisting` `// of N-2 xâ€™s and 2 yâ€™s` `using` `System;`   `class` `GFG{`   `// Function to find the Kth String` `// in lexicographical order which` `// consists of N-2 xâ€™s and 2 yâ€™s` `static` `void` `kth_String(``int` `n, ``int` `k)` `{` `    `  `    ``// Iterate for all possible ` `    ``// positions of eft most Y` `    ``for``(``int` `i = n - 2; i >= 0; i--) ` `    ``{` `        `  `       ``// If i is the left ` `       ``// most position of Y` `       ``if` `(k <= (n - i - 1))` `       ``{` `           `  `           ``// Put Y in their positions` `           ``for``(``int` `j = 0; j < n; j++)` `           ``{` `               ``if` `(j == i || j == n - k)` `                   ``Console.Write(``'Y'``);` `               ``else` `                   ``Console.Write(``'X'``);` `           ``}` `           ``break``;` `       ``}` `       ``k -= (n - i - 1);` `    ``}` `}`   `// Driver code` `public` `static` `void` `Main(String[] args)` `{` `    ``int` `n = 5, k = 7;`   `    ``// Function call` `    ``kth_String(n, k);` `}` `}`   `// This code is contributed by Amit Katiyar`

## Javascript

 ``

Output:

`YXXXY`

Time Complexity: O(N2) // two loops are used nested inside each other so the complexity is n*n

Auxiliary Space: O (1) // no extra array is used so constant space