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-
- we have to find the position of the leftmost occurrence of ‘Y’ by iterating over all positions from n-2 to 0.
- 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.
- 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 x’s and 2 y’s#include <bits/stdc++.h>using namespace std;// Function to find the Kth string// in lexicographical order which// consists of N-2 x’s and 2 y’svoid 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 codeint 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 x’s and 2 y’sclass GFG{// Function to find the Kth String// in lexicographical order which// consists of N-2 x’s and 2 y’sstatic 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 codepublic 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 x’s and 2 y’s # Function to find the Kth string # in lexicographical order which # consists of N-2 x’s and 2 y’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 = 5k = 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 x’s and 2 y’susing System;class GFG{// Function to find the Kth String// in lexicographical order which// consists of N-2 x’s and 2 y’sstatic 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 codepublic static void Main(String[] args){ int n = 5, k = 7; // Function call kth_String(n, k);}}// This code is contributed by Amit Katiyar |
Javascript
<script>// javascript program find the Kth String in// lexicographical order consisting// of N-2 x’s and 2 y’s// Function to find the Kth String// in lexicographical order which// consists of N-2 x’s and 2 y’sfunction kth_String(n , k){ // Iterate for all possible // positions of eft most Y for(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(j = 0; j < n; j++) { if (j == i || j == n - k) document.write('Y'); else document.write('X'); } break; } k -= (n - i - 1); }}// Driver codevar n = 5, k = 7;// Function callkth_String(n, k);// This code is contributed by Amit Katiyar </script> |
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
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