Given a string of length m containing lowercase alphabets only. You have to find the n-th permutation of string lexicographically.
Examples:
Input : str[] = "abc", n = 3 Output : Result = "bac" Explanation : All possible permutation in sorted order: abc, acb, bac, bca, cab, cba Input : str[] = "aba", n = 2 Output : Result = "aba" Explanation : All possible permutation in sorted order: aab, aba, baa
Prerequisite : Permutations of a given string using STL
Idea behind printing n-th permutation is quite simple we should use STL (explained in above link) for finding next permutation and do it till the nth permutation. After n-th iteration, we should break from the loop and then print the string which is our nth permutation.
long int i = 1;
do
{
// check for nth iteration
if (i == n)
break;
i++; // keep incrementing the iteration
} while (next_permutation(str.begin(), str.end()));
// print string after nth iteration
print str;
C++
// C++ program to print nth permutation with // using next_permute() #include <bits/stdc++.h> using namespace std; // Function to print nth permutation // using next_permute() void nPermute(string str, long int n) { // Sort the string in lexicographically // ascending order sort(str.begin(), str.end()); // Keep iterating until // we reach nth position long int i = 1; do { // check for nth iteration if (i == n) break; i++; } while (next_permutation(str.begin(), str.end())); // print string after nth iteration cout << str; } // Driver code int main() { string str = "GEEKSFORGEEKS"; long int n = 100; nPermute(str, n); return 0; } |
Java
// Java program to print nth permutation with // using next_permute() import java.util.*; class GFG { // Function to print nth permutation // using next_permute() static void nPermute(char[] str, int n) { // Sort the string in lexicographically // ascending order Arrays.sort(str); // Keep iterating until // we reach nth position int i = 1; do { // check for nth iteration if (i == n) break; i++; } while (next_permutation(str)); // print string after nth iteration System.out.println(String.valueOf(str)); } static boolean next_permutation(char[] p) { for (int a = p.length - 2; a >= 0; --a) if (p[a] < p[a + 1]) for (int b = p.length - 1;; --b) if (p[b] > p[a]) { char t = p[a]; p[a] = p[b]; p[b] = t; for (++a, b = p.length - 1; a < b; ++a, --b) { t = p[a]; p[a] = p[b]; p[b] = t; } return true; } return false; } // Driver code public static void main(String[] args) { String str = "GEEKSFORGEEKS"; int n = 100; nPermute(str.toCharArray(), n); } } // This code contributed by Rajput-Ji |
Python3
# Python3 program to print nth permutation # with using next_permute() # next_permutation method implementation def next_permutation(L): n = len(L) i = n - 2 while i >= 0 and L[i] >= L[i + 1]: i -= 1 if i == -1: return False j = i + 1 while j < n and L[j] > L[i]: j += 1 j -= 1 L[i], L[j] = L[j], L[i] left = i + 1 right = n - 1 while left < right: L[left], L[right] = L[right], L[left] left += 1 right -= 1 return True # Function to print nth permutation # using next_permute() def nPermute(string, n): string = list(string) new_string = [] # Sort the string in lexicographically # ascending order string.sort() j = 2 # Keep iterating until # we reach nth position while next_permutation(string): new_string = string # check for nth iteration if j == n: break j += 1 # print string after nth iteration print(''.join(new_string)) # Driver Code if __name__ == "__main__": string = "GEEKSFORGEEKS" n = 100 nPermute(string, n) # This code is contributed by # sanjeev2552 |
C#
// C# program to print nth permutation with // using next_permute() using System; class GFG { // Function to print nth permutation // using next_permute() static void nPermute(char[] str, int n) { // Sort the string in lexicographically // ascending order Array.Sort(str); // Keep iterating until // we reach nth position int i = 1; do { // check for nth iteration if (i == n) break; i++; } while (next_permutation(str)); // print string after nth iteration Console.WriteLine(String.Join("",str)); } static bool next_permutation(char[] p) { for (int a = p.Length - 2; a >= 0; --a) if (p[a] < p[a + 1]) for (int b = p.Length - 1;; --b) if (p[b] > p[a]) { char t = p[a]; p[a] = p[b]; p[b] = t; for (++a, b = p.Length - 1; a < b; ++a, --b) { t = p[a]; p[a] = p[b]; p[b] = t; } return true; } return false; } // Driver code public static void Main() { String str = "GEEKSFORGEEKS"; int n = 100; nPermute(str.ToCharArray(), n); } } /* This code contributed by PrinciRaj1992 */ |
Output:
EEEEFGGRKSOSK
Find n-th lexicographically permutation of a string | Set 2
This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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