Lexicographically n-th permutation of a string
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; } |
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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 |
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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 */ |
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Output:
EEEEFGGRKSOSK
Find n-th lexicographically permutation of a string | Set 2
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