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

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

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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My Personal Notes

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

C++

Java

Python3

C#

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