Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Permutation present at the middle of lexicographic ordering of permutations of at most length N made up integers up to K

  • Last Updated : 05 Nov, 2021

Given two positive integers K and N, the task is to find the permutation present at the middle of all permutations of at most length N, consisting of integers from the range [1, K], arranged lexicographically.

Examples:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: K = 3, N = 2
Output: 2 1
Explanation: Lexicographical order of all possible permutations are:



  1. {1}.
  2. {1, 1}
  3. {1, 2}
  4. {1, 3}
  5. {2}
  6. {2, 1}
  7. {2, 2}
  8. {2, 3}
  9. {3}
  10. {3, 1}
  11. {3, 2}
  12. {3, 3}

Therefore, the middle lexicographically the smallest sequence is (N/2)th(= 6th) sequence, which is {2, 1}.

Input: K = 2, N = 4
Output: 1 2 2 2

Naive Approach: The simplest approach to solve the given problem is to generate all possible subsequences of a length [1, N], consisting of integers from [1, K]. Store these elements in an array. After generating all the subsequences, sort the stored list of subsequences and print the middle element of the list.

Time Complexity: O(NK)
Auxiliary Space: O(NK)

Efficient Approach: The above approach can be optimized by checking the parity of K whether K is odd or even and then find the middle lexicographically the smallest sequence accordingly. Follow the steps below to solve the problem:

  • If the value of K is even, then exactly half of the sequences start with an integer K/2 or less. Therefore, the resultant sequence is K/2 followed by (N – 1) occurrence of K.
  • If the value K is odd, then consider B to be a sequence that contains N occurrences of (K + 1)/2.
    • For a sequence X, let f(X) be a sequence such that Xi in X is replaced with (K + 1 − Xi).
    • The only exception happens for prefixes of B.
  • Start with the sequence B, and perform the following steps (N – 1)/2 times:
    • If the last element of the current sequence is 1, then remove it.
    • Otherwise, decrement the last element by 1, and while the sequence contains less than N elements, and insert K at the end of sequence B.
  • After completing the above steps, print the sequence obtained in the array B[].

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function that finds the middle the
// lexicographical smallest sequence
void lexiMiddleSmallest(int K, int N)
{
    // If K is even
    if (K % 2 == 0) {
 
        // First element is K/2
        cout << K / 2 << " ";
 
        // Remaining elements of the
        // sequence are all integer K
        for (int i = 0; i < N - 1; ++i) {
            cout << K << " ";
        }
        cout << "\n";
        exit(0);
    }
 
    // Stores the sequence when K
    // is odd
    vector<int> a(N, (K + 1) / 2);
 
    // Iterate over the range [0, N/2]
    for (int i = 0; i < N / 2; ++i) {
 
        // Check if the sequence ends
        // with in 1 or not
        if (a.back() == 1) {
 
            // Remove the sequence
            // ending in 1
            a.pop_back();
        }
 
        // If it doesn't end in 1
        else {
 
            // Decrement by 1
            --a.back();
 
            // Insert K to the sequence
            // till its size is N
            while ((int)a.size() < N) {
                a.push_back(K);
            }
        }
    }
 
    // Print the sequence stored
    // in the vector
    for (auto i : a) {
        cout << i << " ";
    }
    cout << "\n";
}
 
// Driver Code
int main()
{
    int K = 2, N = 4;
    lexiMiddleSmallest(K, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Function that finds the middle the
    // lexicographical smallest sequence
    static void lexiMiddleSmallest(int K, int N)
    {
       
        // If K is even
        if (K % 2 == 0) {
 
            // First element is K/2
            System.out.print(K / 2 + " ");
 
            // Remaining elements of the
            // sequence are all integer K
            for (int i = 0; i < N - 1; ++i) {
                System.out.print(K + " ");
            }
            System.out.println();
            return;
        }
 
        // Stores the sequence when K
        // is odd
        ArrayList<Integer> a = new ArrayList<Integer>();
 
        // Iterate over the range [0, N/2]
        for (int i = 0; i < N / 2; ++i) {
 
            // Check if the sequence ends
            // with in 1 or not
            if (a.get(a.size() - 1) == 1) {
 
                // Remove the sequence
                // ending in 1
                a.remove(a.size() - 1);
            }
 
            // If it doesn't end in 1
            else {
 
                // Decrement by 1
                int t = a.get(a.size() - 1) - 1;
                a.set(a.get(a.size() - 1), t);
 
                // Insert K to the sequence
                // till its size is N
                while (a.size() < N) {
                    a.add(K);
                }
            }
        }
 
        // Print the sequence stored
        // in the vector
        for (int i : a) {
            System.out.print(i + " ");
        }
        System.out.println();
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int K = 2, N = 4;
        lexiMiddleSmallest(K, N);
    }
}
 
// This code is contributed by Dharanendra L V.

Python3




# Python3 program for the above approach
 
# Function that finds the middle the
# lexicographical smallest sequence
def lexiMiddleSmallest(K, N):
    # If K is even
    if (K % 2 == 0):
 
        # First element is K/2
        print(K // 2,end=" ")
 
        # Remaining elements of the
        # sequence are all integer K
        for i in range(N - 1):
            print(K, end = " ")
        print()
        return
 
    # Stores the sequence when K
    # is odd
    a = [(K + 1) // 2]*(N)
 
    # Iterate over the range [0, N/2]
    for i in range(N//2):
        # Check if the sequence ends
        # with in 1 or not
        if (a[-1] == 1):
 
            # Remove the sequence
            # ending in 1
            del a[-1]
 
        # If it doesn't end in 1
        else:
 
            # Decrement by 1
            a[-1] -= 1
 
            # Insert K to the sequence
            # till its size is N
            while (len(a) < N):
                a.append(K)
 
    # Print sequence stored
    # in the vector
    for i in a:
        print(i, end = " ")
    print()
 
# Driver Code
if __name__ == '__main__':
    K, N = 2, 4
    lexiMiddleSmallest(K, N)
 
# This code is contributed by mohit kumar 29.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
 
    // Function that finds the middle the
    // lexicographical smallest sequence
    static void lexiMiddleSmallest(int K, int N)
    {
        // If K is even
        if (K % 2 == 0) {
 
            // First element is K/2
            Console.Write(K / 2 + " ");
 
            // Remaining elements of the
            // sequence are all integer K
            for (int i = 0; i < N - 1; ++i) {
                Console.Write(K + " ");
            }
            Console.WriteLine();
            return;
        }
 
        // Stores the sequence when K
        // is odd
        List<int> a = new List<int>();
 
        // Iterate over the range [0, N/2]
        for (int i = 0; i < N / 2; ++i) {
 
            // Check if the sequence ends
            // with in 1 or not
            if (a[a.Count - 1] == 1) {
 
                // Remove the sequence
                // ending in 1
                a.Remove(a.Count - 1);
            }
 
            // If it doesn't end in 1
            else {
 
                // Decrement by 1
                a[a.Count - 1] -= 1;
 
                // Insert K to the sequence
                // till its size is N
                while ((int)a.Count < N) {
                    a.Add(K);
                }
            }
        }
 
        // Print the sequence stored
        // in the vector
        foreach(int i in a) { Console.Write(i + " "); }
        Console.WriteLine();
    }
 
    // Driver Code
    public static void Main()
    {
        int K = 2, N = 4;
        lexiMiddleSmallest(K, N);
    }
}
 
// This code is contributed by ukasp.

Javascript




<script>
 
// javascript program for the above approach
 
    // Function that finds the middle the
    // lexicographical smallest sequence
    function lexiMiddleSmallest(K, N)
    {
        // If K is even
        if (K % 2 == 0) {
  
            // First element is K/2
            document.write(K / 2 + " ");
  
            // Remaining elements of the
            // sequence are all integer K
            for (let i = 0; i < N - 1; ++i) {
                document.write(K + " ");
            }
            document.write("<br/>");
            return;
        }
  
        // Stores the sequence when K
        // is odd
        let a = [];
  
        // Iterate over the range [0, N/2]
        for (let i = 0; i < N / 2; ++i) {
  
            // Check if the sequence ends
            // with in 1 or not
            if (a[a.length - 1] == 1) {
  
                // Remove the sequence
                // ending in 1
                a.pop(a.length - 1);
            }
  
            // If it doesn't end in 1
            else {
  
                // Decrement by 1
                a[a.length - 1] -= 1;
  
                // Insert K to the sequence
                // till its size is N
                while (a.length < N) {
                    a.push(K);
                }
            }
        }
  
        // Print the sequence stored
        // in the vector
        for(let i in a) { document.write(i + " "); }
        document.write("<br/>");
    }
 
// Driver Code
 
    let K = 2, N = 4;
    lexiMiddleSmallest(K, N);
     
 
</script>
Output: 
1 2 2 2

 

Time Complexity: O(N)
Auxiliary Space: O(N)

 




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!