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Smallest Derangement of Sequence

Given the sequence   S = {1, 2, 3 dots N}  find the lexicographically smallest (earliest in dictionary order) derangement of  S  .

A derangement of S is as any permutation of S such that no two elements in S and its permutation occur at same position.

Examples:

Input: 3
Output : 2 3 1
Explanation: The Sequence is 1 2 3.
Possible permutations are (1, 2, 3), (1, 3, 2),
          (2, 1, 3), (2, 3, 1), (3, 1, 2) (3, 2, 1).
Derangements are (2, 3, 1), (3, 1, 2).
Smallest Derangement: (2, 3, 1)

Input : 5
Output : 2 1 4 5 3.
Explanation: Out of all the permutations of 
(1, 2, 3, 4, 5), (2, 1, 4, 5, 3) is the first derangement.

Method 1:
We can modify the method shown in this article: Largest Derangement
Using a min heap we can successively get the least element and place them in more significant positions, taking care that the property of derangement is maintained.
Complexity: O(N * log N)

Below is the C++ implementation.

// CPP program to generate smallest derangement
// using priority queue.
#include <bits/stdc++.h>
using namespace std;
  
void generate_derangement(int N)
{
    // Generate Sequence and insert into a 
    // priority queue.
    int S[N + 1];
    priority_queue<int, vector<int>, greater<int> > PQ;
    for (int i = 1; i <= N; i++) {
        S[i] = i;
        PQ.push(S[i]);
    }
  
    // Generate Least Derangement
    int D[N + 1];
    for (int i = 1; i <= N; i++) {
        int d = PQ.top();
        PQ.pop();
        if (d != S[i] || i == N) {
            D[i] = d;
        }
        else {
            D[i] = PQ.top();
            PQ.pop();
            PQ.push(d);
        }
    }
  
    if (D[N] == S[N])
       swap(S[N], D[N]);
  
    // Print Derangement
    for (int i = 1; i <= N; i++) 
        printf("%d ", D[i]);    
    printf(" ");
}
  
// Driver code
int main()
{
    generate_derangement(10);
    return 0;
}

Output:

2 1 4 3 6 5 8 7 10 9 

Method 2
Since we are given a very specific sequence i.e  S_i = i   forall i <= N we can calculate the answer even more efficiently.

Divide the original sequence into pairs of two elements, and then swap the elements of each pair.

If N is odd then the last pair needs to be swapped again.

Pictorial Representation

Complexity: We perform at most N/2 + 1 swaps, so the complexity is O(N).

Why does this method work
This method is a very specific application of method 1 and is based on observation. Given the nature of sequence, at position i we already know the least element that can be put, which is either i+1 or i-1. Since we are already given the least permutation of S it is clear that the derangement must start from 2 and not 1 ie of the form i+1 (i = 1). The next element will be of the form i – 1 . The element after this will be i + 1 and then next i – 1. This pattern will continue until the end.

This operation is most easily understood as the swapping of adjacent elements of pairs of elements of S.

If we can determine the least element in constant time, then the complexity overhead from heap is eliminated. Hence from O(N * log N) the complexity reduces to O(N).

Below is the implementation of above approach:



C++

// Efficient C++ program to find smallest
// derangement.
#include <stdio.h>
  
void generate_derangement(int N)
{        
    // Generate Sequence S
    int S[N + 1];
    for (int i = 1; i <= N; i++)
        S[i] = i;
  
    // Generate Derangement
    int D[N + 1];
    for (int i = 1; i <= N; i += 2) {
        if (i == N) {
  
            // Only if i is odd
            // Swap S[N-1] and S[N]
            D[N] = S[N - 1];
            D[N - 1] = S[N];
        }
        else {
            D[i] = i + 1;
            D[i + 1] = i;
        }
    }
  
    // Print Derangement
    for (int i = 1; i <= N; i++) 
        printf("%d ", D[i]);    
    printf(" ");
}
  
// Driver Program
int main()
{
    generate_derangement(10);
    return 0;
}

Python3

# Efficient Python3 program to find 
# smallest derangement. 
  
def generate_derangement(N):
      
    # Generate Sequence S 
    S = [0] * (N + 1)
    for i in range(1, N + 1):
        S[i] =
  
    # Generate Derangement 
    D = [0] * (N + 1)
    for i in range(1, N + 1, 2):
        if i == N: 
  
            # Only if i is odd 
            # Swap S[N-1] and S[N] 
            D[N] = S[N - 1
            D[N - 1] = S[N]
        else
            D[i] = i + 1
            D[i + 1] = i
  
    # Print Derangement 
    for i in range(1, N + 1):
        print(D[i], end = " ")
    print()
  
# Driver Code 
if __name__ == '__main__':
    generate_derangement(10)
      
# This code is contributed by PranchalK

C#

// Efficient C# program to find 
// smallest derangement.
using System;
  
class GFG
{
      
static void generate_derangement(int N)
    // Generate Sequence S
    int[] S = new int[N + 1];
    for (int i = 1; i <= N; i++)
        S[i] = i;
  
    // Generate Derangement
    int[] D = new int[N + 1];
    for (int i = 1; i <= N; i += 2) 
    {
        if (i == N) 
        {
  
            // Only if i is odd
            // Swap S[N-1] and S[N]
            D[N] = S[N - 1];
            D[N - 1] = S[N];
        }
        else
        {
            D[i] = i + 1;
            D[i + 1] = i;
        }
    }
  
    // Print Derangement
    for (int i = 1; i <= N; i++) 
        Console.Write(D[i] + " "); 
    Console.WriteLine();
}
  
// Driver Program
public static void Main()
{
    generate_derangement(10);
}
}
  
// This code is contributed 
// by Akanksha Rai

PHP

<?php
// Efficient PHP program to find smallest
// derangement.
  
function generate_derangement($N)
{     
    // Generate Sequence S
    $S = array();
    for ($i = 1; $i <= $N; $i++)
        $S[$i] = $i;
  
    // Generate Derangement
    $D = array();
    for ($i = 1; $i <= $N; $i += 2)
    {
        if ($i == $N)
        {
  
            // Only if i is odd
            // Swap S[N-1] and S[N]
            $D[$N] = $S[$N - 1];
            $D[$N - 1] = $S[$N];
        }
        else
        {
            $D[$i] = $i + 1;
            $D[$i + 1] = $i;
        }
    }
  
    // Print Derangement
    for ($i = 1; $i <= $N; $i++) 
        echo $D[$i] . " "
    echo " ";
}
  
// Driver Program
generate_derangement(10);
  
// This code is contributed
// by Akanksha Rai
?>


Output:

2 1 4 3 6 5 8 7 10 9 

Note : The auxiliary space can be reduced to O(1) if we perform the swapping operations on S itself.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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