Lexicographically smallest permutation of {1, .. n} such that no. and position do not match

Given a positive integer n, find the lexicographically smallest permutation p of {1, 2, .. n} such that p_i != i. i.e., i should not be there at i-th position where i varies from 1 to n.

Examples:

Input : 5
Output : 2 1 4 5 3
Consider the two permutations that follow
the requirement that position and numbers
should not be same.
p = (2, 1, 4, 5, 3) and q = (2, 4, 1, 5, 3).  
Since p is lexicographically smaller, our 
output is p.

Input  : 6
Output : 2 1 4 3 6 5

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

Since we need lexicographically smallest (and 1 cannot come at position 1), we put 2 at first position. After 2, we put the next smallest element i.e., 1. After that the next smallest considering it does not violates our condition of pi != i.
Now, if our n is even we simply take two variables one which will contain our count of even numbers and one which will contain our count of odd numbers and then we will keep them adding in the vector till we reach n.
But, if our n is odd, we do the same task till we reach n-1 because if we add till n then in the end we will end up having p_i = i. So when we reach n-1, we first add n to the position n-1 and then on position n we will put n-2.
The implementation of the above program is given below.

C++

// C++ implementation of the above approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to print the permutation 
void findPermutation(vector<int> a, int n) 
{ 
    vector<int> res;    
  
    // Initial numbers to be pushed to result 
    int en = 2, on = 1;  
  
    // If n is even 
    if (n % 2 == 0) { 
        for (int i = 0; i < n; i++) { 
            if (i % 2 == 0) { 
                res.push_back(en); 
                en += 2; 
            } else { 
                res.push_back(on); 
                on += 2; 
            } 
        } 
    }  
  
    // If n is odd 
    else { 
        for (int i = 0; i < n - 2; i++) { 
            if (i % 2 == 0) { 
                res.push_back(en); 
                en += 2; 
            } else { 
                res.push_back(on); 
                on += 2; 
            } 
        } 
        res.push_back(n); 
        res.push_back(n - 2); 
    } 
  
    // Print result 
    for (int i = 0; i < n; i++)  
        cout << res[i] << " ";     
    cout << "\n"; 
} 
  
// Driver Code 
int main() 
{ 
    long long int n = 9; 
    findPermutation(n); 
    return 0; 
} 

Java

// Java implementation of the above approach 
import java.util.Vector; 
  
class GFG { 
  
// Function to print the permutation 
    static void findPermutation(int n) { 
        Vector<Integer> res = new Vector<Integer>(); 
  
        // Initial numbers to be pushed to result 
        int en = 2, on = 1; 
  
        // If n is even 
        if (n % 2 == 0) { 
            for (int i = 0; i < n; i++) { 
                if (i % 2 == 0) { 
                    res.add(en); 
                    en += 2; 
                } else { 
                    res.add(on); 
                    on += 2; 
                } 
            } 
        } // If n is odd 
        else { 
            for (int i = 0; i < n - 2; i++) { 
                if (i % 2 == 0) { 
                    res.add(en); 
                    en += 2; 
                } else { 
                    res.add(on); 
                    on += 2; 
                } 
            } 
            res.add(n); 
            res.add(n - 2); 
        } 
  
        // Print result 
        for (int i = 0; i < n; i++) { 
            System.out.print(res.get(i) + " "); 
        } 
        System.out.println(""); 
    } 
  
// Driver Code 
    public static void main(String[] args) { 
        int n = 9; 
        findPermutation(n); 
    } 
} 
// This code is contributed by PrinciRaj1992 

Python3

# Python3 implementation of the above approach 
  
# Function to print the permutation  
def findPermutation(n) : 
  
    res = [] 
  
    # Initial numbers to be pushed to result  
    en, on = 2, 1
  
    # If n is even  
    if (n % 2 == 0) : 
        for i in range(n) :  
            if (i % 2 == 0) :  
                res.append(en) 
                en += 2 
            else : 
                res.append(on)  
                on += 2 
           
  
    # If n is odd  
    else :  
        for i in range(n-2) : 
            if (i % 2 == 0) :  
                res.append(en)  
                en += 2
            else :  
                res.append(on) 
                on += 2
              
           
        res.append(n) 
        res.append(n - 2)  
       
  
    # Print result  
    for i in range(n) : 
        print(res[i] ,end = " ")      
    print()  
  
  
# Driver Code  
if __name__ == "__main__" : 
   
    n = 9;  
    findPermutation(n)  
  
# This code is contributed by Ryuga 

C#

// C# implementation of the above approach  
using System; 
using System.Collections; 
public class GFG {  
  
// Function to print the permutation  
    static void findPermutation(int n) {  
        ArrayList res = new ArrayList();  
  
        // Initial numbers to be pushed to result  
        int en = 2, on = 1;  
  
        // If n is even  
        if (n % 2 == 0) {  
            for (int i = 0; i < n; i++) {  
                if (i % 2 == 0) {  
                    res.Add(en);  
                    en += 2;  
                } else {  
                    res.Add(on);  
                    on += 2;  
                }  
            }  
        } // If n is odd  
        else {  
            for (int i = 0; i < n - 2; i++) {  
                if (i % 2 == 0) {  
                    res.Add(en);  
                    en += 2;  
                } else {  
                    res.Add(on);  
                    on += 2;  
                }  
            }  
            res.Add(n);  
            res.Add(n - 2);  
        }  
  
        // Print result  
        for (int i = 0; i < n; i++) {  
            Console.Write(res[i] + " ");  
        }  
        Console.WriteLine("");  
    }  
  
// Driver Code  
    public static void Main() {  
        int n = 9;  
        findPermutation(n);  
    }  
}  
// This code is contributed by 29AjayKumar 

PHP

<?php  
// PHP implementation of the above approach 
  
// Function to print the permutation 
function findPermutation($n) 
{ 
    $res = array();  
  
    // Initial numbers to be pushed  
    // to result 
    $en = 2; 
    $on = 1;  
  
    // If n is even 
    if ($n % 2 == 0) 
    { 
        for ($i = 0; $i < $n; $i++)  
        { 
            if (i % 2 == 0)  
            { 
                array_push($res, $en); 
                $en += 2; 
            } else 
            { 
                array_push($res, $on); 
                $on += 2; 
            } 
        } 
    }  
  
    // If n is odd 
    else
    { 
        for ($i = 0; $i < $n - 2; $i++)  
        { 
            if ($i % 2 == 0)  
            { 
                array_push($res, $en); 
                $en += 2; 
            }  
            else 
            { 
                array_push($res, $on); 
                $on += 2; 
            } 
        } 
        array_push($res, $n); 
        array_push($res, $n - 2); 
    } 
  
    // Print result 
    for ($i = 0; $i < $n; $i++)  
        echo $res[$i] . " ";  
    echo "\n"; 
} 
  
// Driver Code 
$n = 9; 
findPermutation($n); 
  
// This code is contributed by ita_c 
?> 

Output:

2 1 4 3 6 5 8 9 7

Time Complexity: O(n)

This article is contributed by Sarthak Kohli. 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.

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

My Personal Notes

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

C++

Java

Python3

C#

PHP

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