# 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 pi != 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 pi = 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 ` `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 res = ``new` `Vector(); ` ` `  `        ``// 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

 ` `

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.

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.

My Personal Notes arrow_drop_up

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.