# Lexicographically smallest permutation with no digits at Original Index

Given an integer N. The task is to find the lexicographically smallest permutation of integer of the form: 12345…N such that no digit occurs at the index as in the original number, i.e. if P1P2P3…PN is our permutation then Pi must not be equal to i.

Note : N is greater than 1 and less than 10.

Examples :

```Input : N = 5
Output : 21435

Input : N = 2
Output : 21
```

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

For smallest permutation, the smaller digits should be placed in starting. So, there are two cases to handle this problem.

1. N is even, i.e. number of digits is even. In such case, if all odd digits got placed to next even index and all even digits got placed to their preceding indexes, we will have the smallest permutation satisfying the above condition.
2. N is odd, i.e. the number of digits is odd. In this, all are similar to above case only change is that the last three digits are shuffled in a way such that their permutation is smallest. For example, if we have 123 as the last three digits then 231 is the smallest possible permutation.

Algorithm :

```If N is even:place all even digits (upto N) in increasing order at odd index.place all odd digits in increasing order at even index.elseplace all even digits (upto N-3) in increasing order at odd index.place all odd digits (upto N-4) in increasing order at even index.Place N at (N-1)th place, N-1 at (N-2)th and N-2 at Nth place.
```

Below is the implementation of the above approach:

## C++

 `// C++ program to find the smallest permutaion ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to print the smallest permutation ` `string smallestPermute(``int` `n) ` `{ ` `    ``char` `res[n + 1]; ` ` `  `    ``// when n is even ` `    ``if` `(n % 2 == 0) { ` `        ``for` `(``int` `i = 0; i < n; i++) { ` `            ``if` `(i % 2 == 0) ` `                ``res[i] = 48 + i + 2; ` `            ``else` `                ``res[i] = 48 + i; ` `        ``} ` `    ``} ` ` `  `    ``// when n is odd ` `    ``else` `{ ` `        ``for` `(``int` `i = 0; i < n - 2; i++) { ` `            ``if` `(i % 2 == 0) ` `                ``res[i] = 48 + i + 2; ` `            ``else` `                ``res[i] = 48 + i; ` `        ``} ` `        ``// handling last 3 digit ` `        ``res[n - 1] = 48 + n - 2; ` `        ``res[n - 2] = 48 + n; ` `        ``res[n - 3] = 48 + n - 1; ` `    ``} ` ` `  `    ``// add EOL and print result ` `    ``res[n] = ``'\0'``; ` ` `  `    ``return` `res; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 7; ` ` `  `    ``cout << smallestPermute(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find the smallest permutaion ` `class` `GFG ` `{ ` ` `  `// Function to print the smallest permutation ` `static` `void` `smallestPermute(``int` `n) ` `{ ` `    ``char` `res[] = ``new` `char``[n + ``1``]; ` ` `  `    ``// when n is even ` `    ``if` `(n % ``2` `== ``0``) { ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{ ` `            ``if` `(i % ``2` `== ``0``) ` `                ``res[i] = (``char``)(``48` `+ i + ``2``); ` `            ``else` `                ``res[i] = (``char``)(``48` `+ i); ` `        ``} ` `    ``} ` ` `  `    ``// when n is odd ` `    ``else`  `    ``{ ` `        ``for` `(``int` `i = ``0``; i < n - ``2``; i++)  ` `        ``{ ` `            ``if` `(i % ``2` `== ``0``) ` `                ``res[i] = (``char``)(``48` `+ i + ``2``); ` `            ``else` `                ``res[i] = (``char``)(``48` `+ i); ` `        ``} ` `        ``// handling last 3 digit ` `        ``res[n - ``1``] = (``char``)(``48` `+ n - ``2``); ` `        ``res[n - ``2``] = (``char``)(``48` `+ n); ` `        ``res[n - ``3``] = (``char``)(``48` `+ n - ``1``); ` `    ``} ` ` `  `    ``// add EOL and print result ` `    ``res[n] = ``'\0'``; ` ` `  `    ``for` `(``int` `i = ``0``; i < n ; i++) ` `    ``{ ` `        ``System.out.print(res[i]); ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String []args) ` `{ ` `    ``int` `n = ``7``; ` ` `  `    ``smallestPermute(n); ` `} ` `} ` ` `  `// This code is contributed by ANKITRAI1 `

## Python 3

 `# Python 3 program to find the  ` `# smallest permutaion ` ` `  `# Function to print the smallest ` `# permutation ` `def` `smallestPermute( n): ` ` `  `    ``res ``=` `[""] ``*` `(n ``+` `1``) ` `     `  `    ``# when n is even ` `    ``if` `(n ``%` `2` `=``=` `0``) : ` `        ``for` `i ``in` `range``(n): ` `            ``if` `(i ``%` `2` `=``=` `0``): ` `                ``res[i] ``=` `chr``(``48` `+` `i ``+` `2``) ` `            ``else``: ` `                ``res[i] ``=` `chr``(``48` `+` `i) ` ` `  `    ``# when n is odd ` `    ``else` `: ` `        ``for` `i ``in` `range``(n ``-` `2` `): ` `            ``if` `(i ``%` `2` `=``=` `0``): ` `                ``res[i] ``=` `chr``(``48` `+` `i ``+` `2``) ` `            ``else``: ` `                ``res[i] ``=` `chr``(``48` `+` `i) ` `         `  `        ``# handling last 3 digit ` `        ``res[n ``-` `1``] ``=` `chr``(``48` `+` `n ``-` `2``) ` `        ``res[n ``-` `2``] ``=` `chr``(``48` `+` `n) ` `        ``res[n ``-` `3``] ``=` `chr``(``48` `+` `n ``-` `1``) ` ` `  `    ``# add EOL and print result ` `    ``res ``=` `''.join(res) ` `    ``return` `res ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `     `  `    ``n ``=` `7` `    ``print``(smallestPermute(n)) ` ` `  `# This code is contributed by ita_c `

## C#

 `// C# program to find the smallest  ` `// permutaion ` `using` `System; ` `class` `GFG ` `{ ` ` `  `// Function to print the smallest  ` `// permutation ` `static` `void` `smallestPermute(``int` `n) ` `{ ` `    ``char``[] res = ``new` `char``[n + 1]; ` ` `  `    ``// when n is even ` `    ``if` `(n % 2 == 0)  ` `    ``{ ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{ ` `            ``if` `(i % 2 == 0) ` `                ``res[i] = (``char``)(48 + i + 2); ` `            ``else` `                ``res[i] = (``char``)(48 + i); ` `        ``} ` `    ``} ` ` `  `    ``// when n is odd ` `    ``else` `    ``{ ` `        ``for` `(``int` `i = 0; i < n - 2; i++)  ` `        ``{ ` `            ``if` `(i % 2 == 0) ` `                ``res[i] = (``char``)(48 + i + 2); ` `            ``else` `                ``res[i] = (``char``)(48 + i); ` `        ``} ` `        ``// handling last 3 digit ` `        ``res[n - 1] = (``char``)(48 + n - 2); ` `        ``res[n - 2] = (``char``)(48 + n); ` `        ``res[n - 3] = (``char``)(48 + n - 1); ` `    ``} ` ` `  `    ``// add EOL and print result ` `    ``res[n] = ``'\0'``; ` ` `  `    ``for` `(``int` `i = 0; i < n ; i++) ` `    ``{ ` `        ``Console.Write(res[i]); ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `n = 7; ` ` `  `    ``smallestPermute(n); ` `} ` `} ` ` `  `// This code is contributed  ` `// by Akanksha Rai `

## PHP

 ` `

Output:

```2143675
```

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