Number of moves required to guess a permutation.

Given an integer N and there is a hidden permutation (of numbers from 1 to N, each occurring exactly once) that you need to guess. You can do the following: 

Choose a number at 1st position: 

• If it is correct, you guess the next position.
• If it is wrong, the whole permutation resets and you go back to guessing the first position.

You can perform trial and error to arrive at the correct permutation, you can also use your previous knowledge for the next guesses. i.e if you know the number at first position correctly, and get 2nd position wrong, in the next move you can input the first position correctly and move on to the second position. 

Find the minimum number of moves that it would take in the worst case scenario to get the entire permutation correct.

Examples:

Input: N = 2 
Output: 3 
You choose 2 for 1st position, and the permutation resets. 
You choose 1 for 1st position, the guess is correct and now you are to guess for the 2nd position. 
You choose 2 for the 2nd position since that is the only remaining option you have.

Input: N = 3 
Output: 7 

Approach: To guess the ith position correctly, it would take (n-i) guesses. And for each guess you would need to make total of i moves( (i-1) moves to enter the correct prefix that you already know and 1 move to guess the current one). In the final step, it would take you N more moves to enter the correct permutation.

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function that returns the required moves
int countMoves(int n)
{
    int ct = 0;
    for (int i = 1; i <= n; i++)
        ct += i * (n - i);

    // Final move
    ct += n;
    return ct;
}

// Driver Program to test above function
int main()
{
    int n = 3;
    cout << countMoves(n);
    return 0;
}

Java

// Java implementation of the approach

import java.io.*;

class GFG {


// Function that returns the required moves
static int countMoves(int n)
{
    int ct = 0;
    for (int i = 1; i <= n; i++)
        ct += i * (n - i);

    // Final move
    ct += n;
    return ct;
}

// Driver Program to test above function


    public static void main (String[] args) {
        int n = 3;
    System.out.println( countMoves(n));
    }
}
// This code is contributed by anuj_67..

Python3

# Python3 implementation of the approach

# Function that returns the
# required moves
def countMoves(n):

    ct = 0
    for i in range(1, n + 1):
        ct += i * (n - i)

    # Final move
    ct += n
    return ct

# Driver Code
if __name__ == "__main__":
    n = 3
    print(countMoves(n))

# This code is contributed
# by ChitraNayal

C#

// C# implementation of the approach
using System;
class GFG {

    // Function that returns the required moves
    static int countMoves(int n)
    {
        int ct = 0;
        for (int i = 1; i <= n; i++)
            ct += i * (n - i);

        // Final move
        ct += n;
        return ct;
    }

    // Driver Program to test above function
    static void Main()
    {
        int n = 3;
        Console.WriteLine(countMoves(n));
    }

    // This code is contributed by Ryuga.

}

PHP

<?php
// PHP implementation of the approach

// Function that returns the
// required moves
function countMoves($n)
{
    $ct = 0;
    for ($i = 1; $i <= $n; $i++)
        $ct += $i * ($n - $i);

    // Final move
    $ct += $n;
    return $ct;
}

// Driver Code
$n = 3;
echo countMoves($n);

// This code is contributed
// by Akanksha Rai
?>

Javascript

<script>

// Javascript implementation of the approach

// Function that returns the required moves
function countMoves(n)
{
    let ct = 0;
    for(let i = 1; i <= n; i++)
        ct += i * (n - i);

    // Final move
    ct += n;
    return ct;
}

// Driver code
let n = 3;

document.write(countMoves(n));

// This code is contributed by Mayank Tyagi

</script>

Output:

7

Time Complexity: O(n)

Auxiliary Space: O(1)