# Minimize product of first N – 1 natural numbers by swapping same positioned bits of pairs

Given an integer N, the task is to find the minimum positive product of first N – 1 natural numbers, i.e. [1, (N – 1)], by swapping any ith bit of any two numbers any number of times.

Note: N is always a perfect power of 2. Since the product can be very large, print the answer modulo 109 + 7.

Examples:

Input: N = 4
Output: 6
Explanation:
No swapping of bits is required. Therefore, the minimum product is 1*2*3 = 6.

Input: N = 8
Output: 1512
Explanation:
Let the array arr[] stores all the value from 1 to N as {1, 2, 3, 4, 5, 6, 7}
Step 1: In elements 2 = (0010) and 5 = (0101), swap 0th and 1st bit. Therefore, replace 2 with 1 and 5 with 6. arr[] = {1, 1, 3, 4, 6, 6, 7}.
Step 2: In elements 3 = (0011) and 4 = (0100), swap 1th bit. Therefore, replace 3 with 1 and 4 with 6. arr[] = {1, 1, 1, 6, 6, 6, 7}.
Hence, the minimum product = 1*1*1*6*6*6*7 = 1512 % 1e9+7 = 1512.

Approach: The idea is to make some observations. For example, if N = 8 and arr[] = {1, 2, 3, 4, 5, 6, 7}, observe that for the product to be minimum there must be three sixes i.e., there must be an element having value (N – 2) with the frequency of occurrence as (1 + (N – 4)/2) and there must be three ones i.e., there must be (1 + (N – 4)/2) ones. And at last multiply the current product with (N – 1). Hence, the formula becomes:

Minimum product for any value N = ((N – 1) * (N – 2)(N – 4)/2 + 1) % 1e9 + 7

Follow the below steps to solve the problem:

1. Initialize the ans as 1.
2. Iterate over the range [0, 1 + (N – 4)/2].
3. In each traversal, multiply ans with N – 2 and update the ans to ans mod 1e9+7.
4. After the above steps, print the value of ans*(N – 1) mod 1e9+7 as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `int` `mod = 1e9 + 7;`   `// Function to find the minimum product` `// of 1 to N - 1 after performing the` `// given operations` `void` `minProduct(``int` `n)` `{` `    ``// Initialize ans with 1` `    ``int` `ans = 1;`   `    ``// Multiply ans with N-2` `    ``// ((N - 4)/2) times` `    ``for` `(``int` `i = 1;` `         ``i <= (n - 4) / 2; i++) {` `        ``ans = (1LL * ans` `               ``* (n - 2))` `              ``% mod;` `    ``}`   `    ``// Multiply ans with N - 1` `    ``// and N - 2 once` `    ``ans = (1LL * ans` `           ``* (n - 2) * (n - 1))` `          ``% mod;`   `    ``// Print ans` `    ``cout << ans << endl;` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given Number N` `    ``int` `N = 8;`   `    ``// Function Call` `    ``minProduct(N);`   `    ``return` `0;` `}`

## Java

 `// Java program for the ` `// above approach` `import` `java.util.*;` `class` `GFG{`   `static` `int` `mod = (``int``)1e9 + ``7``;`   `// Function to find the ` `// minimum product of 1 ` `// to N - 1 after performing ` `// the given operations` `static` `void` `minProduct(``int` `n)` `{` `  ``// Initialize ans with 1` `  ``int` `ans = ``1``;`   `  ``// Multiply ans with N-2` `  ``// ((N - 4)/2) times` `  ``for` `(``int` `i = ``1``;` `           ``i <= (n - ``4``) / ``2``; i++) ` `  ``{` `    ``ans = (``int``)(1L * ans * ` `               ``(n - ``2``)) % mod;` `  ``}`   `  ``// Multiply ans with N - 1` `  ``// and N - 2 once` `  ``ans = (``int``)(1L * ans * ` `             ``(n - ``2``) * (n - ``1``)) % mod;`   `  ``// Print ans` `  ``System.out.print(ans + ``"\n"``);` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `  ``// Given Number N` `  ``int` `N = ``8``;`   `  ``// Function Call` `  ``minProduct(N);` `}` `}`   `// This code is contributed by gauravrajput1`

## Python3

 `# Python3 program for the above approach` `mod ``=` `1e9` `+` `7`   `# Function to find the minimum product` `# of 1 to N - 1 after performing the` `# given operations` `def` `minProduct(n):` `    `  `    ``# Initialize ans with 1` `    ``ans ``=` `1`   `    ``# Multiply ans with N-2` `    ``# ((N - 4)/2) times` `    ``for` `i ``in` `range``(``1``, (n ``-` `4``) ``/``/` `2` `+` `1``):` `        ``ans ``=` `(ans ``*` `(n ``-` `2``)) ``%` `mod`   `    ``# Multiply ans with N - 1` `    ``# and N - 2 once` `    ``ans ``=` `(ans ``*` `(n ``-` `2``) ``*` `(n ``-` `1``)) ``%` `mod`   `    ``# Print ans` `    ``print``(``int``(ans))`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``# Given number N` `    ``N ``=` `8`   `    ``# Function call` `    ``minProduct(N)`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the ` `// above approach` `using` `System;` `class` `GFG{`   `static` `int` `mod = (``int``)1e9 + 7;`   `// Function to find the ` `// minimum product of 1 ` `// to N - 1 after performing ` `// the given operations` `static` `void` `minProduct(``int` `n)` `{` `  ``// Initialize ans with 1` `  ``int` `ans = 1;`   `  ``// Multiply ans with N-2` `  ``// ((N - 4)/2) times` `  ``for` `(``int` `i = 1;` `           ``i <= (n - 4) / 2; i++) ` `  ``{` `    ``ans = (``int``)(1L * ans * ` `               ``(n - 2)) % mod;` `  ``}`   `  ``// Multiply ans with N - 1` `  ``// and N - 2 once` `  ``ans = (``int``)(1L * ans * ` `             ``(n - 2) * ` `             ``(n - 1)) % mod;`   `  ``// Print ans` `  ``Console.Write(ans + ``"\n"``);` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `  ``// Given Number N` `  ``int` `N = 8;`   `  ``// Function Call` `  ``minProduct(N);` `}` `}`   `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output

`1512`

Time Complexity: O(N) where N is the given integer.
Auxiliary Space: O(1)

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