# 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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