Evaluate the expression ( N1 * (N – 1)2 * … * 1N) % (109 + 7)

Given an integer N, the task is to find the value of the expression ( N¹ * (N – 1)² * … * 1^N) % (10⁹ + 7).

Input: N = 1 
Output: 1 
Explanation:
1¹ = 1

Input: N = 4 
Output: 288 
Explanation:
4¹ * (4 – 1)² * (4 – 2)³ * (4-3)⁴
= 4 * 9 * 8 * 1 
= 288

Naive Approach: The simplest approach to solve this problem is to iterate over the range [1, N]. For every i^th iteration, calculate the value of (N – i + 1)ⁱ. Finally, print the product of all the calculated values from each iteration. 
Time Complexity: O(N² * log₂(N))
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following observations:

F(N) = N¹ * (N – 1)² * … * 1^N
= N * (N – 1) * (N – 1) * (N – 2) * (N – 2) * (N – 2)* … 1 * 1 * 1 
= N * (N – 1) * (N – 2)*… * 1 * (N -1) * (N – 2) * …* 1 * … 
= N! * (N – 1)! * (N – 2)! * … * 1!

Follow the steps below to solve the problem:

• Precompute the value of the factorial from 1 to N using factorial(N) = N * factorial(N – 1).
• Iterate over the range [1, N] and find the product of all the factorials over the range [1, N] using the above observations
• Finally, print the value of the expression.

C++

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;
#define mod 1000000007

// Function to find the value of the expression
// ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
int ValOfTheExpression(int n)
{

    // factorial[i]: Stores factorial of i
    int factorial[n] = { 0 };

    // Base Case for factorial
    factorial[0] = factorial[1] = 1;

    // Precompute the factorial
    for (int i = 2; i <= n; i++) {
        factorial[i] = ((factorial[i - 1] % mod)
                        * (i % mod))
                       % mod;
    }

    // dp[N]: Stores the value of the expression
    // ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
    int dp[n] = { 0 };
    dp[1] = 1;

    for (int i = 2; i <= n; i++) {

        // Update dp[i]
        dp[i] = ((dp[i - 1] % mod)
                 * (factorial[i] % mod))
                % mod;
    }

    // Return the answer.
    return dp[n];
}

// Driver Code
int main()
{

    int n = 4;
    // Function call
    cout << ValOfTheExpression(n) << "\n";
}

Java

// Java program to implement
// the above approach
class GFG
{
  static int mod = 1000000007;

  // Function to find the value of the expression
  // ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
  static int ValOfTheExpression(int n)
  {

    // factorial[i]: Stores factorial of i
    int[] factorial = new int[n + 1];

    // Base Case for factorial
    factorial[0] = factorial[1] = 1;

    // Precompute the factorial
    for (int i = 2; i <= n; i++)
    {
      factorial[i] = ((factorial[i - 1] % mod)
                      * (i % mod)) % mod;
    }

    // dp[N]: Stores the value of the expression
    // ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
    int[] dp = new int[n + 1];
    dp[1] = 1;

    for (int i = 2; i <= n; i++)
    {

      // Update dp[i]
      dp[i] = ((dp[i - 1] % mod)
               * (factorial[i] % mod)) % mod;
    }

    // Return the answer.
    return dp[n];
  }

  // Driver code
  public static void main(String[] args) {
    int n = 4;

    // Function call
    System.out.println(ValOfTheExpression(n));
  }
}

// This code is contributed by divyesh072019

Python3

# Python 3 program to implement
# the above approach
mod = 1000000007

# Function to find the value of the expression
# ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
def ValOfTheExpression(n):
    global mod

    # factorial[i]: Stores factorial of i
    factorial = [0 for i in range(n + 1)]

    # Base Case for factorial
    factorial[0] = 1
    factorial[1] = 1

    # Precompute the factorial
    for i in range(2, n + 1, 1):
        factorial[i] = ((factorial[i - 1] % mod) * (i % mod))%mod

    # dp[N]: Stores the value of the expression
    # ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
    dp = [0 for i in range(n+1)]
    dp[1] = 1
    for i in range(2, n + 1, 1):

        # Update dp[i]
        dp[i] = ((dp[i - 1] % mod)*(factorial[i] % mod)) % mod

    # Return the answer.
    return dp[n]

# Driver Code
if __name__ == '__main__':
    n = 4

    # Function call
    print(ValOfTheExpression(n))

    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# program to implement
// the above approach
using System;
class GFG
{

  static int mod = 1000000007;

  // Function to find the value of the expression
  // ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
  static int ValOfTheExpression(int n)
  {

    // factorial[i]: Stores factorial of i
    int[] factorial = new int[n + 1];

    // Base Case for factorial
    factorial[0] = factorial[1] = 1;

    // Precompute the factorial
    for (int i = 2; i <= n; i++)
    {
      factorial[i] = ((factorial[i - 1] % mod)
                      * (i % mod))
        % mod;
    }

    // dp[N]: Stores the value of the expression
    // ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
    int[] dp = new int[n + 1];
    dp[1] = 1;

    for (int i = 2; i <= n; i++)
    {

      // Update dp[i]
      dp[i] = ((dp[i - 1] % mod)
               * (factorial[i] % mod))
        % mod;
    }

    // Return the answer.
    return dp[n];
  }

  // Driver code
  static void Main()
  {
    int n = 4;

    // Function call
    Console.WriteLine(ValOfTheExpression(n));
  }
}

// This code is contributed by divyeshrabadiya07

Javascript

<script>

    // Javascript program to implement
    // the above approach

    let mod = 1000000007;

    // Function to find the value of the expression
    // ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
    function ValOfTheExpression(n)
    {

      // factorial[i]: Stores factorial of i
      let factorial = new Array(n + 1);

      // Base Case for factorial
      factorial[0] = factorial[1] = 1;

      // Precompute the factorial
      for (let i = 2; i <= n; i++)
      {
        factorial[i] = ((factorial[i - 1] % mod) *
                        (i % mod)) % mod;
      }

      // dp[N]: Stores the value of the expression
      // ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
      let dp = new Array(n + 1);
      dp[1] = 1;

      for (let i = 2; i <= n; i++)
      {

        // Update dp[i]
        dp[i] = ((dp[i - 1] % mod) *
                 (factorial[i] % mod)) % mod;
      }

      // Return the answer.
      return dp[n];
    }

    let n = 4;

    // Function call
    document.write(ValOfTheExpression(n));

</script>

Output:

288

Time Complexity: O(N )
Auxiliary Space: O(N)