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}
Follow the below steps:
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:
- Initialize the ans as 1.
- Iterate over the range [0, 1 + (N – 4)/2].
- In each traversal, multiply ans with N – 2 and update the ans to ans mod 1e9+7.
- 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 <bits/stdc++.h>using namespace std;int mod = 1e9 + 7;// Function to find the minimum product// of 1 to N - 1 after performing the// given operationsvoid 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 Codeint main(){ // Given Number N int N = 8; // Function Call minProduct(N); return 0;} |
Java
// Java program for the// above approachimport 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 operationsstatic 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 Codepublic 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 approachmod = 1e9 + 7# Function to find the minimum product# of 1 to N - 1 after performing the# given operationsdef 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 Codeif __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 approachusing System;class GFG{static int mod = (int)1e9 + 7;// Function to find the// minimum product of 1// to N - 1 after performing// the given operationsstatic 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 Codepublic static void Main(String[] args){ // Given Number N int N = 8; // Function Call minProduct(N);}}// This code is contributed by Rajput-Ji |
Javascript
<script>// JavaScript program for the above approachlet mod = 1e9 + 7;// Function to find the minimum product// of 1 to N - 1 after performing the// given operationsfunction minProduct(n){ // Initialize ans with 1 let ans = 1; // Multiply ans with N-2 // ((N - 4)/2) times for (let i = 1; i <= Math.floor((n - 4) / 2); i++) { ans = (1 * ans * (n - 2)) % mod; } // Multiply ans with N - 1 // and N - 2 once ans = (1 * ans * (n - 2) * (n - 1)) % mod; // Print ans document.write(ans + "<br>");}// Driver Code // Given Number N let N = 8; // Function Call minProduct(N);// This code is contributed by Manoj.</script> |
1512
Time Complexity: O(N) where N is the given integer.
Auxiliary Space: O(1)
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