Minimize product of first N – 1 natural numbers by swapping same positioned bits of pairs
Last Updated :
12 Apr, 2021
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++
#include <bits/stdc++.h>
using namespace std;
int mod = 1e9 + 7;
void minProduct( int n)
{
int ans = 1;
for ( int i = 1;
i <= (n - 4) / 2; i++) {
ans = (1LL * ans
* (n - 2))
% mod;
}
ans = (1LL * ans
* (n - 2) * (n - 1))
% mod;
cout << ans << endl;
}
int main()
{
int N = 8;
minProduct(N);
return 0;
}
|
Java
import java.util.*;
class GFG{
static int mod = ( int )1e9 + 7 ;
static void minProduct( int n)
{
int ans = 1 ;
for ( int i = 1 ;
i <= (n - 4 ) / 2 ; i++)
{
ans = ( int )(1L * ans *
(n - 2 )) % mod;
}
ans = ( int )(1L * ans *
(n - 2 ) * (n - 1 )) % mod;
System.out.print(ans + "\n" );
}
public static void main(String[] args)
{
int N = 8 ;
minProduct(N);
}
}
|
Python3
mod = 1e9 + 7
def minProduct(n):
ans = 1
for i in range ( 1 , (n - 4 ) / / 2 + 1 ):
ans = (ans * (n - 2 )) % mod
ans = (ans * (n - 2 ) * (n - 1 )) % mod
print ( int (ans))
if __name__ = = '__main__' :
N = 8
minProduct(N)
|
C#
using System;
class GFG{
static int mod = ( int )1e9 + 7;
static void minProduct( int n)
{
int ans = 1;
for ( int i = 1;
i <= (n - 4) / 2; i++)
{
ans = ( int )(1L * ans *
(n - 2)) % mod;
}
ans = ( int )(1L * ans *
(n - 2) *
(n - 1)) % mod;
Console.Write(ans + "\n" );
}
public static void Main(String[] args)
{
int N = 8;
minProduct(N);
}
}
|
Javascript
<script>
let mod = 1e9 + 7;
function minProduct(n)
{
let ans = 1;
for (let i = 1; i <= Math.floor((n - 4) / 2); i++)
{
ans = (1 * ans * (n - 2)) % mod;
}
ans = (1 * ans * (n - 2) * (n - 1)) % mod;
document.write(ans + "<br>" );
}
let N = 8;
minProduct(N);
</script>
|
Time Complexity: O(N) where N is the given integer.
Auxiliary Space: O(1)
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