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Minimize product of first 2^K–1 Natural Numbers by swapping bits for any pair any number of times

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  • Last Updated : 18 Oct, 2021
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Given a positive integer K, the task is to minimize the positive product of the first (2K – 1) Natural Numbers by swapping the bits at the corresponding position of any two numbers any number of times.

Examples:

Input: K = 3
Output: 1512
Explanation: The original product is 5040. The given array in binary notation is {001, 010, 011, 100, 101, 110, 111}

  • In the first operation swap the leftmost bit of the second and fifth elements. The resulting array is [001, 110, 011, 100, 001, 110, 111].
  • In the second operation swap the middle bit of the third and fourth elements. The resulting array is [001, 110, 001, 110, 001, 110, 111].

After the above operations, the array elements are {1, 6, 1, 6, 1, 6, 7}. Therefore, the product is 1 * 6 * 1 * 6 * 1 * 6 * 7 = 1512, which is the minimum possible positive product.

Input: K = 2
Output: 6
Explanation: The original permutation in binary notation is {’00’, ’01’, ’10’}. Any swap will lead to product zero or does not change at all. Hence 6 is the correct output.

Approach: The given problem can be solved based on the observation that to get the minimum positive product the frequency of the number 1 should be maximum by swapping the bits of any two numbers. Follow the steps below to solve the given problem:

  • Find the value of the range as (2K – 1).
  • Convert range/2 elements to 1 by shifting all bits of odd numbers to even numbers except the 0th bit.
  • Therefore, range/2 numbers will be 1 and range/2 numbers will be range – 1, and the last number in the array will remain the same as the value of the range.
  • Find the resultant product of all the numbers formed in the above step as the resultant minimum positive product obtained.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum positive
// product after swapping bits at the
// similar position for any two numbers
int minimumPossibleProduct(int K)
{
    // Stores the resultant number
    int res = 1;
 
    int range = (1 << K) - 1;
 
    // range/2 numbers will be 1 and
    // range/2 numbers will be range-1
    for (int i = 0; i < K; i++) {
        res *= (range - 1);
    }
 
    // Multiply with the final number
    res *= range;
 
    // Return the answer
    return res;
}
// Driver Code
int main()
{
    int K = 3;
    cout << minimumPossibleProduct(K);
 
    return 0;
}

Java




// Java program for the above approach
 
import java.io.*;
 
class GFG {
   
// Function to find the minimum positive
// product after swapping bits at the
// similar position for any two numbers
static int minimumPossibleProduct(int K)
{
    // Stores the resultant number
    int res = 1;
 
    int range = (1 << K) - 1;
 
    // range/2 numbers will be 1 and
    // range/2 numbers will be range-1
    for (int i = 0; i < K; i++) {
        res *= (range - 1);
    }
 
    // Multiply with the final number
    res *= range;
 
    // Return the answer
    return res;
}
   
// Driver Code
public static void main (String[] args) {
   
    int K = 3;
    System.out.println(minimumPossibleProduct(K));
}
}
 
// This code is contributed by Dharanendra L V.

Python3




# python program for the above approach
 
# Function to find the minimum positive
# product after swapping bits at the
# similar position for any two numbers
def minimumPossibleProduct(K):
   
    # Stores the resultant number
    res = 1
    r = (1 << K) - 1
     
    # range/2 numbers will be 1 and
    # range/2 numbers will be range-1
    for i in range(0, K):
        res *= (r - 1)
 
    # Multiply with the final number
    res *= r
 
    # Return the answer
    return res
 
# Driver Code
K = 3
print(minimumPossibleProduct(K))
 
# This code is contributed by amreshkumar3.

C#




// C# program for the above approach
using System;
 
class GFG {
 
    // Function to find the minimum positive
    // product after swapping bits at the
    // similar position for any two numbers
    static int minimumPossibleProduct(int K)
    {
       
        // Stores the resultant number
        int res = 1;
 
        int range = (1 << K) - 1;
 
        // range/2 numbers will be 1 and
        // range/2 numbers will be range-1
        for (int i = 0; i < K; i++) {
            res *= (range - 1);
        }
 
        // Multiply with the final number
        res *= range;
 
        // Return the answer
        return res;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
 
        int K = 3;
        Console.WriteLine(minimumPossibleProduct(K));
    }
}
 
// This code is contributed by ukasp.

Javascript




<script>
        // JavaScript Program to implement
        // the above approach
 
        // Function to find the minimum positive
        // product after swapping bits at the
        // similar position for any two numbers
        function minimumPossibleProduct(K)
        {
         
            // Stores the resultant number
            let res = 1;
 
            let range = (1 << K) - 1;
 
            // range/2 numbers will be 1 and
            // range/2 numbers will be range-1
            for (let i = 0; i < K; i++) {
                res *= (range - 1);
            }
 
            // Multiply with the final number
            res *= range;
 
            // Return the answer
            return res;
        }
         
        // Driver Code
 
        let K = 3;
        document.write(minimumPossibleProduct(K));
 
     // This code is contributed by Potta Lokesh
 
    </script>

Output: 

1512

 

Time Complexity: O(K)
Auxiliary Space: O(1)


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