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Minimum Non-Zero Product of the Array Elements

You are given a positive integer p. Consider an array nums (1-indexed) that consists of the integers in the inclusive range [1, 2p - 1] in their binary representations. You are allowed to do the following operation any number of times:

  • Choose two elements x and y from nums.
  • Choose a bit in x and swap it with its corresponding bit in y. Corresponding bit refers to the bit that is in the same position in the other integer.

For example, if x = 1101 and y = 0011, after swapping the 2nd bit from the right, we have x = 1111 and y = 0001.

Find the minimum non-zero product of nums after performing the above operation any number of times. Return this product modulo 109 + 7.

Note: The answer should be the minimum product before the modulo operation is done.

 

Example 1:

Input: p = 1
Output: 1
Explanation: nums = [1].
There is only one element, so the product equals that element.

Example 2:

Input: p = 2
Output: 6
Explanation: nums = [01, 10, 11].
Any swap would either make the product 0 or stay the same.
Thus, the array product of 1 * 2 * 3 = 6 is already minimized.

Example 3:

Input: p = 3
Output: 1512
Explanation: nums = [001, 010, 011, 100, 101, 110, 111]
- In the first operation we can 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 we can swap the middle bit of the third and fourth elements.
    - The resulting array is [001, 110, 001, 110, 001, 110, 111].
The array product is 1 * 6 * 1 * 6 * 1 * 6 * 7 = 1512, which is the minimum possible product.

 

Constraints:

  • 1 <= p <= 60

Solution Explanation for Minimum Non-Zero Product of the Array Elements

This problem asks to find the minimum non-zero product of an array nums containing integers from 1 to 2p - 1, after performing any number of bit swaps between pairs of elements.

Core Idea: The key observation is that to minimize the product without making it zero, we should aim to have as many smaller numbers as possible. We can achieve this by leveraging the bitwise operations. Instead of performing individual swaps, we can strategically manipulate the bits to achieve the desired distribution of numbers.

Optimal Strategy:

  1. Pair up numbers: We pair numbers symmetrically around the middle. For example, if p=3, we have numbers [1, 2, 3, 4, 5, 6, 7]. We pair (1,6), (2,5), (3,4). The number 7 (2p-1) remains unpaired.

  2. Bit manipulation (implicit): The problem allows us to swap corresponding bits between any two numbers. This means we can effectively redistribute the set bits across the numbers. The optimal distribution is to make as many numbers equal to 1 as possible while still having a non-zero product. By pairing, we end up with several 1s and several 2<sup>p</sup>-2s.

  3. Calculate the product: The resulting product will be (2p - 1) * (2p - 2)(2p-1 - 1). The exponent (2p-1 - 1) represents the number of pairs.

Time and Space Complexity:

  • Time Complexity: O(p). The dominant operation is calculating the power using fast exponentiation (using pow function or similar optimized techniques), which has a logarithmic time complexity O(log(2p-1-1)) approximately O(p).

  • Space Complexity: O(1). The algorithm uses a constant amount of extra space regardless of the input size.

Code Implementation (Python):

class Solution:
    def minNonZeroProduct(self, p: int) -> int:
        mod = 10**9 + 7
        return (pow(2, p, mod) - 1) * pow(pow(2, p, mod) - 2, pow(2, p - 1, mod) - 1, mod) % mod

Explanation of Python Code:

  1. mod = 10**9 + 7: This line sets the modulo value to prevent integer overflow.

  2. return (pow(2, p, mod) - 1) * pow(pow(2, p, mod) - 2, pow(2, p - 1, mod) - 1, mod) % mod: This line calculates the minimum non-zero product using the formula derived above and applies the modulo operation at each step to prevent integer overflow. The pow(a, b, mod) function calculates (ab) mod mod efficiently.

The code in other languages (Java, C++, Go, TypeScript) follows the same logic, adapting the syntax and built-in functions as needed for efficient modular exponentiation. The key is to use a fast exponentiation method (often provided in standard libraries) to calculate the powers efficiently.