Formatted question description: https://leetcode.ca/all/1969.html

# 1969. Minimum Non-Zero Product of the Array Elements

Medium

## Description

You are given a positive integer p. Consider an array nums (1-indexed) that consists of the integers in the inclusive range [1, 2^p - 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 10^9 + 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

The maximum element is 2^p - 1. To make the product minimum, the elements should have the greatest difference possible. That is, for the remaining elements, half of the elements are 1 and half of the elements are 2^p - 2. Then calculate the product.

class Solution {
static final long MODULO = 1000000007;

public int minNonZeroProduct(int p) {
long maxNum = (1L << p) - 1;
long remain = maxNum - 1;
long pairs = remain / 2;
long minProduct = quickMul(remain, pairs) * (maxNum % MODULO) % MODULO;
return (int) (minProduct % MODULO);
}

public long quickMul(long x, long n) {
x %= MODULO;
long power = 1;
while (n > 0) {
if (n % 2 == 1)
power = power * x % MODULO;
x = x * x % MODULO;
n /= 2;
}
return power;
}
}