1969. Minimum Non-Zero Product of the Array Elements

Problem

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 **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 (Java)

class Solution {
    public int minNonZeroProduct(int p) {
        int m = (int) (1e9 + 7);
        long n = ((1L << p) - 2) % m;
        long ans = n + 1;
        long cnt = (1L << (p - 1)) - 1;
        while (cnt > 0) {
            if ((cnt & 1) == 1) {
                ans = ans * n % m;
            }
            cnt >>= 1;
            n = n * n % m;
        }
        return (int) ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).