Formatted question description: https://leetcode.ca/all/1969.html
1969. Minimum Non-Zero Product of the Array Elements
Level
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
andy
fromnums
. - Choose a bit in
x
and swap it with its corresponding bit iny
. 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;
}
}