LeetCode-in-Java
1969. Minimum Non-Zero Product of the Array Elements
Medium
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
xandyfromnums. - Choose a bit in
xand 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 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
public 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;
}
}