- Difficulty: Medium
- Tags: LeetCode, Medium, Greedy, Recursion, Math, leetcode-1969, O(min(p, logM)), O(1)
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
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 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