Package g1901_2000.s1969_minimum_non_zero_product_of_the_array_elements

Class Solution

java.lang.Object
g1901_2000.s1969_minimum_non_zero_product_of_the_array_elements.Solution
public class Solution extends Object
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 `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**_ 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`

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • minNonZeroProduct

      public int minNonZeroProduct(int p)