LeetCode Problem Workspace
Minimum Non-Zero Product of the Array Elements
The problem asks to minimize the product of an array after performing bit-swapping operations on its binary representations.
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Practice Focus
Medium · Greedy choice plus invariant validation
Answer-first summary
The problem asks to minimize the product of an array after performing bit-swapping operations on its binary representations.
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The problem involves minimizing the product of an array of integers in binary form. It can be solved by swapping bits between numbers to reach the smallest non-zero product, modulo 10^9 + 7. The correct approach ensures elements are minimized optimally with a greedy strategy.
Problem Statement
You are given a positive integer p. Consider an array nums of integers in the inclusive range [1, 2^p - 1] in their binary representations. You can swap any bits between the elements in the array any number of times. The goal is to minimize the product of all elements in the array, ensuring the product is non-zero. Return the product modulo 10^9 + 7.
For example, given p = 1, the array consists of a single number, 1. For p = 2, the array is [1, 2, 3], and any swaps either keep the product the same or make it 0. With p = 3, the array consists of [1, 2, 3, 4, 5, 6, 7], and swaps are performed to minimize the product.
Examples
Example 1
Input: p = 1
Output: 1
nums = [1]. There is only one element, so the product equals that element.
Example 2
Input: p = 2
Output: 6
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
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 Approach
Greedy Approach for Minimizing Product
To minimize the product, a greedy strategy can be used. Swap the bits between elements to ensure the smaller elements remain in positions where their multiplication is less impactful. The key observation is that minimizing each number optimally reduces the overall product.
Validating Invariant During Swaps
While swapping bits, it's crucial to ensure that the invariant (non-zero product) is maintained. This validation ensures that we do not accidentally make the product zero while still minimizing the array elements effectively.
Modulo Arithmetic to Avoid Overflow
Since the product can grow very large, use modulo 10^9 + 7 throughout the computation to prevent overflow and ensure the result is within bounds as specified by the problem statement.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time and space complexity depend on the approach chosen. A greedy approach, involving bit swapping and invariant validation, might lead to a solution with complexity based on the number of elements in the array, typically O(p^2) for the worst case. Using modulo arithmetic ensures we never exceed bounds.
What Interviewers Usually Probe
- Tests understanding of greedy strategies and invariant validation.
- Evaluates knowledge of bit manipulation and modulo arithmetic.
- Assesses the ability to optimize mathematical problems through simple operations.
Common Pitfalls or Variants
Common pitfalls
- Failing to maintain the invariant and inadvertently setting the product to zero.
- Mismanaging bit-swapping operations, leading to incorrect results.
- Not applying modulo arithmetic, resulting in overflow or incorrect answers.
Follow-up variants
- Allowing a larger range for
p. - Testing with arrays that contain duplicates or a more diverse range of binary numbers.
- Modifying the operation to include swaps between non-adjacent elements.
FAQ
How do greedy strategies help solve the Minimum Non-Zero Product of the Array Elements?
Greedy strategies help by ensuring the array elements are minimized as much as possible through bit swaps, thus reducing the product effectively.
What is the time complexity of solving this problem?
The time complexity depends on the chosen approach but could typically be O(p^2) for a greedy solution that checks each pair of elements.
Why do we need to validate the invariant during swaps?
Validating the invariant ensures that we don't accidentally create a product of zero, which would invalidate the problem's conditions.
What is the significance of modulo 10^9 + 7 in this problem?
The modulo 10^9 + 7 ensures that the result remains within the required bounds and prevents overflow when dealing with large products.
What is the pattern used in the Minimum Non-Zero Product of the Array Elements?
The problem utilizes a greedy choice strategy combined with invariant validation to minimize the product effectively.
Solution
Solution 1: Greedy + Fast Power
We notice that each operation does not change the sum of the elements. When the sum of the elements remains unchanged, to minimize the product, we should maximize the difference between the elements as much as possible.
class Solution:
def minNonZeroProduct(self, p: int) -> int:
mod = 10**9 + 7
return (2**p - 1) * pow(2**p - 2, 2 ** (p - 1) - 1, mod) % modContinue Practicing
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