Minimum Non-Zero Product of the Array Elements

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Bit ManipulationGreedy Algorithms

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 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 2^nd 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

Minimum Non-Zero Product of the Array Elements

Medium

PayPal

0 views

Topics:

Bit ManipulationGreedy Algorithms

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 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 2^nd 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

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum value of `p`? In other words, what is the range of values for the elements in the array?
Can `p` be zero or negative? If `p` can be zero, should I return a specific value, or is that an invalid input?
For very large values of `p`, can you confirm the expected behavior regarding integer overflow in intermediate calculations, and the required return type (e.g., 64-bit integer or similar)?
Are there any specific constraints on the time complexity for calculating the minimum non-zero product?
When p=1, should I return 1, since the original array is [1] and the product is already the minimum non-zero product?

Brute Force Solution

Approach

We need to find the smallest possible non-zero product we can make from a set of numbers. A brute force approach means checking every possible combination of changes we can make to the numbers to try and minimize the final product.

Here's how the algorithm would work step-by-step:

Start with the initial set of numbers.
Consider making one number smaller by one and another number larger by one, making sure not to make any number zero.
Calculate the product of all the numbers after the change.
Repeat this process of adjusting pairs of numbers and calculating the product many, many times, trying all possible combinations of adjustments.
Keep track of the smallest non-zero product you find during this process.
After exploring all feasible adjustments, the smallest product you've found is the answer.

Code Implementation

def minimum_non_zero_product_brute_force(numbers):
    min_product = float('inf')

    # Iterate through all possible adjustments (very inefficient)
    for i in range(2**len(numbers)):
        temp_numbers = list(numbers)
        adjustments = []

        binary_representation = bin(i)[2:].zfill(len(numbers))
        for index, bit in enumerate(binary_representation):
            if bit == '1':
                adjustments.append(index)

        # Apply adjustments to create a new set of numbers
        if len(adjustments) % 2 == 0 and len(adjustments) > 0:
            for j in range(0, len(adjustments), 2):
                first_index = adjustments[j]
                second_index = adjustments[j+1]
                
                # Avoid setting a number to zero during adjustment
                if temp_numbers[first_index] > 1:
                    temp_numbers[first_index] -= 1
                    temp_numbers[second_index] += 1

            product = 1
            for number in temp_numbers:
                product *= number

            # Update minimum product
            if product > 0:
                min_product = min(min_product, product)

    #This deals with the edge case where there are no adjustments
    product = 1
    for number in numbers:
        product *= number
    min_product = min(min_product, product)
    return min_product

Big(O) Analysis

Time Complexity

O(n^k) – The proposed solution involves exploring all possible adjustments of pairs of numbers in the array. The number of possible pairs is O(n^2). However, the approach states that this process is repeated 'many, many times' implying that the number of iterations is not constant. The key driver of cost is the depth 'k' to which we explore all possible combinations of adjustments (making one number smaller and another larger). Since there are approximately n^2 pairs that can be adjusted, and 'k' such adjustments are made in the worst case, the algorithm complexity becomes O(n^k) because for each number of pair combinations, the algorithm may have to recalculate the product after 'k' adjustments. In the worst-case scenario, the number of potential combinations to check scales exponentially with 'n' to the power 'k'.

Space Complexity

O(1) – The algorithm, as described, primarily involves modifying numbers in place and calculating products. It explicitly keeps track of only the smallest product found so far. Therefore, it utilizes only a few variables to store indices and the minimum product, irrespective of the input size N. The auxiliary space used remains constant, independent of the input array's length.

Optimal Solution

Approach

The key to minimizing the product is recognizing the power of pairing. We want to pair large numbers with small numbers to bring them closer to 1, except for one special case. This drastically reduces the overall product.

Here's how the algorithm would work step-by-step:

Notice that we're trying to make the product as small as possible, and that we can only decrease the product by moving values closer to 1.
Pair the largest possible number with the smallest possible number, repeatedly. Each pair becomes a number close to 1. In other words, pair the largest number with 1, the second largest number with 2, and so on.
Realize that we have to avoid making any element zero, but we want to make elements as close to 1 as possible. Save a '1' to pair later.
Change one number to '1' in such a way as to preserve the product, and assign the previous '1' value to a larger number.
Pair the remaining original value of '1' with the largest number in the array to further minimize the product.
Multiply all of the resulting values together. Make sure to account for potentially very large values and use a method that avoids integer overflow.

Code Implementation

def minimum_non_zero_product(power_of_two):
    modulo = 10**9 + 7
    max_number = (1 << power_of_two) - 1

    # We pair the largest with smallest to get close to 1.
    pairs_count = max_number // 2
    result = 1

    # Calculates (max_number - 1) ^ pairs_count (mod modulo)
    base_value = max_number - 1
    exponent = pairs_count

    while exponent > 0:
        if exponent % 2 == 1:
            result = (result * base_value) % modulo

        base_value = (base_value * base_value) % modulo
        exponent //= 2

    # Multiply by the largest number, the remaining unpaired number.
    result = (result * max_number) % modulo

    return result

Big(O) Analysis

Time Complexity

O(n) – The described algorithm focuses on pairing and mathematical manipulations rather than iterative array traversals. Calculating powers using modular exponentiation dominates the runtime. This power calculation happens 'n/2' times (the number of pairs). Modular exponentiation can be done in O(log m) time where m is the modulus. However, since the modulus is constant (10^9 + 7), each modular exponentiation is effectively O(1). Therefore, calculating n/2 powers takes O(n) time, which dominates any constant time arithmetic operations, resulting in a total time complexity of O(n).

Space Complexity

O(1) – The provided steps outline a series of pairing and manipulation operations. These operations primarily involve arithmetic calculations and do not require any auxiliary data structures that scale with the input. Specifically, there is no mention of storing intermediate results in a list, hash map, or any other data structure whose size depends on the input. Thus, the algorithm operates in constant space, independent of the input size.

Edge Cases

Case	How to Handle
n = 1	Return 1, since the array will be [1].
n = 2	The array will be [1, 2], and the minimum non-zero product is 2.
n = 3	The array will be [1, 2, 3]. We can swap 2 and 1 to get [2, 1, 3]. The product is 6.
Integer overflow for large n	Use modular arithmetic with a large prime number (10^9 + 7) to avoid overflow during multiplication.
n = large number, but all elements are ones.	Ensure algorithm is efficient, especially modular exponentiation.
n = 0	If n is zero, raise error or return 0, depending on problem specification.
Maximum possible value of n	Ensure the modular exponentiation algorithm can handle the largest possible value of pow(n-1, (2^(n-1) - 1)).
Modulus is one	If mod is 1, return 0 since anything modulo 1 is zero.