Minimizing Array After Replacing Pairs With Their Product

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ArraysGreedy AlgorithmsDynamic Programming

You are given a 0-indexed array nums of length n containing non-negative integers.

You are allowed to perform the following operation any number of times:

Choose two different indices i and j, such that nums[i] > 0 and nums[j] > 0.
Replace nums[i] and nums[j] with nums[i] * nums[j].

Your task is to minimize the length of the array nums after performing the operation as many times as you want.

Return an integer representing the minimum length of the array after performing the operation as many times as you want.

Example 1:

Input: nums = [1,2,0,3,0]
Output: 2
Explanation: We can perform the following operations:
- Choose indices i = 1 and j = 3. Now, nums[1] becomes nums[1] * nums[3] = 2 * 3 = 6, and nums[3] becomes 6.
- Choose indices i = 1 and j = 4. Now, nums[1] becomes nums[1] * nums[4] = 6 * 0 = 0, and nums[4] becomes 0.
After all operations, the array nums is [1,0,0,0,0].
The length of the array is 5, but we have 3 elements that are equal to 0. We can ignore them and consider the array as [1,6].
It can be proven that the length of the array after performing operations as many times as you want cannot be less than 2.

Example 2:

Input: nums = [0,2,1,0,0]
Output: 1
Explanation: We can perform the following operations:
- Choose indices i = 1 and j = 2. Now, nums[1] becomes nums[1] * nums[2] = 2 * 1 = 2, and nums[2] becomes 2.
After all operations, the array nums is [0,2,2,0,0].
The length of the array is 5, but we have 3 elements that are equal to 0. We can ignore them and consider the array as [2,2].
It can be proven that the length of the array after performing operations as many times as you want cannot be less than 1.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁹

Try coding it in LeetCode

Minimizing Array After Replacing Pairs With Their Product

Medium

Asked by:

1 view

Topics:

ArraysGreedy AlgorithmsDynamic Programming

You are given a 0-indexed array nums of length n containing non-negative integers.

You are allowed to perform the following operation any number of times:

Choose two different indices i and j, such that nums[i] > 0 and nums[j] > 0.
Replace nums[i] and nums[j] with nums[i] * nums[j].

Your task is to minimize the length of the array nums after performing the operation as many times as you want.

Return an integer representing the minimum length of the array after performing the operation as many times as you want.

Example 1:

Input: nums = [1,2,0,3,0]
Output: 2
Explanation: We can perform the following operations:
- Choose indices i = 1 and j = 3. Now, nums[1] becomes nums[1] * nums[3] = 2 * 3 = 6, and nums[3] becomes 6.
- Choose indices i = 1 and j = 4. Now, nums[1] becomes nums[1] * nums[4] = 6 * 0 = 0, and nums[4] becomes 0.
After all operations, the array nums is [1,0,0,0,0].
The length of the array is 5, but we have 3 elements that are equal to 0. We can ignore them and consider the array as [1,6].
It can be proven that the length of the array after performing operations as many times as you want cannot be less than 2.

Example 2:

Input: nums = [0,2,1,0,0]
Output: 1
Explanation: We can perform the following operations:
- Choose indices i = 1 and j = 2. Now, nums[1] becomes nums[1] * nums[2] = 2 * 1 = 2, and nums[2] becomes 2.
After all operations, the array nums is [0,2,2,0,0].
The length of the array is 5, but we have 3 elements that are equal to 0. We can ignore them and consider the array as [2,2].
It can be proven that the length of the array after performing operations as many times as you want cannot be less than 1.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁹

Try coding it in LeetCode

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 size of the input array `nums`?
Are all elements in the array positive integers, or can they be zero or negative?
If the input array is empty, what should I return?
Can you provide an example where the optimal solution requires multiple operations?
Are there any constraints on the maximum value of an element in the array, or the maximum product that might result from the operation?

Brute Force Solution

Approach

The brute force method to minimize an array after product replacements involves trying out every single possible pair replacement and seeing what happens. We keep doing this until we can't replace any more pairs, and then remember the smallest resulting array we ever saw. This is like testing every combination, no matter how inefficient.

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

Start with the original array.
Consider all possible pairs of numbers in the array.
For each pair, imagine replacing those two numbers with their product, creating a new array.
If the new array is shorter than any array we've seen so far, remember its length.
Repeat the process of looking at all pairs and replacing them with their product for this new array, again keeping track of the shortest array length.
Keep doing this until there are no more pairs we can multiply together to make a smaller array. This means all possible combinations of pair replacements have been exhausted.
Finally, pick the smallest array length we encountered during all the replacements. This will be our answer.

Code Implementation

def minimize_array_brute_force(numbers):
    smallest_sum = float('inf')

    # Iterate through all possible pairs of indices
    for first_index in range(len(numbers)):
        for second_index in range(first_index + 1, len(numbers)):

            # Create a copy of the array to modify
            modified_numbers = numbers[:]

            # Replace the pair with their product
            product = modified_numbers[first_index] * modified_numbers[second_index]
            del modified_numbers[second_index]
            del modified_numbers[first_index]
            modified_numbers.insert(first_index, product)

            # Calculate the sum of the modified array
            current_sum = sum(modified_numbers)

            # Update the smallest sum if needed
            if current_sum < smallest_sum:
                smallest_sum = current_sum

    return smallest_sum

Big(O) Analysis

Time Complexity

O(n!) – The brute force approach involves exploring all possible combinations of pair replacements in the array. In the worst-case scenario, we might consider every possible pair replacement until the array is reduced to a single element or no further replacements are possible. The number of such combinations grows factorially with the input size n, as each replacement creates a new state and subsequent replacements depend on these states. This results in the exploration of a decision tree where the number of branches explodes with increasing n. Thus, the time complexity is proportional to the number of possible combinations, leading to O(n!).

Space Complexity

O(N!) – The brute force approach, as described, involves exploring every possible combination of pair replacements. Each potential combination leads to a new array, and in the worst-case scenario, we might need to store a significant portion of these intermediate arrays to track the minimum length encountered. The number of possible combinations can grow factorially with the input size N, where N is the number of elements in the array. Consequently, the auxiliary space required to store these intermediate arrays and track the minimum length becomes proportional to O(N!).

Optimal Solution

Approach

The goal is to repeatedly combine pairs of numbers from a list to get the smallest possible final number. The key idea is to prioritize combining smaller numbers early on, because multiplication makes numbers bigger.

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

First, arrange the numbers from smallest to largest.
Take the two smallest numbers and multiply them together. This creates a new number.
Put the new number back into the list, keeping the list sorted from smallest to largest.
Repeat the process: take the two smallest numbers (which might now include the number you just created), multiply them, and put the result back in the sorted list.
Keep doing this until you only have one number left. That's the smallest possible final result you can achieve.

Code Implementation

def minimize_array_product(input_array):
    mutable_array = input_array[:]

    while len(mutable_array) > 1:
        # Find indices of the two smallest elements.
        first_smallest_index = mutable_array.index(min(mutable_array))
        first_smallest_number = mutable_array.pop(first_smallest_index)

        second_smallest_index = mutable_array.index(min(mutable_array))
        second_smallest_number = mutable_array.pop(second_smallest_index)

        # Replace with the product
        product = first_smallest_number * second_smallest_number
        mutable_array.append(product)

    return mutable_array[0]

Big(O) Analysis

Time Complexity

O(n^2 log n) – We start with n numbers. In each iteration, we extract the two smallest numbers which takes O(1) time with a heap, but involves re-inserting the product back into the sorted list. Re-inserting an element into a sorted list of size 'k' takes O(k) time if we linearly search for the right position, making each iteration O(n). Since we repeat this n-1 times, the overall complexity would seem like O(n^2). However, if we use a heap (priority queue) for sorting, extracting the two smallest is O(1), multiplying and re-inserting will take O(log n), and repeating this n-1 times makes the overall time complexity O(n log n). But, after each multiplication, the array may not be already sorted. Sorting the entire array after each step takes O(n log n). Since we iterate almost n times, the overall time complexity for sorting would be O(n * n log n), simplified to O(n^2 log n).

Space Complexity

O(N) – The plain English explanation details repeatedly inserting a new element back into a sorted list. While the prompt doesn't explicitly state which sorting algorithm is used, inserting into a sorted list generally requires shifting existing elements to accommodate the new value. If the list is implemented using a standard array, this insertion would necessitate creating a new array to hold the updated list, potentially of size N in the worst case where N represents the number of elements. This repeated array creation and copying contributes O(N) auxiliary space. The overall space complexity is therefore O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 if the input array is null or empty, as no operations are possible.
Array with a single element	Return 1 if the array has only one element, as no operations are possible.
Array with two elements	Return 1 if the product of the two elements is computed; otherwise, the array remains with a length of 2.
Array with many small numbers that can quickly result in very large products causing integer overflow	Use a data type with sufficient capacity (e.g., long) to handle potential overflow or consider using an arbitrary-precision arithmetic library.
Array with all elements equal to 1	Repeatedly multiply adjacent 1s to reduce the array length until only one element is left.
Array where multiplying any two numbers will make the final length larger than initial length (e.g. large numbers)	Greedy approach with dynamic programming will solve this scenario by finding a correct set of merging operations to minimize the length.
Maximum size input array (scalability)	The solution should be optimized (e.g., using dynamic programming) to handle large input sizes without exceeding time or memory limits.
A mixed array of small and large numbers that would cause overflow when multiplying pairs.	Use `long` or appropriate data type to handle potentially large intermediate products before applying the merge operations.