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Maximum Strength of a Group

Medium

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ArraysGreedy Algorithms

You are given a 0-indexed integer array nums representing the score of students in an exam. The teacher would like to form one non-empty group of students with maximal strength, where the strength of a group of students of indices i₀, i₁, i₂, ... , i_k is defined as nums[i₀] * nums[i₁] * nums[i₂] * ... * nums[i_k].

Return the maximum strength of a group the teacher can create.

Example 1:

Input: nums = [3,-1,-5,2,5,-9]
Output: 1350
Explanation: One way to form a group of maximal strength is to group the students at indices [0,2,3,4,5]. Their strength is 3 * (-5) * 2 * 5 * (-9) = 1350, which we can show is optimal.

Example 2:

Input: nums = [-4,-5,-4]
Output: 20
Explanation: Group the students at indices [0, 1] . Then, we’ll have a resulting strength of 20. We cannot achieve greater strength.

Constraints:

1 <= nums.length <= 13
-9 <= nums[i] <= 9

Try coding it in LeetCode

Maximum Strength of a Group

Medium

Asked by:

6 views

Topics:

ArraysGreedy Algorithms

Return the maximum strength of a group the teacher can create.

Example 1:

Input: nums = [3,-1,-5,2,5,-9]
Output: 1350
Explanation: One way to form a group of maximal strength is to group the students at indices [0,2,3,4,5]. Their strength is 3 * (-5) * 2 * 5 * (-9) = 1350, which we can show is optimal.

Example 2:

Input: nums = [-4,-5,-4]
Output: 20
Explanation: Group the students at indices [0, 1] . Then, we’ll have a resulting strength of 20. We cannot achieve greater strength.

Constraints:

1 <= nums.length <= 13
-9 <= nums[i] <= 9

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:

To clarify the term 'maximal strength', are we looking for the numerically largest product? For instance, is a strength of 2 considered greater than -1, and is 0 considered greater than -1?
What is the expected behavior for an input like `[-1, 0]`? Should the group `[0]` with a strength of 0 be preferred over the group `[-1]` with a strength of -1?
The example `[-4,-5,-4]` suggests that with an odd number of negatives, we should exclude one to get a positive product. What if the array contains only one negative number, like `[-5]`? Is the answer -5?
What should be the output if the input array consists entirely of zeros, for example `[0, 0]`?
Given that the array can have up to 13 elements, the product could become quite large. Should I assume the result will fit within a standard 64-bit integer type, like a `long long` in C++?

Brute Force Solution

Approach

To solve this, we can use a straightforward method of trying out every single possibility. We will form every conceivable group from the given numbers, calculate the strength of each one, and keep track of the highest strength we find along the way.

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

First, create a list of every single possible group of numbers you can make from the original collection. This includes groups with just one number, groups with two numbers, and so on, all the way up to the group that includes every number.
Now, go through these possible groups one by one.
For each group, calculate its 'strength' by multiplying all the numbers in that group together.
Keep a running record of the highest strength you have seen so far. Let's call this the 'maximum strength found'.
After you calculate a group's strength, compare it to the 'maximum strength found'.
If the new group's strength is bigger, then update the 'maximum strength found' to this new, higher value.
Repeat this process for every single group you listed out at the start.
Once you have checked all possible groups, the final value of your 'maximum strength found' is the answer.

Code Implementation

def max_strength_of_group(numbers_list):
    all_possible_groups = []

    # We must generate all non-empty subsets (groups) to check every single possibility.

    def find_groups_recursive(start_index, current_group):
        if current_group:
            all_possible_groups.append(list(current_group))

        for index in range(start_index, len(numbers_list)):
            current_group.append(numbers_list[index])
            find_groups_recursive(index + 1, current_group)
            current_group.pop()

    find_groups_recursive(0, [])

    # To start, we must calculate the strength of the first group as our initial baseline maximum.

    first_group = all_possible_groups[0]
    highest_strength_seen = 1
    for number in first_group:
        highest_strength_seen *= number

    # We iterate through all other generated groups to find the true maximum strength.

    for group_index in range(1, len(all_possible_groups)):
        current_group = all_possible_groups[group_index]
        current_group_strength = 1
        for number in current_group:
            current_group_strength *= number

        # This comparison and update ensures we are always tracking the largest strength found so far.

        if current_group_strength > highest_strength_seen:
            highest_strength_seen = current_group_strength

    return highest_strength_seen

Big(O) Analysis

Time Complexity

O(n * 2^n) – The time complexity is driven by the process of generating every possible group, or subset, from the initial n numbers. For a collection of size n, there are 2^n possible subsets. For each of these subsets, the algorithm calculates its strength by multiplying all its members, which takes up to O(n) operations for a single subset. The total work involves processing 2^n subsets with up to n operations each, leading to a total runtime of approximately n * 2^n, which simplifies to O(n * 2^n).

Space Complexity

O(N * 2^N) – The dominant factor for space complexity comes from the first step: 'create a list of every single possible group'. For an input of size N, there are 2^N - 1 possible non-empty groups. Storing all these groups in memory requires an auxiliary data structure, like a list of lists, whose total size is proportional to the sum of the lengths of all subgroups, which is N * 2^(N-1). This exponential storage requirement for all generated subgroups completely overshadows the constant space used for tracking the maximum strength. Therefore, the auxiliary space usage is O(N * 2^N).

Optimal Solution

Approach

To find the maximum strength, the best approach is to build the biggest possible positive product. This involves multiplying all positive numbers together and then carefully handling the negative numbers, since pairs of negatives also create a positive product. A zero can be used as a 'safety net' to get a result of 0, which is better than any negative result.

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

First, separate your numbers into three groups: positives, negatives, and zeroes.
Multiply all of the positive numbers together. This is always a good start for building a large product.
Now, look at the negative numbers. An even number of them is great, because when multiplied together, they result in a positive number. If you have an even count, multiply all of them into your running total.
If you have an odd number of negative numbers, multiplying all of them would make the final answer negative. To avoid this, you must exclude one. To have the smallest impact, find the negative number that is closest to zero and leave it out of the calculation.
At this point, you have a candidate for the maximum strength. However, there's a special situation to consider.
If the product you calculated is negative (which can happen if you started with no positives and ended up with just one negative), check if you had any zeroes in your original collection. If so, the best answer is 0.
There's one final check: if the main strategy resulted in not picking any numbers (for example, if the original list only had zeroes and a single negative number), the answer is simply the largest single number from the original list.

Code Implementation

def maximum_strength_of_a_group(nums):
    positive_numbers = []
    negative_numbers = []
    contains_zero = False

    for number in nums:
        if number > 0:
            positive_numbers.append(number)
        elif number < 0:
            negative_numbers.append(number)
        else:
            contains_zero = True

    # This handles special cases where building a large positive product is impossible.

    if not positive_numbers and len(negative_numbers) <= 1:
        if contains_zero or not negative_numbers:
            return 0
        return negative_numbers[0]

    maximum_strength_product = 1
    for number in positive_numbers:
        maximum_strength_product *= number

    # The product of an even number of negatives is positive, maximizing strength.

    if len(negative_numbers) % 2 == 1:
        # To cause the least damage to the product, we exclude the negative number closest to zero.

        negative_numbers.sort()
        for index in range(len(negative_numbers) - 1):
            maximum_strength_product *= negative_numbers[index]
    else:
        for number in negative_numbers:
            maximum_strength_product *= number

    return maximum_strength_product

Big(O) Analysis

Time Complexity

O(n) – The time complexity is primarily driven by the need to process each of the n elements in the input array. The solution first iterates through the entire array once to categorize numbers into positive, negative, and zero groups. Subsequent operations, like calculating products or finding the largest negative number to exclude, also require at most a single pass over these sub-groups. Since these linear-time operations are performed sequentially, the total work is proportional to n, resulting in a final time complexity of O(n).

Space Complexity

O(N) – The algorithm's space complexity is determined by the step that separates numbers into three groups: positives, negatives, and zeroes. To achieve this, three auxiliary lists are created to store these partitioned numbers. The total number of elements across these three lists will be equal to N, where N is the size of the input array. Because the memory required for these lists grows linearly with the input size, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Input array with a single element, e.g., `[5]` or `[-5]`.	The solution must return the element's value, as it is the only possible non-empty group.
The best product from non-zero numbers is negative, but a zero is available in the array, such as in `[-5, 0, 0]`.	The maximum strength is 0, achieved by selecting a group containing only a zero, which is greater than any negative result.
The input contains only negative numbers with an odd count, like `[-2, -3, -5]`.	The solution must exclude the negative number closest to zero (e.g., -2) to ensure the final product is positive and maximized.
The array contains no positive numbers and at most one negative number, such as `[-5, 0]` or `[0, 0]`.	The algorithm must return 0 if a zero is present, otherwise it returns the value of the single negative number.
The input array consists entirely of zeros, for example `[0, 0, 0]`.	The maximum strength must be 0, as any non-empty group's product will inevitably be zero.
The input consists of all negative numbers with an even count, such as `[-2, -3, -4, -5]`.	The maximum strength is the product of all numbers, as an even quantity of negative factors results in a positive value.
The product of the strongest group can exceed the range of a 32-bit integer, for instance with an input of thirteen 9s.	A 64-bit integer type must be used for the variable storing the product to prevent data overflow.
The input array contains only a single negative number and no zeros, e.g., `[-5]`.	The maximum strength is the value of the negative number itself, as forming an empty group is not allowed.

Input array with a single element, e.g., `[5]` or `[-5]`.

How to Handle:

The solution must return the element's value, as it is the only possible non-empty group.

The best product from non-zero numbers is negative, but a zero is available in the array, such as in `[-5, 0, 0]`.

How to Handle:

The maximum strength is 0, achieved by selecting a group containing only a zero, which is greater than any negative result.

The input contains only negative numbers with an odd count, like `[-2, -3, -5]`.

How to Handle:

The solution must exclude the negative number closest to zero (e.g., -2) to ensure the final product is positive and maximized.

The array contains no positive numbers and at most one negative number, such as `[-5, 0]` or `[0, 0]`.

How to Handle:

The algorithm must return 0 if a zero is present, otherwise it returns the value of the single negative number.

The input array consists entirely of zeros, for example `[0, 0, 0]`.

How to Handle:

The maximum strength must be 0, as any non-empty group's product will inevitably be zero.

The input consists of all negative numbers with an even count, such as `[-2, -3, -4, -5]`.

How to Handle:

The maximum strength is the product of all numbers, as an even quantity of negative factors results in a positive value.

The product of the strongest group can exceed the range of a 32-bit integer, for instance with an input of thirteen 9s.

How to Handle:

A 64-bit integer type must be used for the variable storing the product to prevent data overflow.

The input array contains only a single negative number and no zeros, e.g., `[-5]`.

How to Handle:

The maximum strength is the value of the negative number itself, as forming an empty group is not allowed.

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:

To clarify the term 'maximal strength', are we looking for the numerically largest product? For instance, is a strength of 2 considered greater than -1, and is 0 considered greater than -1?
What is the expected behavior for an input like `[-1, 0]`? Should the group `[0]` with a strength of 0 be preferred over the group `[-1]` with a strength of -1?
The example `[-4,-5,-4]` suggests that with an odd number of negatives, we should exclude one to get a positive product. What if the array contains only one negative number, like `[-5]`? Is the answer -5?
What should be the output if the input array consists entirely of zeros, for example `[0, 0]`?
Given that the array can have up to 13 elements, the product could become quite large. Should I assume the result will fit within a standard 64-bit integer type, like a `long long` in C++?

Brute Force Solution

Approach

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

First, create a list of every single possible group of numbers you can make from the original collection. This includes groups with just one number, groups with two numbers, and so on, all the way up to the group that includes every number.
Now, go through these possible groups one by one.
For each group, calculate its 'strength' by multiplying all the numbers in that group together.
Keep a running record of the highest strength you have seen so far. Let's call this the 'maximum strength found'.
After you calculate a group's strength, compare it to the 'maximum strength found'.
If the new group's strength is bigger, then update the 'maximum strength found' to this new, higher value.
Repeat this process for every single group you listed out at the start.
Once you have checked all possible groups, the final value of your 'maximum strength found' is the answer.

Code Implementation

def max_strength_of_group(numbers_list):
    all_possible_groups = []

    # We must generate all non-empty subsets (groups) to check every single possibility.

    def find_groups_recursive(start_index, current_group):
        if current_group:
            all_possible_groups.append(list(current_group))

        for index in range(start_index, len(numbers_list)):
            current_group.append(numbers_list[index])
            find_groups_recursive(index + 1, current_group)
            current_group.pop()

    find_groups_recursive(0, [])

    # To start, we must calculate the strength of the first group as our initial baseline maximum.

    first_group = all_possible_groups[0]
    highest_strength_seen = 1
    for number in first_group:
        highest_strength_seen *= number

    # We iterate through all other generated groups to find the true maximum strength.

    for group_index in range(1, len(all_possible_groups)):
        current_group = all_possible_groups[group_index]
        current_group_strength = 1
        for number in current_group:
            current_group_strength *= number

        # This comparison and update ensures we are always tracking the largest strength found so far.

        if current_group_strength > highest_strength_seen:
            highest_strength_seen = current_group_strength

    return highest_strength_seen

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, separate your numbers into three groups: positives, negatives, and zeroes.
Multiply all of the positive numbers together. This is always a good start for building a large product.
Now, look at the negative numbers. An even number of them is great, because when multiplied together, they result in a positive number. If you have an even count, multiply all of them into your running total.
If you have an odd number of negative numbers, multiplying all of them would make the final answer negative. To avoid this, you must exclude one. To have the smallest impact, find the negative number that is closest to zero and leave it out of the calculation.
At this point, you have a candidate for the maximum strength. However, there's a special situation to consider.
If the product you calculated is negative (which can happen if you started with no positives and ended up with just one negative), check if you had any zeroes in your original collection. If so, the best answer is 0.
There's one final check: if the main strategy resulted in not picking any numbers (for example, if the original list only had zeroes and a single negative number), the answer is simply the largest single number from the original list.

Code Implementation

def maximum_strength_of_a_group(nums):
    positive_numbers = []
    negative_numbers = []
    contains_zero = False

    for number in nums:
        if number > 0:
            positive_numbers.append(number)
        elif number < 0:
            negative_numbers.append(number)
        else:
            contains_zero = True

    # This handles special cases where building a large positive product is impossible.

    if not positive_numbers and len(negative_numbers) <= 1:
        if contains_zero or not negative_numbers:
            return 0
        return negative_numbers[0]

    maximum_strength_product = 1
    for number in positive_numbers:
        maximum_strength_product *= number

    # The product of an even number of negatives is positive, maximizing strength.

    if len(negative_numbers) % 2 == 1:
        # To cause the least damage to the product, we exclude the negative number closest to zero.

        negative_numbers.sort()
        for index in range(len(negative_numbers) - 1):
            maximum_strength_product *= negative_numbers[index]
    else:
        for number in negative_numbers:
            maximum_strength_product *= number

    return maximum_strength_product

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input array with a single element, e.g., `[5]` or `[-5]`.	The solution must return the element's value, as it is the only possible non-empty group.
The best product from non-zero numbers is negative, but a zero is available in the array, such as in `[-5, 0, 0]`.	The maximum strength is 0, achieved by selecting a group containing only a zero, which is greater than any negative result.
The input contains only negative numbers with an odd count, like `[-2, -3, -5]`.	The solution must exclude the negative number closest to zero (e.g., -2) to ensure the final product is positive and maximized.
The array contains no positive numbers and at most one negative number, such as `[-5, 0]` or `[0, 0]`.	The algorithm must return 0 if a zero is present, otherwise it returns the value of the single negative number.
The input array consists entirely of zeros, for example `[0, 0, 0]`.	The maximum strength must be 0, as any non-empty group's product will inevitably be zero.
The input consists of all negative numbers with an even count, such as `[-2, -3, -4, -5]`.	The maximum strength is the product of all numbers, as an even quantity of negative factors results in a positive value.
The product of the strongest group can exceed the range of a 32-bit integer, for instance with an input of thirteen 9s.	A 64-bit integer type must be used for the variable storing the product to prevent data overflow.
The input array contains only a single negative number and no zeros, e.g., `[-5]`.	The maximum strength is the value of the negative number itself, as forming an empty group is not allowed.

Input array with a single element, e.g., `[5]` or `[-5]`.

How to Handle:

The solution must return the element's value, as it is the only possible non-empty group.

The best product from non-zero numbers is negative, but a zero is available in the array, such as in `[-5, 0, 0]`.

How to Handle:

The maximum strength is 0, achieved by selecting a group containing only a zero, which is greater than any negative result.

The input contains only negative numbers with an odd count, like `[-2, -3, -5]`.

How to Handle:

The solution must exclude the negative number closest to zero (e.g., -2) to ensure the final product is positive and maximized.

The array contains no positive numbers and at most one negative number, such as `[-5, 0]` or `[0, 0]`.

How to Handle:

The algorithm must return 0 if a zero is present, otherwise it returns the value of the single negative number.

The input array consists entirely of zeros, for example `[0, 0, 0]`.

How to Handle:

The maximum strength must be 0, as any non-empty group's product will inevitably be zero.

The input consists of all negative numbers with an even count, such as `[-2, -3, -4, -5]`.

How to Handle:

The maximum strength is the product of all numbers, as an even quantity of negative factors results in a positive value.

The product of the strongest group can exceed the range of a 32-bit integer, for instance with an input of thirteen 9s.

How to Handle:

A 64-bit integer type must be used for the variable storing the product to prevent data overflow.

The input array contains only a single negative number and no zeros, e.g., `[-5]`.

How to Handle:

The maximum strength is the value of the negative number itself, as forming an empty group is not allowed.