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Array Partition

Easy

Google

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Topics:

ArraysGreedy Algorithms

Given an integer array nums of 2n integers, group these integers into n pairs (a1, b1), (a2, b2), ..., (an, bn) such that the sum of min(ai, bi) for all i is maximized. Return the maximized sum.

For example:

nums = [1,4,3,2]

All possible pairings (ignoring the ordering of elements) are:

(1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
(1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
(1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4

So the maximum possible sum is 4.

Here is another example: nums = [6,2,6,5,1,2] The optimal pairing is (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9.

What is the most efficient way to find the maximized sum? What is the time and space complexity of your approach? What are the edge cases?

Array Partition

Easy

Google

1 view

Topics:

ArraysGreedy Algorithms

For example:

nums = [1,4,3,2]

All possible pairings (ignoring the ordering of elements) are:

(1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
(1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
(1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4

So the maximum possible sum is 4.

Here is another example: nums = [6,2,6,5,1,2] The optimal pairing is (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9.

What is the most efficient way to find the maximized sum? What is the time and space complexity of your approach? What are the edge cases?

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 expected data type of the array elements, and what is the range of possible values, including the possibility of negative numbers or zero?
What should I return if no such partition is possible?
Can the input array be empty or null?
Are there any constraints on the relative ordering of elements within each partition after the partitioning process?
Are duplicate numbers allowed in the input array, and if so, how should they be handled during the partitioning?

Brute Force Solution

Approach

The brute force strategy for this problem is all about trying every single combination of how to split a group of numbers into smaller groups. It's like exploring every possible arrangement, no matter how silly, to see if it works. We check if each possible split fulfills some condition, and if it does, we note it down.

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

First, consider the absolute simplest split: put the first number in its own group, and everything else in another group.
Next, try putting the first two numbers together in a group, and the rest in another group. Keep doing this, each time adding one more number to the first group, until the first group contains all the numbers.
For each of these splits, check if some specific conditions hold true for the resulting groups (like, do all the numbers in each group add up to less than some value?).
If a particular split does meet the condition, remember it as a possible solution.
Once we've explored every possible split, compare all the valid solutions we found.
Finally, choose the best solution from the valid ones, based on some criteria defined in the problem.

Code Implementation

def array_partition_brute_force(numbers):
    all_possible_splits = []

    # Iterate through all possible split points.
    for split_point_index in range(1, len(numbers) + 1):

        first_group = numbers[:split_point_index]
        second_group = numbers[split_point_index:]

        all_possible_splits.append((first_group, second_group))

    valid_solutions = []

    # Placeholder condition function.  Replace with actual condition per the prompt
    def check_condition(group_one, group_two):
        return True

    # Iterate through all splits and apply the condition
    for first_group, second_group in all_possible_splits:
        if check_condition(first_group, second_group):
            valid_solutions.append((first_group, second_group))

    # Need to have a selection criteria for "best solution" per prompt definition
    if not valid_solutions:
        return None

    # A placeholder comparison function.
    def compare_solutions(solution_one, solution_two):
        return 0

    best_solution = valid_solutions[0]

    # Iterate through all valid solutions, finding the best one.
    for solution in valid_solutions[1:]:
        if compare_solutions(solution, best_solution) > 0:
            best_solution = solution

    return best_solution

Big(O) Analysis

Time Complexity

O(2^n) – The provided brute force approach involves exploring every possible way to split the array into two groups. For an array of size n, each element can either belong to the first group or the second group, resulting in 2 possibilities for each element. Therefore, the total number of possible splits to explore becomes 2*2*...*2 (n times), which equals 2^n. Consequently, the time complexity grows exponentially with the input size n, making it O(2^n).

Space Complexity

O(1) – The plain English explanation describes checking various splits of an array without explicitly mentioning the creation of new data structures that scale with the input size, N. The algorithm explores combinations and keeps track of the best solution seen so far, presumably using a fixed number of variables to store intermediate results and the best split's value. Therefore, the space complexity is determined by the constant space usage for storing a few variables, regardless of the input size N, resulting in O(1) auxiliary space.

Optimal Solution

Approach

The goal is to pair numbers in a collection to maximize the sum of the smallest number from each pair. The clever idea is to organize the numbers first, then create pairs in a very specific way. This avoids checking every possible pairing.

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

First, arrange all the numbers from smallest to largest.
Then, create pairs by combining the first number with the second, the third with the fourth, and so on.
Finally, add up the smaller number from each of these pairs to get the largest possible total.

Code Implementation

def array_pair_sum(numbers):
    # Sorting ensures minimal values are grouped.
    numbers.sort()

    maximum_sum = 0
    for i in range(0, len(numbers), 2):
        # Sum the smaller element from each pair.
        maximum_sum += numbers[i]

    return maximum_sum

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this algorithm is sorting the input array of size n. Efficient sorting algorithms, such as merge sort or quicksort, typically have a time complexity of O(n log n). Pairing and summing the minimums afterward involves iterating through the sorted array once, which takes O(n) time. Since O(n log n) grows faster than O(n), the overall time complexity is determined by the sorting step.

Space Complexity

O(1) – The provided solution sorts the input array in place and then iterates through it to form pairs and calculate the sum. Sorting in place (if the sorting algorithm used is in-place, like heapsort) does not require additional memory proportional to the input size, N. Furthermore, creating pairs and summing the smaller elements only involves a few constant extra variables to store the running sum and indices, and these do not scale with N. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or undefined input array	Throw an IllegalArgumentException or return null to indicate invalid input.
Array with only one element	Return an empty list/array since a partition requires at least two elements.
Array with an even number of elements, but no valid partition exists that satisfies the condition	Return an empty list to signify that no valid partition was found.
Array contains duplicate values that could lead to multiple possible partitions	The algorithm should return only one valid partition if multiple exist, or be designed to enumerate all possible answers depending on requirements.
Array contains negative numbers or zero	The solution should correctly handle negative numbers and zero without causing arithmetic errors or logical flaws.
Array with a very large number of elements that could lead to memory or performance issues	Consider optimizing the algorithm to use less memory and have a better time complexity, possibly using in-place sorting or more efficient data structures.
Integer overflow issues when summing or multiplying array elements	Use appropriate data types (e.g., long) to prevent integer overflow and handle extremely large numbers.
Extreme boundary values, such as Integer.MAX_VALUE or Integer.MIN_VALUE	Ensure calculations involving these values do not cause unexpected behavior or overflow.

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 expected data type of the array elements, and what is the range of possible values, including the possibility of negative numbers or zero?
What should I return if no such partition is possible?
Can the input array be empty or null?
Are there any constraints on the relative ordering of elements within each partition after the partitioning process?
Are duplicate numbers allowed in the input array, and if so, how should they be handled during the partitioning?

Brute Force Solution

Approach

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

First, consider the absolute simplest split: put the first number in its own group, and everything else in another group.
Next, try putting the first two numbers together in a group, and the rest in another group. Keep doing this, each time adding one more number to the first group, until the first group contains all the numbers.
For each of these splits, check if some specific conditions hold true for the resulting groups (like, do all the numbers in each group add up to less than some value?).
If a particular split does meet the condition, remember it as a possible solution.
Once we've explored every possible split, compare all the valid solutions we found.
Finally, choose the best solution from the valid ones, based on some criteria defined in the problem.

Code Implementation

def array_partition_brute_force(numbers):
    all_possible_splits = []

    # Iterate through all possible split points.
    for split_point_index in range(1, len(numbers) + 1):

        first_group = numbers[:split_point_index]
        second_group = numbers[split_point_index:]

        all_possible_splits.append((first_group, second_group))

    valid_solutions = []

    # Placeholder condition function.  Replace with actual condition per the prompt
    def check_condition(group_one, group_two):
        return True

    # Iterate through all splits and apply the condition
    for first_group, second_group in all_possible_splits:
        if check_condition(first_group, second_group):
            valid_solutions.append((first_group, second_group))

    # Need to have a selection criteria for "best solution" per prompt definition
    if not valid_solutions:
        return None

    # A placeholder comparison function.
    def compare_solutions(solution_one, solution_two):
        return 0

    best_solution = valid_solutions[0]

    # Iterate through all valid solutions, finding the best one.
    for solution in valid_solutions[1:]:
        if compare_solutions(solution, best_solution) > 0:
            best_solution = solution

    return best_solution

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, arrange all the numbers from smallest to largest.
Then, create pairs by combining the first number with the second, the third with the fourth, and so on.
Finally, add up the smaller number from each of these pairs to get the largest possible total.

Code Implementation

def array_pair_sum(numbers):
    # Sorting ensures minimal values are grouped.
    numbers.sort()

    maximum_sum = 0
    for i in range(0, len(numbers), 2):
        # Sum the smaller element from each pair.
        maximum_sum += numbers[i]

    return maximum_sum

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or undefined input array	Throw an IllegalArgumentException or return null to indicate invalid input.
Array with only one element	Return an empty list/array since a partition requires at least two elements.
Array with an even number of elements, but no valid partition exists that satisfies the condition	Return an empty list to signify that no valid partition was found.
Array contains duplicate values that could lead to multiple possible partitions	The algorithm should return only one valid partition if multiple exist, or be designed to enumerate all possible answers depending on requirements.
Array contains negative numbers or zero	The solution should correctly handle negative numbers and zero without causing arithmetic errors or logical flaws.
Array with a very large number of elements that could lead to memory or performance issues	Consider optimizing the algorithm to use less memory and have a better time complexity, possibly using in-place sorting or more efficient data structures.
Integer overflow issues when summing or multiplying array elements	Use appropriate data types (e.g., long) to prevent integer overflow and handle extremely large numbers.
Extreme boundary values, such as Integer.MAX_VALUE or Integer.MIN_VALUE	Ensure calculations involving these values do not cause unexpected behavior or overflow.