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Maximize Score After Pair Deletions

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ArraysGreedy Algorithms

You are given a 0-indexed integer array nums of length n. Each element in nums is between 0 and n-1, inclusive.

You can perform the following operation any number of times:

  • Choose two indices i and j such that i != j and nums[i] == nums[j].
  • Delete both nums[i] and nums[j] from the array.

You want to maximize the score obtained after performing any number of operations. The score is the total number of deleted elements.

Return the maximum possible score.

Example 1:

Input: nums = [3,1,3,2,2,2]
Output: 4
Explanation: 
We can perform the following operations:
1. Choose indices 0 and 2. nums[0] == nums[2] == 3. Delete both. nums becomes [1,2,2,2]. Score is 2.
2. Choose indices 1 and 2. nums[1] == nums[2] == 2. Delete both. nums becomes [1,2]. Score is 2 + 2 = 4.
We cannot perform any more operations. The maximum score is 4.

Example 2:

Input: nums = [0,0,1,2,0,1,0,1]
Output: 6
Explanation: 
We can perform the following operations:
1. Choose indices 0 and 4. nums[0] == nums[4] == 0. Delete both. nums becomes [0,1,2,0,1,0,1]. Score is 2.
2. Choose indices 0 and 6. nums[0] == nums[6] == 0. Delete both. nums becomes [1,2,0,1,0,1]. Score is 2 + 2 = 4.
3. Choose indices 1 and 5. nums[1] == nums[5] == 1. Delete both. nums becomes [2,0,1,0,1]. Score is 4 + 2 = 6.
We cannot perform any more operations. The maximum score is 6.

Constraints:

  • n == nums.length
  • 1 <= n <= 105
  • 0 <= nums[i] < n
Try coding it in LeetCodeLeetCode

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:

  1. The problem states 'distinct integers'. Does this mean all elements in the input array `nums` are guaranteed to be unique, or should I consider cases where duplicates might exist despite the wording?
  2. What is the expected range of values for the integers in the `nums` array, and should I consider negative numbers or zero?
  3. If it's possible to perform zero operations, what should be returned in that case?
  4. Are there any constraints on the size of the input array `nums` that I should be aware of?
  5. Can an activity be used in multiple pairs? Or once an activity is selected and deleted, can it be used again?

Brute Force Solution

Approach

We have a collection of numbers and want to find the best way to group them into pairs to achieve the highest possible score. The brute force method explores every single way to form these pairs.

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

  1. Consider all possible ways to pick the very first pair of numbers from the collection.
  2. Once a pair is chosen, set it aside and then consider all possible ways to pick the next pair from the remaining numbers.
  3. Continue this process, picking pairs one by one, until all numbers are used up.
  4. For each complete arrangement of pairs, calculate the total score achieved.
  5. After trying every single combination of pairings, find the arrangement that resulted in the largest total score.

Code Implementation

import itertoolsdef maximize_score_after_pair_deletions_brute_force(numbers):
    if not numbers or len(numbers) % 2 != 0:
        return 0

    maximum_achieved_score = 0

    # The core idea is to explore every single possible way to form pairs.
    # This ensures that no combination is missed and we can find the absolute maximum.
    for pair_combination in itertools.combinations(range(len(numbers)), 2):
        remaining_indices = list(range(len(numbers)))
        current_score = 0
        used_indices = set()
        
        # We must start by considering the first pair selected from all possibilities.
        # This sets the stage for recursively exploring subsequent pairings.
        if pair_combination[0] in remaining_indices and pair_combination[1] in remaining_indices:
            current_score += abs(numbers[pair_combination[0]] - numbers[pair_combination[1]])
            used_indices.add(pair_combination[0])
            used_indices.add(pair_combination[1])
            remaining_indices.remove(pair_combination[0])
            remaining_indices.remove(pair_combination[1])

            # Continue forming pairs until no numbers are left, simulating the process.
            # This ensures that each full partition of numbers into pairs is evaluated.
            while remaining_indices:
                next_pair_found = False
                for next_pair in itertools.combinations(remaining_indices, 2):
                    if next_pair[0] not in used_indices and next_pair[1] not in used_indices:
                        current_score += abs(numbers[next_pair[0]] - numbers[next_pair[1]])
                        used_indices.add(next_pair[0])
                        used_indices.add(next_pair[1])
                        remaining_indices.remove(next_pair[0])
                        remaining_indices.remove(next_pair[1])
                        next_pair_found = True
                        break # Move to find the next pair once one is formed
                if not next_pair_found and remaining_indices: # Should not happen if logic is correct for even length
                    break
            
            # After evaluating a complete set of pairs, update the maximum score if this combination is better.
            # This step ensures we keep track of the highest score found across all explored pairings.
            maximum_achieved_score = max(maximum_achieved_score, current_score)

    return maximum_achieved_score

Big(O) Analysis

Time Complexity
O(n!) – The core of the brute force approach involves exploring all possible pairings. For the first pair, we have n choices for the first element and (n-1) for the second, but since order doesn't matter for the pair itself, it's n * (n-1) / 2. After picking one pair, we're left with n-2 elements and repeat the process. This recursive selection of pairs leads to a factorial number of combinations. Specifically, the number of ways to partition n items into n/2 pairs is (n! / (2^(n/2) * (n/2)!)). This complexity grows extremely rapidly, approximated as O(n!).
Space Complexity
O(n^2 * log(n)) or O(n!) depending on implementation details – The brute force approach involves exploring all possible pairings. This typically requires recursion or a systematic generation of permutations. The recursion depth can go up to N/2 (for pairs), and at each level, we might store intermediate pairings or available numbers. If we consider generating permutations of numbers to form pairs, the process can lead to storing intermediate permutations which can take O(N) space for each permutation. Alternatively, if we model this as a decision tree where each node represents picking a pair, the number of nodes can be very large, and the depth of the recursion stack contributes to space complexity. The worst-case auxiliary space complexity is often dominated by the recursion stack depth and the storage needed to keep track of available elements at each recursive call, which can be O(N) in a well-optimized recursive traversal. However, if explicitly storing all possible pairings or permutations, it can reach O(N!) in naive implementations, or O(N^2) for dynamic programming states or memoization. Given the description of exploring 'every single way', a recursive backtracking approach is implied, which would use O(N) for the recursion stack and O(N) for auxiliary data structures to track available numbers at each step, leading to O(N) auxiliary space. However, if we consider the state space explored by a brute force permutation generator, it could be O(N!) permutations, each potentially requiring O(N) space to store, making O(N * N!) space in the worst naive implementations. A more reasonable brute-force implementation might involve generating permutations of indices which takes O(N) stack space for recursion and O(N) space for the current permutation being built. However, to consider 'all possible pairings' and 'all complete arrangements', the act of generating and evaluating these might implicitly require storing or managing intermediate states which could be more significant. The number of ways to form pairs from N elements is roughly N! / ((N/2)! * 2^(N/2)), which suggests a combinatorial explosion. If we are generating these combinations, the space needed to hold these combinations or the recursion depth to generate them will be the dominant factor. A common brute-force recursive approach would involve N/2 levels of recursion, with each level potentially copying parts of the remaining numbers, leading to O(N^2) auxiliary space if not optimized, or O(N) for the recursion stack plus O(N) for tracking remaining elements. Considering the description, a recursive approach is most fitting, with the recursion stack depth being proportional to N/2, and auxiliary data structures to manage remaining elements at each step. The dominant space usage comes from the recursion stack frames and potentially storing subsets of numbers for pairing, leading to an auxiliary space complexity related to the depth of the recursion, which is O(N). However, if we are explicitly generating and storing all permutations to evaluate them, the space could be much higher. Let's assume a recursive backtracking approach where we pass remaining elements or use a boolean array to mark used elements. In this case, the recursion depth is N/2 and each recursive call might carry a copy of remaining numbers or a state array, leading to O(N) space for the recursion stack and O(N) for auxiliary state tracking, resulting in O(N) auxiliary space. The complexity of generating all pairings can be seen as picking the first pair in N*(N-1)/2 ways, then the next in (N-2)*(N-3)/2 ways and so on. This recursive structure implies a space usage proportional to the depth of recursion, which is N/2, and the space used at each level to manage remaining elements. Thus, the auxiliary space complexity is O(N) for the recursion stack and for temporary storage of available elements.

Optimal Solution

Approach

The best way to solve this is to realize that the order in which you consider deleting pairs of numbers matters significantly. Instead of checking every single pair deletion, we can be strategic about which pairs to prioritize, leading to the maximum possible score.

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

  1. Imagine you have a collection of numbers. We want to find pairs of numbers that are close to each other in value.
  2. The core idea is to always look for the pair of numbers that are closest in value among all available numbers.
  3. When you find such a closest pair, you form a group, get a score based on the difference between them, and then you are left with a new, smaller collection of numbers.
  4. The clever part is that by always picking the closest available pair, you are essentially setting up the remaining numbers in a way that will allow you to make the best future pairings.
  5. You keep repeating this process: find the closest pair, make a group, get a score, and reduce your collection of numbers.
  6. You continue this until you can't make any more pairs.
  7. The total score you accumulated from all the pairs you formed is your maximum possible score.

Code Implementation

import collections
def maximum_score_from_pair_deletions(numbers_list):
    numbers_list.sort()
    total_score = 0
    number_counts = collections.Counter(numbers_list)
    unique_sorted_numbers = sorted(number_counts.keys())

    # Process numbers from smallest to largest to find closest pairs first.
    for current_number in unique_sorted_numbers:
        while number_counts[current_number] >= 2:
            # Find the next available number that is closest in value.
            next_closest_number = -1
            for potential_next_number in unique_sorted_numbers:
                if number_counts[potential_next_number] > 0 and potential_next_number != current_number:
                    next_closest_number = potential_next_number
                    break
                elif number_counts[potential_next_number] > 0 and potential_next_number == current_number and number_counts[current_number] >= 2:
                    next_closest_number = current_number
                    break

            if next_closest_number != -1:
                # Form a pair and update counts.
                total_score += abs(current_number - next_closest_number)
                number_counts[current_number] -= 1
                number_counts[next_closest_number] -= 1
                if number_counts[current_number] == 0:
                    del number_counts[current_number]
                if number_counts[next_closest_number] == 0 and next_closest_number in number_counts:
                    del number_counts[next_closest_number]
            else:
                break
    return total_score

Big(O) Analysis

Time Complexity
O(n log n) – The strategy involves repeatedly finding the closest pair of numbers. If we sort the input array of size n, which takes O(n log n) time, subsequent searches for closest pairs can be done efficiently. In a sorted array, the closest pairs will always be adjacent elements. Iterating through the sorted array to find adjacent pairs and potentially removing them can be done in O(n) time. Since we might perform these pairing operations up to n/2 times, and each operation benefits from the initial sort, the dominant cost is the initial sorting. Therefore, the overall time complexity is O(n log n).
Space Complexity
O(n) – The solution involves sorting the input collection of numbers, which typically requires auxiliary space proportional to the input size N for the sorting algorithm itself (e.g., merge sort). After sorting, the process of repeatedly finding the closest pair and forming groups does not require additional data structures that grow with N; it can be done in-place or with a constant amount of extra space relative to N. Therefore, the dominant factor in auxiliary space complexity is the sorting step.

Edge Cases

Empty input array nums
How to Handle:
Return 0 as no operations can be performed to gain any score.
Input array nums with only one element
How to Handle:
Return 0 because a pair of distinct activities cannot be formed.
Input array nums with two elements
How to Handle:
The only possible operation is to sum these two elements and return their sum.
Input array nums with distinct negative numbers
How to Handle:
The strategy of pairing the largest two remaining numbers still maximizes the score, even if negative.
Input array nums containing zeros
How to Handle:
Zeros can be paired with any other number, and their contribution to the sum is zero, so they are handled naturally by the sorting and pairing approach.
Input array nums with a very large number of elements (scalability)
How to Handle:
A solution involving sorting (O(N log N)) and then a linear scan (O(N)) should be efficient enough for typical FAANG constraints.
All elements in the input array are identical
How to Handle:
The problem statement specifies distinct integers, so this case is technically invalid according to the constraints.
Extremely large positive or negative integer values
How to Handle:
Ensure that the sum of two elements does not cause integer overflow, which might require using larger data types if available in the language.
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