Maximum Number of Operations With the Same Score I

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ArraysTwo Pointers

You are given an array of integers nums. Consider the following operation:

Delete the first two elements nums and define the score of the operation as the sum of these two elements.

You can perform this operation until nums contains fewer than two elements. Additionally, the same score must be achieved in all operations.

Return the maximum number of operations you can perform.

Example 1:

Input: nums = [3,2,1,4,5]

Output: 2

Explanation:

We can perform the first operation with the score 3 + 2 = 5. After this operation, nums = [1,4,5].
We can perform the second operation as its score is 4 + 1 = 5, the same as the previous operation. After this operation, nums = [5].
As there are fewer than two elements, we can't perform more operations.

Example 2:

Input: nums = [1,5,3,3,4,1,3,2,2,3]

Output: 2

Explanation:

We can perform the first operation with the score 1 + 5 = 6. After this operation, nums = [3,3,4,1,3,2,2,3].
We can perform the second operation as its score is 3 + 3 = 6, the same as the previous operation. After this operation, nums = [4,1,3,2,2,3].
We cannot perform the next operation as its score is 4 + 1 = 5, which is different from the previous scores.

Example 3:

Input: nums = [5,3]

Output: 1

Constraints:

2 <= nums.length <= 100
1 <= nums[i] <= 1000

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 are the constraints on the size of the input array `nums`?
Can the input array `nums` contain negative numbers, zeros, or floating-point numbers?
If no operations can be performed with the same score, what should the function return?
Are there any constraints on the range of values within the `nums` array?
If multiple pairs of numbers can be chosen to achieve the maximum number of operations, is the order in which I choose them relevant?

Brute Force Solution

Approach

The brute force approach to solving this problem involves examining all possible pairs of numbers in the list and checking if we can make valid operations. We simply try every possible combination of pairs and see which one gives us the most successful operations where each operation has the same score.

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

Start by picking the first two numbers in the list and add them together to get their score.
Now, go through all the remaining pairs of numbers in the list.
For each pair, check if their sum equals the score we calculated in the first step.
If the sum matches, count this as a successful operation and remove these two numbers from our list to consider the next operation.
If the sum doesn't match, move on to the next pair.
After checking all possible pairs with the initial score, count how many successful operations we had.
Repeat this process, starting by picking a *different* first pair of numbers to get a different initial score.
Keep track of the number of successful operations for each different initial score you try.
Finally, compare the number of successful operations from all the different starting pairs, and choose the highest number as the answer.

Code Implementation

def maximum_number_of_operations_with_same_score_brute_force(numbers):
    maximum_operations = 0
    
    # Iterate through all possible starting pairs to determine score.
    for first_pair_index in range(len(numbers) - 1):
        for second_pair_index in range(first_pair_index + 1, len(numbers)):           
            initial_score = numbers[first_pair_index] + numbers[second_pair_index]
            current_operations = 1
            remaining_numbers = numbers[:]

            # Remove the numbers of initial score pair
            del remaining_numbers[max(first_pair_index, second_pair_index)]
            del remaining_numbers[min(first_pair_index, second_pair_index)]

            # Iterate through remaining pairs to find matching score
            index = 0
            while index < len(remaining_numbers) - 1:
                found_match = False
                for second_index in range(index + 1, len(remaining_numbers)):  
                    if remaining_numbers[index] + remaining_numbers[second_index] == initial_score:

                        # Found a matching pair
                        current_operations += 1
                        del remaining_numbers[second_index]

                        # Have to delete from end to beginning
                        del remaining_numbers[index]
                        found_match = True
                        break
                if not found_match:
                    index += 1

            # Update the maximum operations found
            maximum_operations = max(maximum_operations, current_operations)

    return maximum_operations

Big(O) Analysis

Time Complexity

O(n^3) – The brute force approach iterates through all possible initial pairs in the input array of size n. This leads to O(n) initial pair selections. For each initial pair, the algorithm iterates through the remaining pairs to find matches, which is O(n^2) in the worst case because it needs to check all possible pairs of numbers after picking the initial pair. Thus, the total time complexity is O(n * n^2), which simplifies to O(n^3).

Space Complexity

O(N) – The brute force approach, as described, involves repeatedly removing elements from the original list. While the explanation doesn't explicitly create a new list, the repeated removal operations implicitly modify the input list, which for space complexity analysis, is considered in-place modification. However, to *simulate* the removal and keep track of processed elements, a common implementation involves marking elements as 'visited' or creating a copy of the list and filtering out used pairs. Both approaches will likely require auxiliary space proportional to the input size N, where N is the number of elements in the input list. Thus, the space complexity will be O(N).

Optimal Solution

Approach

The problem requires us to find the maximum number of pairs with the same sum. We can achieve this by calculating the sum of the first pair, then iterating through the rest of the numbers to find other pairs that match this sum. This approach is efficient since it directly focuses on finding matching pairs.

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

Calculate the sum of the first two numbers.
Keep track of how many pairs have this sum.
Move through the list of numbers in steps of two, checking each pair.
For each pair, determine if their sum equals the sum of the first pair.
If the sums are equal, increase the count of pairs with the same sum.
If the sums are not equal, you can stop because we need all pairs to have same sum. If there is less than 2 numbers then stop.
The final count represents the maximum number of operations (pairs) that have the same sum.

Code Implementation

def max_operations_with_same_score(numbers):
    if len(numbers) < 2:
        return 0

    initial_sum = numbers[0] + numbers[1]

    operations_count = 1

    # Only check for pairs if the list contains at least two numbers.

    current_index = 2
    while current_index < len(numbers):
        if current_index + 1 >= len(numbers):
            break

        current_sum = numbers[current_index] + numbers[current_index + 1]

        # If the current pair's sum is equal, increment operations.
        if current_sum == initial_sum:
            operations_count += 1
        else:
            # If sum is not equal, we can't form equal pairs.
            break

        current_index += 2

    return operations_count

Big(O) Analysis

Time Complexity

O(n) – The dominant operation is iterating through the input array nums with a size of n, checking pairs of elements. This iteration proceeds with a step of 2, examining approximately n/2 pairs. Therefore, the number of operations scales linearly with the input size n. Hence, the time complexity is O(n).

Space Complexity

O(1) – The algorithm only uses a fixed number of variables: one to store the initial sum, one to count the operations, and an index variable to traverse the input array. These variables occupy a constant amount of space regardless of the input size N (the number of elements in the array). Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array	Return 0 since no operations are possible with an empty array.
Input array with only one element	Return 0 since at least two elements are required for an operation.
Input array with two elements.	Compute the sum, store the score, and return 1 if the subsequent operation matches the score.
Array with all elements being zero.	The sum will always be zero, so the solution should correctly count pairs summing to zero.
Array with a single valid pair and many invalid elements.	The algorithm should stop after finding the first optimal score and removing those elements.
Array containing negative and positive numbers.	The solution should handle negative numbers correctly when calculating sums.
Array with a large number of elements (scalability).	A linear-time solution (O(n)) using a hashmap or similar data structure is desirable for large inputs.
Array where no two elements sum to the same value as any other pair	Return 0 if no consistent sum can be found across multiple pairs.

Empty input array

How to Handle:

Return 0 since no operations are possible with an empty array.

Input array with only one element

How to Handle:

Return 0 since at least two elements are required for an operation.

Input array with two elements.

How to Handle:

Compute the sum, store the score, and return 1 if the subsequent operation matches the score.

Array with all elements being zero.

How to Handle:

The sum will always be zero, so the solution should correctly count pairs summing to zero.

Array with a single valid pair and many invalid elements.

How to Handle:

The algorithm should stop after finding the first optimal score and removing those elements.

Array containing negative and positive numbers.

How to Handle:

The solution should handle negative numbers correctly when calculating sums.

Array with a large number of elements (scalability).

How to Handle:

A linear-time solution (O(n)) using a hashmap or similar data structure is desirable for large inputs.

Array where no two elements sum to the same value as any other pair

How to Handle:

Return 0 if no consistent sum can be found across multiple pairs.