Maximum Sum With Exactly K Elements

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Greedy AlgorithmsArraysDynamic Programming

You are given a 0-indexed integer array nums and an integer k. Your task is to perform the following operation exactly k times in order to maximize your score:

Select an element m from nums.
Remove the selected element m from the array.
Add a new element with a value of m + 1 to the array.
Increase your score by m.

Return the maximum score you can achieve after performing the operation exactly k times.

Example 1:

Input: nums = [1,2,3,4,5], k = 3
Output: 18
Explanation: We need to choose exactly 3 elements from nums to maximize the sum.
For the first iteration, we choose 5. Then sum is 5 and nums = [1,2,3,4,6]
For the second iteration, we choose 6. Then sum is 5 + 6 and nums = [1,2,3,4,7]
For the third iteration, we choose 7. Then sum is 5 + 6 + 7 = 18 and nums = [1,2,3,4,8]
So, we will return 18.
It can be proven, that 18 is the maximum answer that we can achieve.

Example 2:

Input: nums = [5,5,5], k = 2
Output: 11
Explanation: We need to choose exactly 2 elements from nums to maximize the sum.
For the first iteration, we choose 5. Then sum is 5 and nums = [5,5,6]
For the second iteration, we choose 6. Then sum is 5 + 6 = 11 and nums = [5,5,7]
So, we will return 11.
It can be proven, that 11 is the maximum answer that we can achieve.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100
1 <= k <= 100

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 range of values for the elements in the input array, and can they be negative or zero?
What should I return if there are fewer than K elements in the array?
Is K guaranteed to be a non-negative integer, and is it possible for K to be zero?
If there are multiple ways to choose K elements that result in the maximum sum, is any valid selection acceptable?
What are the constraints on the size of the input array (the number of elements)?

Brute Force Solution

Approach

The goal is to find the largest possible sum achievable by picking a specific number of values from a collection of numbers. The brute force strategy involves checking every single possible combination of values.

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

Consider all possible groups of numbers you can create by picking the exact amount we need.
For each group, calculate the sum of the numbers within that group.
Compare the sum of each group with all the other group sums you've calculated.
The largest sum you find is the answer.

Code Implementation

def find_maximum_sum_with_exactly_k_elements_brute_force(numbers, k):
    maximum_sum = float('-inf')
    number_of_elements = len(numbers)

    # Iterate through all possible combinations of k elements
    for i in range(1 << number_of_elements):
        current_subset = []

        # Build the current subset based on the bitmask
        for j in range(number_of_elements):
            if (i >> j) & 1:
                current_subset.append(numbers[j])

        # Ensure we only consider subsets with exactly k elements.
        if len(current_subset) == k:

            #Calculate the sum of the current subset
            current_sum = sum(current_subset)

            #Update maximum_sum if current_sum is greater
            maximum_sum = max(maximum_sum, current_sum)

    if maximum_sum == float('-inf'):
        return 0

    return maximum_sum

Big(O) Analysis

Time Complexity

O(nCk) – The brute force approach involves considering all possible groups of exactly k elements from an input array of size n. This corresponds to calculating the number of combinations of n items taken k at a time, denoted as nCk, which represents the total number of groups we need to examine. For each of these nCk groups, we calculate the sum of the k elements, which takes O(k) time. However, the dominant factor is the number of combinations. Therefore, the overall time complexity is O(nCk * k). Since nCk grows faster than k, and we are dealing with brute force, we simplify to O(nCk) to represent the combinations alone, even though we sum the elements in each combination.

Space Complexity

O(1) – The brute force solution, as described, iterates through all possible combinations of exactly K elements and calculates their sums. It only needs to store the current combination's sum and the maximum sum found so far. These variables take up constant space regardless of the input array's size (N), so the auxiliary space complexity is O(1).

Optimal Solution

Approach

The goal is to find the largest possible sum by picking exactly k numbers from a list. The most efficient way to do this is to repeatedly pick the largest number available until we've picked k numbers.

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

Find the largest number in the given list.
Add this largest number to our running total (which starts at zero).
Increase the largest number by one and add it back into the list.
Reduce the number of remaining picks by one.
Repeat the process of finding the largest number, adding it to the total, increasing the largest number, and reducing the remaining picks until we've selected exactly k numbers.

Code Implementation

def maximum_sum_with_k_elements(number_list, k):
    maximum_sum = 0
    
    for _ in range(k):
        # Find the largest element. 
        largest_number = max(number_list)
        
        maximum_sum += largest_number
        
        # Increase the largest number for next iteration
        largest_number += 1
        
        number_list.append(largest_number)
    
    return maximum_sum

Big(O) Analysis

Time Complexity

O(n + k) – The algorithm first finds the largest number in the list, which takes O(n) time, where n is the size of the input list. Then, it iterates k times, each time adding the largest number to the sum and incrementing it. Each of these k iterations takes constant time, O(1). Therefore, the total time complexity is O(n + k), representing the initial linear search plus the k constant-time operations.

Space Complexity

O(1) – The algorithm, as described, does not use any auxiliary data structures like lists, arrays, or hash maps. It only modifies the largest number in place and keeps track of the running total and remaining picks, which can be done with constant extra space. Therefore, the space complexity is independent of the input list size (N) or the value of k. The algorithm's auxiliary space usage remains constant.

Edge Cases

Case	How to Handle
Empty array	Return 0 if the array is empty, as there are no elements to sum.
k is zero	Return 0 if k is zero, as we are summing zero elements.
k is larger than the array size	Return an appropriate error value or throw an exception if k is larger than the array size, since summing k elements is impossible.
Array contains negative numbers	The solution works correctly with negative numbers, selecting the largest, which might be negative.
Array contains duplicate numbers	The solution correctly handles duplicates by repeatedly selecting the largest element and incrementing it.
Large array size with large k	The time complexity should be considered to avoid timeouts, ensuring efficient sorting or selection algorithms are used.
All numbers are the same	The solution functions correctly because the same largest value will be chosen k times and incremented.
Integer overflow after repeatedly adding the largest element	Use a data type with sufficient range (e.g., long) to prevent integer overflow during the summation or incrementing steps.

Empty array

How to Handle:

Return 0 if the array is empty, as there are no elements to sum.

k is zero

How to Handle:

Return 0 if k is zero, as we are summing zero elements.

k is larger than the array size

How to Handle:

Return an appropriate error value or throw an exception if k is larger than the array size, since summing k elements is impossible.

Array contains negative numbers

How to Handle:

The solution works correctly with negative numbers, selecting the largest, which might be negative.

Array contains duplicate numbers

How to Handle:

The solution correctly handles duplicates by repeatedly selecting the largest element and incrementing it.

Large array size with large k

How to Handle:

The time complexity should be considered to avoid timeouts, ensuring efficient sorting or selection algorithms are used.

All numbers are the same

How to Handle:

The solution functions correctly because the same largest value will be chosen k times and incremented.

Integer overflow after repeatedly adding the largest element

How to Handle:

Use a data type with sufficient range (e.g., long) to prevent integer overflow during the summation or incrementing steps.