Maximum Sum With at Most K Elements

Medium

Asked by:

8 views

Topics:

ArraysDynamic Programming

You are given a 2D integer matrix grid of size n x m, an integer array limits of length n, and an integer k. The task is to find the maximum sum of at most k elements from the matrix grid such that:

The number of elements taken from the i^th row of grid does not exceed limits[i].

Return the maximum sum.

Example 1:

Input: grid = [[1,2],[3,4]], limits = [1,2], k = 2

Output: 7

Explanation:

From the second row, we can take at most 2 elements. The elements taken are 4 and 3.
The maximum possible sum of at most 2 selected elements is 4 + 3 = 7.

Example 2:

Input: grid = [[5,3,7],[8,2,6]], limits = [2,2], k = 3

Output: 21

Explanation:

From the first row, we can take at most 2 elements. The element taken is 7.
From the second row, we can take at most 2 elements. The elements taken are 8 and 6.
The maximum possible sum of at most 3 selected elements is 7 + 8 + 6 = 21.

Constraints:

n == grid.length == limits.length
m == grid[i].length
1 <= n, m <= 500
0 <= grid[i][j] <= 10⁵
0 <= limits[i] <= m
0 <= k <= min(n * m, sum(limits))

Try coding it in LeetCode

Maximum Sum With at Most K Elements

Medium

Asked by:

8 views

Topics:

ArraysDynamic Programming

You are given a 2D integer matrix grid of size n x m, an integer array limits of length n, and an integer k. The task is to find the maximum sum of at most k elements from the matrix grid such that:

The number of elements taken from the i^th row of grid does not exceed limits[i].

Return the maximum sum.

Example 1:

Input: grid = [[1,2],[3,4]], limits = [1,2], k = 2

Output: 7

Explanation:

From the second row, we can take at most 2 elements. The elements taken are 4 and 3.
The maximum possible sum of at most 2 selected elements is 4 + 3 = 7.

Example 2:

Input: grid = [[5,3,7],[8,2,6]], limits = [2,2], k = 3

Output: 21

Explanation:

From the first row, we can take at most 2 elements. The element taken is 7.
From the second row, we can take at most 2 elements. The elements taken are 8 and 6.
The maximum possible sum of at most 3 selected elements is 7 + 8 + 6 = 21.

Constraints:

n == grid.length == limits.length
m == grid[i].length
1 <= n, m <= 500
0 <= grid[i][j] <= 10⁵
0 <= limits[i] <= m
0 <= k <= min(n * m, sum(limits))

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 possible ranges for the integer values within the input array and the value of K?
Can the input array be empty, or can K be zero or larger than the size of the input array?
If there are multiple subsets with the same maximum sum that use at most K elements, is any one of them acceptable?
Is the input array sorted or unsorted?
If no subset of at most K elements exists (e.g., the array is empty, or K is zero and all numbers are negative), what should the return value be (e.g., null, 0, negative infinity, or an error)?
Are all the values in the array integers?

Brute Force Solution

Approach

The brute force way to solve this problem is like trying every single combination of items to see which one gives us the biggest sum, but with a limit on how many items we can pick. We will explore every possible subset of items, as long as that subset doesn't have too many elements.

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

Start by considering every possible group of elements, from the smallest possible group to the largest allowed (at most K).
For each group, add up the values of the elements in that group.
Keep track of the largest sum you've found so far.
After considering all the possible groups, the largest sum you kept track of is the answer.

Code Implementation

def maximum_sum_with_k_elements_brute_force(elements, max_elements_allowed):
    maximum_sum_found = 0

    for i in range(1 << len(elements)):
        current_subset = []
        current_sum = 0

        for j in range(len(elements)):
            # Check if j-th bit is set in the current subset
            if (i >> j) & 1:
                current_subset.append(elements[j])

        # Check if the current subset has at most max_elements_allowed elements
        if len(current_subset) <= max_elements_allowed:

            # The subset is valid, calculate sum of subset.
            for element in current_subset:
                current_sum += element

            # Update the maximum sum found so far
            maximum_sum_found = max(maximum_sum_found, current_sum)

    return maximum_sum_found

Big(O) Analysis

Time Complexity

O(n^k) – The brute force approach involves examining all possible subsets of the input array of size n with at most k elements. For each possible subset size from 1 to k, we need to consider all combinations of elements. The number of such combinations is on the order of n choose 1 + n choose 2 + ... + n choose k. In the worst case, k could be close to n, making the runtime exponential. However, if we treat k as a fixed parameter, the dominant term becomes n^k, leading to a time complexity of O(n^k) because it represents iterating through all possible sub arrays of sizes up to k.

Space Complexity

O(1) – The brute force approach, as described, primarily involves iterating and calculating sums. While the description mentions exploring subsets, it doesn't explicitly state the creation of auxiliary data structures to store these subsets. The algorithm keeps track of the largest sum found so far using a single variable, which takes constant space, irrespective of the input size N (where N is the total number of elements). Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The goal is to find the largest possible sum by picking at most a certain number of elements from a list. The smart approach is to sort the list and then simply add up the largest possible values to get the maximum sum.

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

First, arrange all the numbers in the list from largest to smallest.
Then, take the top 'k' numbers from the sorted list, where 'k' is the maximum number of elements you are allowed to pick.
Finally, add up those 'k' largest numbers. This sum will be the maximum possible sum you can achieve.

Code Implementation

def maximum_sum_with_k_elements(number_list, k):
    # Sort the list in descending order.
    sorted_number_list = sorted(number_list, reverse=True)

    maximum_sum = 0
    # Iterate through the first k elements.
    for i in range(min(k, len(sorted_number_list))):
        # Adding the element to the maximum sum
        maximum_sum += sorted_number_list[i]

    # Returning final result
    return maximum_sum

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this approach is sorting the input list of n numbers. Efficient sorting algorithms like merge sort or quicksort have a time complexity of O(n log n). After sorting, summing the top k elements only takes O(k) time, which is at most O(n) but is less significant than the sorting time. Therefore, the overall time complexity is determined by the sorting step, resulting in O(n log n).

Space Complexity

O(N) – The dominant space usage comes from sorting the input list of size N. Most sorting algorithms, like merge sort or quicksort (in worst-case scenarios), require O(N) auxiliary space for temporary storage during the sorting process. While some in-place sorting algorithms exist, the plain English explanation doesn't specify one, and common implementations will use O(N) temporary space. Therefore, the space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 or throw an exception (depending on requirements) to indicate invalid input.
k is 0	Return 0, as no elements can be selected.
k is greater than the array length	Take the sum of all array elements if k exceeds array length.
Array contains negative numbers	The algorithm should correctly handle negative numbers by either including them in the sum if they increase the overall sum or excluding them if they don't.
Array contains only negative numbers and k > 0	Return the largest (least negative) k numbers' sum, or 0 if k is larger than array length (and all negative).
Large array size and large k, near array size	The solution needs to be efficient and not time out, suggesting efficient selection using sorting or priority queue.
Array contains very large positive numbers that can lead to integer overflow when summed.	Use a data type with a larger range (e.g., long) to store the sum or check for overflow conditions during accumulation.
Array contains duplicates and needs to select some, but not all of them	Sorting approach or priority queue will treat duplicates independently, so algorithm should naturally select the largest k elements, even if duplicate numbers are present.