Largest Values From Labels

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Greedy Algorithms

You are given n item's value and label as two integer arrays values and labels. You are also given two integers numWanted and useLimit.

Your task is to find a subset of items with the maximum sum of their values such that:

The number of items is at most numWanted.
The number of items with the same label is at most useLimit.

Return the maximum sum.

Example 1:

Input: values = [5,4,3,2,1], labels = [1,1,2,2,3], numWanted = 3, useLimit = 1

Output: 9

Explanation:

The subset chosen is the first, third, and fifth items with the sum of values 5 + 3 + 1.

Example 2:

Input: values = [5,4,3,2,1], labels = [1,3,3,3,2], numWanted = 3, useLimit = 2

Output: 12

Explanation:

The subset chosen is the first, second, and third items with the sum of values 5 + 4 + 3.

Example 3:

Input: values = [9,8,8,7,6], labels = [0,0,0,1,1], numWanted = 3, useLimit = 1

Output: 16

Explanation:

The subset chosen is the first and fourth items with the sum of values 9 + 7.

Constraints:

n == values.length == labels.length
1 <= n <= 2 * 10⁴
0 <= values[i], labels[i] <= 2 * 10⁴
1 <= numWanted, useLimit <= n

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 values in the `values` and `labels` arrays, and are negative values allowed?
What are the constraints on `numWanted` and `useLimit`? Can they be zero or negative?
If there aren't `numWanted` items available after applying the `useLimit` for each label, what should I return?
Are the `values` and `labels` arrays guaranteed to be the same length, and what should I do if they are not?
Is there a constraint on the number of unique labels, and could the `useLimit` ever exceed this number?

Brute Force Solution

Approach

The brute force method for finding the largest values from labels is like trying out every single possible combination of items and labels. We examine each combination to see if it meets the given restrictions, and then select the combination that results in the largest total value.

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

Consider every possible selection of items from the given set of values and labels.
For each selection, check if the number of times each label appears is within the allowed limit.
Also, verify that the total number of items selected does not exceed the allowed limit.
If a selection satisfies both the label limit and the item limit, calculate the sum of the values of the selected items.
Keep track of the largest sum seen so far.
After considering all possible selections, return the largest sum that was found.

Code Implementation

def largest_values_from_labels_brute_force(
    values, labels, number_of_allowed_items, use_limit
):
    number_of_items = len(values)
    max_sum = 0

    # Iterate through all possible combinations of items
    for i in range(1 << number_of_items):
        current_sum = 0
        number_of_items_selected = 0
        label_counts = {}
        items_selected = []

        for j in range(number_of_items):
            # Check if the j-th item is included in the current combination
            if (i >> j) & 1:
                items_selected.append(j)

        for item_index in items_selected:
            current_sum += values[item_index]
            number_of_items_selected += 1

            label = labels[item_index]
            label_counts[label] = label_counts.get(label, 0) + 1

        # Check if the current combination satisfies the limits
        is_valid = True
        
        # Verify that the number of items is within the allowed limit.
        if number_of_items_selected > number_of_allowed_items:
            is_valid = False
        
        # Verify that we are not exceeding the number of possible uses for labels
        for label, count in label_counts.items():
            if count > use_limit:
                is_valid = False

        if is_valid:
            # Update the maximum sum if the current combination is valid
            max_sum = max(max_sum, current_sum)

    return max_sum

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach considers every possible subset of the input values and labels. Given n items, there are 2^n possible subsets. For each subset, we iterate through its elements to check label limits and item limits, which takes at most O(n) time. Thus, the overall time complexity is O(n * 2^n). However, as 2^n dominates n, we can simplify this to O(2^n), representing exponential growth with respect to the input size n.

Space Complexity

O(1) – The brute force method, as described, explores all possible combinations of items. While it conceptually considers combinations, the provided description does not explicitly mention the creation of auxiliary data structures to store each combination. The algorithm keeps track of the 'largest sum seen so far', which uses constant space. Therefore, the auxiliary space complexity is O(1) because the algorithm only uses a constant amount of extra memory regardless of the input size.

Optimal Solution

Approach

The goal is to select items that give the largest total value, while respecting limits on the number of items and the usage of each label. We'll pick the items with the highest values first, but we must carefully track how many of each label we've used to avoid exceeding the allowed limits.

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

Combine the values and labels into pairs so we can keep track of which value goes with which label.
Sort these pairs from highest value to lowest value. This way, we consider the best options first.
Create a way to count how many times we've used each label.
Go through the sorted pairs. If we haven't used too many of a specific label, then add that item's value to our total and increase the count for that label.
If we've reached the limit of how many items we can pick, or if using the current item would exceed the usage limit for its label, skip that item and move to the next one.
Repeat until we've either picked the maximum number of items or have gone through all the value-label pairs.
The total value we've accumulated is the largest possible total that respects the limits.

Code Implementation

def largest_vals_from_labels(
        values, labels, num_wanted, use_limit):

    value_label_pairs = list(zip(values, labels))

    # Sort pairs by value in descending order to pick largest values first.
    value_label_pairs.sort(key=lambda x: x[0], reverse=True)

    label_counts = {}
    total_value = 0
    items_picked = 0

    for value, label in value_label_pairs:
        # Ensure we don't exceed the number of items we want.
        if items_picked == num_wanted:
            break

        # Ensure each label is used within its limit.
        if label not in label_counts:
            label_counts[label] = 0

        if label_counts[label] < use_limit:
            total_value += value
            label_counts[label] += 1
            items_picked += 1

    return total_value

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation is sorting the value-label pairs, which takes O(n log n) time, where n is the number of value-label pairs. We then iterate through the sorted pairs, checking the label counts. This iteration takes O(n) time. Since O(n log n) dominates O(n), the overall time complexity is O(n log n).

Space Complexity

O(N) – The algorithm creates pairs of values and labels, effectively creating a new list of size N, where N is the number of values (and labels) in the input. It also uses a dictionary (or hash map) to count the occurrences of each label; in the worst case, all labels are unique, leading to a dictionary of size N. Therefore, the auxiliary space used is proportional to the input size N, giving a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty values or labels array	Return 0 immediately as there are no values to select.
num_wanted is 0	Return 0 since no values are wanted.
use_limit is 0 for all labels	Return 0 as no label can be used even once.
values and labels arrays have different lengths	Throw an IllegalArgumentException or return an appropriate error code to indicate invalid input.
values array contains negative values	Handle negative values as valid values that can be selected based on the other constraints.
num_wanted is greater than the length of the values array	Return the sum of all valid values since you can select at most all the values.
Large input size: very long arrays for values and labels	Ensure the solution has O(n log n) time complexity for sorting, or potentially O(n) if using a selection algorithm, to avoid exceeding time limits.
Values are all the same, labels are also all the same, and use_limit is 1	Select only one value since the label limit is only one even though all values are identical.

Empty values or labels array

How to Handle:

Return 0 immediately as there are no values to select.

num_wanted is 0

How to Handle:

Return 0 since no values are wanted.

use_limit is 0 for all labels

How to Handle:

Return 0 as no label can be used even once.

values and labels arrays have different lengths

How to Handle:

Throw an IllegalArgumentException or return an appropriate error code to indicate invalid input.

values array contains negative values

How to Handle:

Handle negative values as valid values that can be selected based on the other constraints.

num_wanted is greater than the length of the values array

How to Handle:

Return the sum of all valid values since you can select at most all the values.

Large input size: very long arrays for values and labels

How to Handle:

Ensure the solution has O(n log n) time complexity for sorting, or potentially O(n) if using a selection algorithm, to avoid exceeding time limits.

Values are all the same, labels are also all the same, and use_limit is 1

How to Handle:

Select only one value since the label limit is only one even though all values are identical.