LeetCode 3052: Maximize Items Solution

Master LeetCode problem 3052 (Maximize Items), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3052. Maximize Items

Hard

Database

Problem Explanation

Explanation:

To solve this problem, we can use a greedy approach. We can sort the items based on their value-to-weight ratio and then start picking items starting from the one with the highest ratio until we exceed the weight limit. This ensures that we pick the items with the maximum value possible.

Sort the items based on their value-to-weight ratio in descending order.
Initialize a variable maxValue = 0 to keep track of the total value of picked items.
Iterate through the sorted items and pick items until the weight limit is exceeded.
Calculate the total value of the picked items.

Time Complexity: O(n log n) - Sorting the items based on their ratio. Space Complexity: O(1) - Constant extra space is used.

Solution Code

import java.util.Arrays;

class Item {
    int weight;
    int value;

    public Item(int weight, int value) {
        this.weight = weight;
        this.value = value;
    }
}

public class MaximizeItems {
    public static int maximizeItems(Item[] items, int weightLimit) {
        Arrays.sort(items, (a, b) -> Double.compare((double) b.value / b.weight, (double) a.value / a.weight));
        int maxValue = 0;
        int currentWeight = 0;

        for (Item item : items) {
            if (currentWeight + item.weight <= weightLimit) {
                maxValue += item.value;
                currentWeight += item.weight;
            } else {
                int remainingWeight = weightLimit - currentWeight;
                maxValue += item.value * ((double) remainingWeight / item.weight);
                break;
            }
        }

        return maxValue;
    }
}

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