26. Fractional Knapsack

The problem can be found at the following link: Question Link

My Approach

• I implemented a greedy strategy to maximize the value per weight.

• I sorted the items in descending order based on their value-to-weight ratios.

• I iterated through the sorted items, adding them to the knapsack until it's full or there are no more items.

Time and Auxiliary Space Complexity

• Time Complexity: O(n log n) - Sorting the items.

• Auxiliary Space Complexity: O(1) - No additional space is used.

Code (C++)

class Solution {
public:
    double fractionalKnapsack(int W, Item arr[], int n) {
        sort(arr, arr + n, [](auto &a, auto &b) {
            return a.value * b.weight > b.value * a.weight;
        });

        double value = 0;

        // Filling the knapsack
        for (int i = 0; i < n; i++) {
            auto &item = arr[i];

            if (item.weight <= W) {
                value += item.value;
                W -= item.weight;
            } else {
                value += double(item.value) * W / item.weight;
                break;
            }
        }

        return value;
    }
};

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