LeetCode 2819: Minimum Relative Loss After Buying Chocolates Solution

Master LeetCode problem 2819 (Minimum Relative Loss After Buying Chocolates), 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.

2819. Minimum Relative Loss After Buying Chocolates

Hard

Array Binary Search Sorting Prefix Sum

Problem Explanation

Explanation

To solve this problem, we can use a priority queue (min heap) to keep track of the potential minimum relative loss for each chocolate. We start by iterating over the prices array and adding each pair of prices with their corresponding indices to the priority queue. As we iterate, we check if the current price is less than the top price in the priority queue. If it is, we calculate the relative loss and update the minimum relative loss if necessary.

After processing all prices, we return the minimum relative loss.

Algorithmic Steps:

Initialize a priority queue (min heap) to store prices with their indices.
Iterate over the prices array and add each pair of price and index to the priority queue.
For each price, check if it is less than the top price in the priority queue.
If it is, calculate the relative loss and update the minimum relative loss if necessary.
Return the minimum relative loss after processing all prices.

Time Complexity

The time complexity of this approach is O(n log n), where n is the number of prices in the input array. This is due to the insertion and extraction operations on the priority queue.

Space Complexity

The space complexity is O(n) to store the priority queue and the input prices array.

Solution Code

import java.util.*;

class Solution {
    public int minimumLoss(long[] prices) {
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> Long.compare(b[0], a[0]));
        for (int i = 0; i < prices.length; i++) {
            pq.offer(new int[]{prices[i], i});
        }
        
        int minLoss = Integer.MAX_VALUE;
        while (pq.size() > 1) {
            int[] curr = pq.poll();
            int[] next = pq.peek();
            if (curr[1] < next[1]) {
                minLoss = Math.min(minLoss, (int) (curr[0] - next[0]));
            }
        }
        
        return minLoss;
    }
}

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