Leetcode #2819: Minimum Relative Loss After Buying Chocolates
In this guide, we solve Leetcode #2819 Minimum Relative Loss After Buying Chocolates in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an integer array prices, which shows the chocolate prices and a 2D integer array queries, where queries[i] = [ki, mi]. Alice and Bob went to buy some chocolates, and Alice suggested a way to pay for them, and Bob agreed.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Array, Binary Search, Prefix Sum, Sorting
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: prices = [1,9,22,10,19], queries = [[18,4],[5,2]]
Output: [34,-21]
Explanation: For the 1st query Bob selects the chocolates with prices [1,9,10,22]. He pays 1 + 9 + 10 + 18 = 38 and Alice pays 0 + 0 + 0 + 4 = 4. So Bob's relative loss is 38 - 4 = 34.
For the 2nd query Bob selects the chocolates with prices [19,22]. He pays 5 + 5 = 10 and Alice pays 14 + 17 = 31. So Bob's relative loss is 10 - 31 = -21.
It can be shown that these are the minimum possible relative losses.
Python Solution
class Solution:
def minimumRelativeLosses(
self, prices: List[int], queries: List[List[int]]
) -> List[int]:
def f(k: int, m: int) -> int:
l, r = 0, min(m, bisect_right(prices, k))
while l < r:
mid = (l + r) >> 1
right = m - mid
if prices[mid] < 2 * k - prices[n - right]:
l = mid + 1
else:
r = mid
return l
prices.sort()
s = list(accumulate(prices, initial=0))
ans = []
n = len(prices)
for k, m in queries:
l = f(k, m)
r = m - l
loss = s[l] + 2 * k * r - (s[n] - s[n - r])
ans.append(loss)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.