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2819. Minimum Relative Loss After Buying Chocolates
Description
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.
The terms for each query are as follows:
- If the price of a chocolate is less than or equal to
ki, Bob pays for it. - Otherwise, Bob pays
kiof it, and Alice pays the rest.
Bob wants to select exactly mi chocolates such that his relative loss is minimized, more formally, if, in total, Alice has paid ai and Bob has paid bi, Bob wants to minimize bi - ai.
Return an integer array ans where ans[i] is Bob's minimum relative loss possible for queries[i].
Example 1:
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.
Example 2:
Input: prices = [1,5,4,3,7,11,9], queries = [[5,4],[5,7],[7,3],[4,5]] Output: [4,16,7,1] Explanation: For the 1st query Bob selects the chocolates with prices [1,3,9,11]. He pays 1 + 3 + 5 + 5 = 14 and Alice pays 0 + 0 + 4 + 6 = 10. So Bob's relative loss is 14 - 10 = 4. For the 2nd query Bob has to select all the chocolates. He pays 1 + 5 + 4 + 3 + 5 + 5 + 5 = 28 and Alice pays 0 + 0 + 0 + 0 + 2 + 6 + 4 = 12. So Bob's relative loss is 28 - 12 = 16. For the 3rd query Bob selects the chocolates with prices [1,3,11] and he pays 1 + 3 + 7 = 11 and Alice pays 0 + 0 + 4 = 4. So Bob's relative loss is 11 - 4 = 7. For the 4th query Bob selects the chocolates with prices [1,3,7,9,11] and he pays 1 + 3 + 4 + 4 + 4 = 16 and Alice pays 0 + 0 + 3 + 5 + 7 = 15. So Bob's relative loss is 16 - 15 = 1. It can be shown that these are the minimum possible relative losses.
Example 3:
Input: prices = [5,6,7], queries = [[10,1],[5,3],[3,3]] Output: [5,12,0] Explanation: For the 1st query Bob selects the chocolate with price 5 and he pays 5 and Alice pays 0. So Bob's relative loss is 5 - 0 = 5. For the 2nd query Bob has to select all the chocolates. He pays 5 + 5 + 5 = 15 and Alice pays 0 + 1 + 2 = 3. So Bob's relative loss is 15 - 3 = 12. For the 3rd query Bob has to select all the chocolates. He pays 3 + 3 + 3 = 9 and Alice pays 2 + 3 + 4 = 9. So Bob's relative loss is 9 - 9 = 0. It can be shown that these are the minimum possible relative losses.
Constraints:
1 <= prices.length == n <= 1051 <= prices[i] <= 1091 <= queries.length <= 105queries[i].length == 21 <= ki <= 1091 <= mi <= n
Solutions
Solution 1: Sorting + Binary Search + Prefix Sum
Based on the problem description, we know:
If $prices[i] \leq k$, then Bob needs to pay $prices[i]$, and Alice doesn’t need to pay. Therefore, Bob’s relative loss is $prices[i]$. In this case, Bob should choose the chocolate with a lower price to minimize the relative loss.
If $prices[i] > k$, then Bob needs to pay $k$, and Alice needs to pay $prices[i] - k$. Therefore, Bob’s relative loss is $k - (prices[i] - k) = 2k - prices[i]$. In this case, Bob should choose the chocolate with a higher price to minimize the relative loss.
Therefore, we first sort the price array $prices$, and then preprocess the prefix sum array $s$, where $s[i]$ represents the sum of the prices of the first $i$ chocolates.
Next, for each query $[k, m]$, we first use binary search to find the index $r$ of the first chocolate with a price greater than $k$. Then, we use binary search again to find the number of chocolates $l$ that should be chosen on the left, so the number of chocolates that should be chosen on the right is $m - l$. At this point, Bob’s relative loss is $s[l] + 2k(m - l) - (s[n] - s[n - (m - l)])$.
In the second binary search process mentioned above, we need to judge whether $prices[mid] < 2k - prices[n - (m - mid)]$, where $right$ represents the number of chocolates that should be chosen on the right. If this inequality holds, it means that choosing the chocolate at position $mid$ has a lower relative loss, so we update $l = mid + 1$. Otherwise, it means that the chocolate at position $mid$ has a higher relative loss, so we update $r = mid$.
The time complexity is $O((n + m) \times \log n)$, and the space complexity is $O(n)$. Where $n$ and $m$ are the lengths of the arrays $prices$ and $queries$, respectively.
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class Solution { private int n; private int[] prices; public long[] minimumRelativeLosses(int[] prices, int[][] queries) { n = prices.length; Arrays.sort(prices); this.prices = prices; long[] s = new long[n + 1]; for (int i = 0; i < n; ++i) { s[i + 1] = s[i] + prices[i]; } int q = queries.length; long[] ans = new long[q]; for (int i = 0; i < q; ++i) { int k = queries[i][0], m = queries[i][1]; int l = f(k, m); int r = m - l; ans[i] = s[l] + 2L * k * r - (s[n] - s[n - r]); } return ans; } private int f(int k, int m) { int l = 0, r = Arrays.binarySearch(prices, k); if (r < 0) { r = -(r + 1); } r = Math.min(m, r); while (l < r) { int mid = (l + r) >> 1; int right = m - mid; if (prices[mid] < 2L * k - prices[n - right]) { l = mid + 1; } else { r = mid; } } return l; } } -
class Solution { public: vector<long long> minimumRelativeLosses(vector<int>& prices, vector<vector<int>>& queries) { int n = prices.size(); sort(prices.begin(), prices.end()); long long s[n + 1]; s[0] = 0; for (int i = 1; i <= n; ++i) { s[i] = s[i - 1] + prices[i - 1]; } auto f = [&](int k, int m) { int l = 0, r = upper_bound(prices.begin(), prices.end(), k) - prices.begin(); r = min(r, m); while (l < r) { int mid = (l + r) >> 1; int right = m - mid; if (prices[mid] < 2LL * k - prices[n - right]) { l = mid + 1; } else { r = mid; } } return l; }; vector<long long> ans; for (auto& q : queries) { int k = q[0], m = q[1]; int l = f(k, m); int r = m - l; ans.push_back(s[l] + 2LL * k * r - (s[n] - s[n - r])); } return ans; } }; -
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 -
func minimumRelativeLosses(prices []int, queries [][]int) []int64 { n := len(prices) sort.Ints(prices) s := make([]int, n+1) for i, x := range prices { s[i+1] = s[i] + x } f := func(k, m int) int { l, r := 0, sort.Search(n, func(i int) bool { return prices[i] > k }) if r > m { r = m } for 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 } ans := make([]int64, len(queries)) for i, q := range queries { k, m := q[0], q[1] l := f(k, m) r := m - l ans[i] = int64(s[l] + 2*k*r - (s[n] - s[n-r])) } return ans } -
function minimumRelativeLosses(prices: number[], queries: number[][]): number[] { const n = prices.length; prices.sort((a, b) => a - b); const s: number[] = Array(n).fill(0); for (let i = 0; i < n; ++i) { s[i + 1] = s[i] + prices[i]; } const search = (x: number): number => { let l = 0; let r = n; while (l < r) { const mid = (l + r) >> 1; if (prices[mid] > x) { r = mid; } else { l = mid + 1; } } return l; }; const f = (k: number, m: number): number => { let l = 0; let r = Math.min(search(k), m); while (l < r) { const mid = (l + r) >> 1; const right = m - mid; if (prices[mid] < 2 * k - prices[n - right]) { l = mid + 1; } else { r = mid; } } return l; }; const ans: number[] = []; for (const [k, m] of queries) { const l = f(k, m); const r = m - l; ans.push(s[l] + 2 * k * r - (s[n] - s[n - r])); } return ans; }