Sponsored
Sponsored
In this approach, we perform the following steps:
Time Complexity: O((n + m) log n) where n = number of items and m = number of queries.
Space Complexity: O(n) for storing the processed items data.
1import java.util.*;
2
3class Solution {
4 public int[] maxBeautyForQueries(int[][] items, int[] queries) {
5 Arrays.sort(items, (a, b) -> a[0] - b[0]);
6 int[] maxBeauties = new int[items.length];
7 int maxBeauty = 0;
8
9 for (int i = 0; i < items.length; i++) {
10 maxBeauty = Math.max(maxBeauty, items[i][1]);
11 maxBeauties[i] = maxBeauty;
12 }
13
14 int[] result = new int[queries.length];
15 for (int i = 0; i < queries.length; i++) {
16 int idx = Arrays.binarySearch(items, new int[]{queries[i], Integer.MAX_VALUE}, Comparator.comparingInt(a -> a[0]));
17 if (idx < 0) idx = -idx - 2;
18 result[i] = idx >= 0 ? maxBeauties[idx] : 0;
19 }
20
21 return result;
22 }
23
24 public static void main(String[] args) {
25 Solution sol = new Solution();
26 int[][] items = {{1,2}, {3,2}, {2,4}, {5,6}, {3,5}};
27 int[] queries = {1, 2, 3, 4, 5, 6};
28 int[] result = sol.maxBeautyForQueries(items, queries);
29 System.out.println(Arrays.toString(result));
30 }
31}This Java solution operates by first sorting the items by their prices then iteratively building a list of maximum beauties at each price point. The binary search facilitates efficient queries, dramatically reducing potential computational overhead.
Coordinate compression can be used to reduce problem complexity when dealing with large range values like prices. In this approach:
Time Complexity: O((n + m) log(n + m)) due to sorting during compression.
Space Complexity: O(n + m) for storing all price points and decision arrays.
This solution first compresses the range of prices into indexes, reducing the search space. A dynamic programming approach then determines the maximum beauty at each possible price point index, allowing quick lookup when answering queries. This methodology deals efficiently with large query ranges.