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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.
using System.Collections.Generic;
public class Solution {
public int[] MaxBeautyWithCompression(int[][] items, int[] queries) {
var priceSet = new SortedSet<int>();
foreach (var item in items) priceSet.Add(item[0]);
foreach (var query in queries) priceSet.Add(query);
var prices = new List<int>(priceSet);
var dp = new Dictionary<int, int>();
foreach (var price in prices) dp[price] = 0;
foreach (var item in items) {
dp[item[0]] = Math.Max(dp[item[0]], item[1]);
}
int currentMax = 0;
foreach (var price in prices) {
currentMax = Math.Max(currentMax, dp[price]);
dp[price] = currentMax;
}
var result = new int[queries.Length];
for (int i = 0; i < queries.Length; i++) {
var index = prices.BinarySearch(queries[i]);
if (index < 0) index = ~index - 1;
result[i] = index >= 0 ? dp[prices[index]] : 0;
}
return result;
}
public static void Main(string[] args) {
var solution = new Solution();
int[][] items = new int[][] { new int[] {1,2}, new int[] {3,2}, new int[] {2,4}, new int[] {5,6}, new int[] {3,5} };
int[] queries = new int[] { 1, 2, 3, 4, 5, 6 };
int[] result = solution.MaxBeautyWithCompression(items, queries);
Console.WriteLine(string.Join(" ", result));
}
}
C# Hash and SortedSet are employed for index compression, allowing a reduced coordinate system upon which a dynamic array of maximum beauties can efficiently resolve queries using the BinarySearch feature for indexed price identification.