2179. Most Beautiful Item For Each Query¶

Difficulty: Medium

2179. Most Beautiful Item for Each Query

Medium

You are given a 2D integer array items where items[i] = [price_i, beauty_i] denotes the price and beauty of an item respectively.

You are also given a 0-indexed integer array queries. For each queries[j], you want to determine the maximum beauty of an item whose price is less than or equal to queries[j]. If no such item exists, then the answer to this query is 0.

Return an array answer of the same length as queries where answer[j] is the answer to the j^th query.

Example 1:

Input: items = [[1,2],[3,2],[2,4],[5,6],[3,5]], queries = [1,2,3,4,5,6]
Output: [2,4,5,5,6,6]
Explanation:
- For queries[0]=1, [1,2] is the only item which has price <= 1. Hence, the answer for this query is 2.
- For queries[1]=2, the items which can be considered are [1,2] and [2,4]. 
  The maximum beauty among them is 4.
- For queries[2]=3 and queries[3]=4, the items which can be considered are [1,2], [3,2], [2,4], and [3,5].
  The maximum beauty among them is 5.
- For queries[4]=5 and queries[5]=6, all items can be considered.
  Hence, the answer for them is the maximum beauty of all items, i.e., 6.

Example 2:

Input: items = [[1,2],[1,2],[1,3],[1,4]], queries = [1]
Output: [4]
Explanation: 
The price of every item is equal to 1, so we choose the item with the maximum beauty 4. 
Note that multiple items can have the same price and/or beauty.

Example 3:

Input: items = [[10,1000]], queries = [5]
Output: [0]
Explanation:
No item has a price less than or equal to 5, so no item can be chosen.
Hence, the answer to the query is 0.

Constraints:

1 <= items.length, queries.length <= 10⁵
items[i].length == 2
1 <= price_i, beauty_i, queries[j] <= 10⁹

Solution¶

class Solution {
    static class Pair {
        int price;
        int beauty;
        public Pair(int price, int beauty) {
            this.price = price;
            this.beauty = beauty;
        }
        @Override
        public String toString() {
            return "(" + price + " " + beauty + ")";
        }
    }
    static class custom_sort implements Comparator<Pair> {
        @Override
        public int compare(Pair first, Pair second) {
            int op1 = Integer.compare(first.price, second.price);
            return op1;
        }
    }
    public int[] maximumBeauty(int[][] items, int[] queries) {
        int n = items.length;
        ArrayList<Pair> res = new ArrayList<>();
        for (int currItem[] : items) {
            res.add(new Pair(currItem[0] , currItem[1]));
        }
        Collections.sort(res, new custom_sort());
        int maxBeautyPref[] = new int[n + 1];
        int maxi = Integer.MIN_VALUE;
        for (int i = 0; i < res.size(); i++) {
            maxi = Math.max(maxi, res.get(i).beauty);
            maxBeautyPref[i] = maxi;
        }
        int answer[] = new int[queries.length];
        int k = 0;
        for (int current_query : queries) {
            answer[k++] = binary_search(res, current_query, maxBeautyPref);
        }
        return answer;
    }

    static int binary_search(ArrayList<Pair> res, int current_price, int maxBeautyPref[]) {
        int n = res.size();
        int maxi = 0;
        int low = 0;
        int high = n - 1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (res.get(mid).price > current_price) high = mid - 1;
            else if (res.get(mid).price <= current_price) {
                maxi = Math.max(maxi, maxBeautyPref[mid]);
                low = mid + 1;
            }
        }
        return maxi;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here