2070. Most Beautiful Item for Each Query
Difficulty: Medium
Topics: Array
, Binary Search
, Sorting
You are given a 2D integer array items
where items[i] = [pricei, beautyi]
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 jth
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 <= pricei, beautyi, queries[j] <= 109
Hint:
- Can we process the queries in a smart order to avoid repeatedly checking the same items?
- How can we use the answer to a query for other queries?
Solution:
We can use sorting and binary search techniques. Here’s the plan:
Approach
-
Sort the Items by Price:
- First, sort
items
by theirprice
. This way, as we iterate through the items, we can keep track of the maximum beauty seen so far for items up to any given price.
- First, sort
-
Sort the Queries with their Original Indices:
- Create an array of queries paired with their original indices, then sort this array by the query values.
- Sorting helps because we can process queries in increasing order of
price
and avoid recalculating beauty values for lower prices repeatedly.
-
Iterate through Items and Queries Simultaneously:
- Using two pointers, process each query:
- For each query, move the pointer through items with a price less than or equal to the query’s price.
- Track the maximum beauty as you go through these items, and use this value to answer the current query.
- This avoids repeatedly checking items for multiple queries.
- Using two pointers, process each query:
-
Store and Return Results:
- Once processed, store the maximum beauty result for each query based on the original index to maintain the order.
- Return the
answer
array.
Let's implement this solution in PHP: 2070. Most Beautiful Item for Each Query
<?php
/**
* @param Integer[][] $items
* @param Integer[] $queries
* @return Integer[]
*/
function maximumBeauty($items, $queries) {
...
...
...
/**
* go to ./solution.php
*/
}
// Example usage
$items = [[1,2],[3,2],[2,4],[5,6],[3,5]];
$queries = [1,2,3,4,5,6];
print_r(maximumBeauty($items, $queries));
// Output: [2,4,5,5,6,6]
?>
Explanation:
-
Sorting
items
andqueries
: This allows efficient processing without redundant calculations. -
Two-pointer technique: Moving through
items
only once for each query avoids excessive computations. -
Track
maxBeauty
: We updatemaxBeauty
progressively, allowing each query to access the highest beauty seen so far.
Complexity
-
Time Complexity: O(n log n + m log m) for sorting
items
andqueries
, and O(n + m) for processing, where n is the length ofitems
and m is the length ofqueries
. - Space Complexity: O(m) for storing the results.
This solution is efficient and meets the constraints of the problem.
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