LeetCode in Kotlin
2070. Most Beautiful Item for Each Query
Medium
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 <= 105items[i].length == 21 <= pricei, beautyi, queries[j] <= 109
Solution
class Solution {
fun maximumBeauty(items: Array<IntArray>, queries: IntArray): IntArray {
val res = IntArray(queries.size)
items.sortWith(compareBy { a: IntArray -> a[1] })
for (i in res.indices) {
res[i] = maxBeauty(items, queries[i])
}
return res
}
private fun maxBeauty(items: Array<IntArray>, query: Int): Int {
for (i in items.indices.reversed()) {
val price = items[i][0]
val beauty = items[i][1]
if (price <= query) {
return beauty
}
}
return 0
}
}