Description#
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
Solutions#
Solution 1#
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| class Solution:
def maximumBeauty(self, items: List[List[int]], queries: List[int]) -> List[int]:
items.sort()
prices = [p for p, _ in items]
mx = [items[0][1]]
for _, b in items[1:]:
mx.append(max(mx[-1], b))
ans = [0] * len(queries)
for i, q in enumerate(queries):
j = bisect_right(prices, q)
if j:
ans[i] = mx[j - 1]
return ans
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| class Solution {
public int[] maximumBeauty(int[][] items, int[] queries) {
Arrays.sort(items, (a, b) -> a[0] - b[0]);
for (int i = 1; i < items.length; ++i) {
items[i][1] = Math.max(items[i - 1][1], items[i][1]);
}
int n = queries.length;
int[] ans = new int[n];
for (int i = 0; i < n; ++i) {
int left = 0, right = items.length;
while (left < right) {
int mid = (left + right) >> 1;
if (items[mid][0] > queries[i]) {
right = mid;
} else {
left = mid + 1;
}
}
if (left > 0) {
ans[i] = items[left - 1][1];
}
}
return ans;
}
}
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| class Solution {
public:
vector<int> maximumBeauty(vector<vector<int>>& items, vector<int>& queries) {
sort(items.begin(), items.end());
for (int i = 1; i < items.size(); ++i) items[i][1] = max(items[i - 1][1], items[i][1]);
int n = queries.size();
vector<int> ans(n);
for (int i = 0; i < n; ++i) {
int left = 0, right = items.size();
while (left < right) {
int mid = (left + right) >> 1;
if (items[mid][0] > queries[i])
right = mid;
else
left = mid + 1;
}
if (left) ans[i] = items[left - 1][1];
}
return ans;
}
};
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| func maximumBeauty(items [][]int, queries []int) []int {
sort.Slice(items, func(i, j int) bool {
return items[i][0] < items[j][0]
})
for i := 1; i < len(items); i++ {
items[i][1] = max(items[i-1][1], items[i][1])
}
n := len(queries)
ans := make([]int, n)
for i, v := range queries {
left, right := 0, len(items)
for left < right {
mid := (left + right) >> 1
if items[mid][0] > v {
right = mid
} else {
left = mid + 1
}
}
if left > 0 {
ans[i] = items[left-1][1]
}
}
return ans
}
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