greedy medium sorting dec order 2d array

2070. Most Beautiful Item for Each Query
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Medium
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Hint
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

Seen this question in a real interview before?
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Accepted
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Submissions
187.8K
Acceptance Rate
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Hint 1
Can we process the queries in a smart order to avoid repeatedly checking the same items?
Hint 2
How can we use the answer to a query for other queries?

Author: Gauranshgoel123

most-beautiful-item-for-each-query-optimized.java

@@ -0,0 +1,24 @@
+class Solution {
+    public int[] maximumBeauty(int[][] items, int[] queries) {
+        Arrays.sort(items,(a,b)->b[1]-a[1]);
+        //items = {{5, 6}, {3, 5}, {2, 4}, {1, 2}, {3, 2}};
+        // sorts a 2D array (items) in descending order based on the second element ([1]) of each sub-array.
+        int[] ans=new int[queries.length];
+        for(int i=0;i<queries.length;i++){
+            boolean flag=false;
+            for(int j=0;j<items.length;j++){
+                int p=items[j][0];
+                int b=items[j][1];
+                if(p<=queries[i]){
+                    flag=true;
+                    ans[i]=b;
+                    break;
+                }
+            }
+            if(flag==false){
+                ans[i]=0;
+            }
+        }
+        return ans;
+    }
+}