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Author: NamVr

2179-most-beautiful-item-for-each-query/README.md

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+<p>You are given a 2D integer array <code>items</code> where <code>items[i] = [price<sub>i</sub>, beauty<sub>i</sub>]</code> denotes the <strong>price</strong> and <strong>beauty</strong> of an item respectively.</p>
+
+<p>You are also given a <strong>0-indexed</strong> integer array <code>queries</code>. For each <code>queries[j]</code>, you want to determine the <strong>maximum beauty</strong> of an item whose <strong>price</strong> is <strong>less than or equal</strong> to <code>queries[j]</code>. If no such item exists, then the answer to this query is <code>0</code>.</p>
+
+<p>Return <em>an array </em><code>answer</code><em> of the same length as </em><code>queries</code><em> where </em><code>answer[j]</code><em> is the answer to the </em><code>j<sup>th</sup></code><em> query</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> items = [[1,2],[3,2],[2,4],[5,6],[3,5]], queries = [1,2,3,4,5,6]
+<strong>Output:</strong> [2,4,5,5,6,6]
+<strong>Explanation:</strong>
+- For queries[0]=1, [1,2] is the only item which has price &lt;= 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.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> items = [[1,2],[1,2],[1,3],[1,4]], queries = [1]
+<strong>Output:</strong> [4]
+<strong>Explanation:</strong> 
+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.  
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> items = [[10,1000]], queries = [5]
+<strong>Output:</strong> [0]
+<strong>Explanation:</strong>
+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.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= items.length, queries.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>items[i].length == 2</code></li>
+	<li><code>1 &lt;= price<sub>i</sub>, beauty<sub>i</sub>, queries[j] &lt;= 10<sup>9</sup></code></li>
+</ul>

2179-most-beautiful-item-for-each-query/solution.cpp

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+class Solution {
+public:
+    vector<int> maximumBeauty(vector<vector<int>>& items, vector<int>& queries) {
+        int n = queries.size();
+        vector<int> ans(n);
+
+        vector<pair<int,int>> queriesPair;
+
+        for (int i = 0; i < n; i++) {
+            queriesPair.push_back({queries[i], i});
+        }
+
+        sort(queriesPair.begin(), queriesPair.end());
+        sort(items.begin(), items.end());
+
+        int itemIndex = 0, maxBeauty = 0;
+        for (int i = 0; i < queriesPair.size(); i++) {
+            int maxPriceAllowed = queriesPair[i].first;
+            int queryOriginalIndex = queriesPair[i].second;
+            
+            while (itemIndex < items.size() && items[itemIndex][0] <= maxPriceAllowed) {
+                maxBeauty = max(maxBeauty, items[itemIndex][1]);
+                itemIndex++;
+            }
+            
+            ans[queryOriginalIndex] = maxBeauty;
+        }
+
+        return ans;
+    }
+};