New Problem Solution - "1828. Queries on Number of Points Inside a Ci…

…rcle"

README.md

@@ -9,6 +9,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|1828|[Queries on Number of Points Inside a Circle](https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/) | [C++](./algorithms/cpp/queriesOnNumberOfPointsInsideACircle/QueriesOnNumberOfPointsInsideACircle.cpp)|Medium|
 |1827|[Minimum Operations to Make the Array Increasing](https://leetcode.com/problems/minimum-operations-to-make-the-array-increasing/) | [C++](./algorithms/cpp/minimumOperationsToMakeTheArrayIncreasing/MinimumOperationsToMakeTheArrayIncreasing.cpp)|Easy|
 |1825|[Finding MK Average](https://leetcode.com/problems/finding-mk-average/) | [C++](./algorithms/cpp/findingMkAverage/FindingMkAverage.cpp)|Hard|
 |1824|[Minimum Sideway Jumps](https://leetcode.com/problems/minimum-sideway-jumps/) | [C++](./algorithms/cpp/minimumSidewayJumps/MinimumSidewayJumps.cpp)|Medium|

algorithms/cpp/queriesOnNumberOfPointsInsideACircle/QueriesOnNumberOfPointsInsideACircle.cpp

@@ -0,0 +1,72 @@
+// Source : https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/
+// Author : Hao Chen
+// Date   : 2021-04-20
+
+/***************************************************************************************************** 
+ *
+ * You are given an array points where points[i] = [xi, yi] is the coordinates of the i^th point on a 
+ * 2D plane. Multiple points can have the same coordinates.
+ * 
+ * You are also given an array queries where queries[j] = [xj, yj, rj] describes a circle centered at 
+ * (xj, yj) with a radius of rj.
+ * 
+ * For each query queries[j], compute the number of points inside the j^th circle. Points on the 
+ * border of the circle are considered inside.
+ * 
+ * Return an array answer, where answer[j] is the answer to the j^th query.
+ * 
+ * Example 1:
+ * 
+ * Input: points = [[1,3],[3,3],[5,3],[2,2]], queries = [[2,3,1],[4,3,1],[1,1,2]]
+ * Output: [3,2,2]
+ * Explanation: The points and circles are shown above.
+ * queries[0] is the green circle, queries[1] is the red circle, and queries[2] is the blue circle.
+ * 
+ * Example 2:
+ * 
+ * Input: points = [[1,1],[2,2],[3,3],[4,4],[5,5]], queries = [[1,2,2],[2,2,2],[4,3,2],[4,3,3]]
+ * Output: [2,3,2,4]
+ * Explanation: The points and circles are shown above.
+ * queries[0] is green, queries[1] is red, queries[2] is blue, and queries[3] is purple.
+ * 
+ * Constraints:
+ * 
+ * 	1 <= points.length <= 500
+ * 	points[i].length == 2
+ * 	0 <= x​​​​​​i, y​​​​​​i <= 500
+ * 	1 <= queries.length <= 500
+ * 	queries[j].length == 3
+ * 	0 <= xj, yj <= 500
+ * 	1 <= rj <= 500
+ * 	All coordinates are integers.
+ * 
+ * Follow up: Could you find the answer for each query in better complexity than O(n)?
+ ******************************************************************************************************/
+
+class Solution {
+private:
+    //refer to: https://stackoverflow.com/a/7227057/13139221
+    bool inCircle( vector<int>& point,  vector<int>& circle ) {
+        int x = point[0], y = point[1];
+        int xo = circle[0], yo = circle[1], r = circle[2];
+
+        int dx = abs(x-xo);
+        if ( dx > r ) return false;
+        int dy = abs(y-yo);
+        if ( dy > r ) return false;
+        if ( dx + dy <= r ) return true;
+        return ( dx*dx + dy*dy <= r*r );
+    }
+public:
+    vector<int> countPoints(vector<vector<int>>& points, vector<vector<int>>& queries) {
+        vector<int> result;
+        for(auto& c : queries) {
+            int cnt = 0;
+            for(auto& p : points) {
+                if ( inCircle(p, c) ) cnt++;
+            }
+            result.push_back(cnt);
+        }
+        return result;
+    }
+};