8. K Closest Points To Origin (HEAP) C++ Solution Added

Heap/C++/8.K_Closet_Points_To_Origin.cpp

@@ -0,0 +1,138 @@
+/*
+-> Problem Statement
+You are given an array of points having x and y co-ordinates (vector of vectors) and an integer k, 
+You need to return k closest points to origin according to euclidean distance.
+
+-> Approach Explanation:
+- First for all points (vector having x and y) , we will map value of x^2 + y^2 with its key as point's index.
+  ( since we will be using int while comparing distance we are not taking square root of distance, 
+    since taking square root can not make comparision exact for int , and not taking sqrt will not affect while comparing
+- We need K closest points i.e. K points having shortest distance which can be found using priority_queue as minheap. )
+- Priority queue is considered as vector of vectors (each having distance and index taken from map),
+  priority_queue will keep vectors in assending order of distance as per given comparision function.
+  And we can get first k closest points from index inside first k priority_queue vectors. 
+*/
+
+#include <iostream>
+#include <map>
+#include <queue>
+using namespace std;
+
+class Solution {
+public:
+
+    bool operator()(vector<int> const& a, vector<int> const& b )     //comparision function to sort according to distance 
+    {
+        return a[0] > b[0];
+    }
+
+    vector<vector<int>> kClosest(vector<vector<int>>& points, int k) {
+
+        map<int,int>mp;
+        for(int i=0;i<points.size();i++)
+        {
+            mp[i] = (points[i][0] * points[i][0]) + (points[i][1] * points[i][1]); //mapping distances and index for points
+        }
+
+        priority_queue<vector<int>, vector<vector<int>>, Solution> pq;             // priority_queue as minheap(for distances)
+
+        for(auto i : mp)
+        {
+            vector<int>v;
+            v.push_back(i.second);
+            v.push_back(i.first);
+            pq.push(v);                                             // pushing distances and indexes from maps to priority_queue
+        }
+
+        vector<vector<int>>ans;
+
+        while(k--)
+        {
+            vector<int>top = pq.top();                              // getting first k vectors having shortest distances
+            vector<int>xy = points[top[1]];                         // getting corrsponding points from index
+            ans.push_back(xy);                                      
+            pq.pop();
+        }
+
+        return ans;                                                 // returning final vector having k closest points from origin
+
+    }
+};
+
+// It can also be easily solved with similar logic just by using vector of vectors or map having distances and indices in sorted manner.
+
+int main() 
+{
+    cout<<"\nEnter number of test cases: ";
+	int t; cin>>t;  //no. of test cases
+
+	/*
+	Constraints:
+    1 <= k <= points array length <= 10^4
+   -10^4 < x,y values < 10^4
+    */
+
+	while (t--)
+	{
+	    cout<<"\n\nEnter number of points you have: ";
+	    int n; cin>>n;
+
+	    vector<vector<int>> points (n);
+
+	    cout<<"\nEnter points as x and y co-ordinates in array : ";
+	    for (int i=0; i<n; i++)
+           {
+                cout<<"\nPoint :"<<i + 1<<endl;
+                cout<<"\nx :";
+                cin>>points[i][0];
+                cout<<"\ny :";
+                cin>>points[i][1];
+           }
+
+        cout<<"\nEnter value of K (No of Closest points you want): ";
+        int k;cin>>k;
+
+	    Solution *ob;
+        cout<<"\nK Closest Poits to Origin are :";
+	    vector<vector<int>> closestpoints =  ob-> kClosest(points,k);
+        for (int i=0; i<k; i++)
+           {
+                cout<<"\nPoint :"<<i + 1<<endl;
+                cout<<"\nx :";
+                cin>>points[i][0];
+                cout<<"\ny :";
+                cin>>points[i][1];
+           }
+
+	}
+	return 0;
+}
+
+/*
+SAMPLE INPUT
+2				         // number of test cases 
+3				         // no.of points for test case:1
+4 3 	                 // points as x and y in vector for test case:1
+-1 2
+-5 3
+2                        // value of k
+5				         // no.of pointsfor test case:2
+-2 5                     // points as x and y in vector for test case:2
+0 0 
+1 -6
+7 -4  
+3                        // value of k
+
+SAMPLE OUTPUT (excluding interactive instructions)
+-1 2    ( 2 clost points to origin for test case-1 )
+4 3
+0 0    ( 3 closest points to origin for test case-2 )
+-2 5 
+1 -6  
+COMPLEXITY ANALYSIS : N= no.of points in array
+Time : O(N) to iterate array/vector and insert in map  . 
+       O(NLogN) for creating minheap or inserting elements in priority_que
+       Hence, O(NlogN) overall.
+Space: O(N) for inputs and aux space: O(N) to create a priority_queue.
+       Hence, O(N) overall.
+*/