Merge pull request #268 from ShwetKhatri2001/6-heap

6. Reduce Array Size To The Half (HEAP) C++ Solution

Author: Debashish-hub

Heap/C++/6.Reduse_Array_Size_To_Half.cpp

@@ -0,0 +1,102 @@
+/*
+-> Problem Statement
+From given integer Array Choose a set of integers and remove all the occurrences of these integers in the array.
+You need to find the minimum size of that set such that at least half of the integers of the array are removed.
+
+-> Approach Explanation:
+- First save each integer present in array and its no of occurrences in array using map 
+- As we need to find minimum size of the set we will keep removing elements having maximum occurrences to make its size less than or equal to half
+- For finding elements having maximum occurrences we will use heap (priority_queue or maxheap) which will be take occurrences from map
+  and will return in descending order , so that we can remove at least half elements from array using minimum set of imtegers
+*/
+
+#include <iostream>
+#include <map>
+#include <queue>
+using namespace std;
+
+class Solution {
+public:
+    
+    int minSetSize(vector<int>& arr) {
+        
+    map<int,int>mp;
+    for(int i=0;i<arr.size();i++)
+    {
+        mp[arr[i]]++;                  // mapping integers and frequencies of integers
+    }
+    
+    priority_queue<int>pq;            
+    
+    for(auto i : mp)
+    {
+        pq.push(i.second);            // frequencies from map are pushed in priority_queue
+    }
+        
+    int remove_count = 0;             // required count of integers in set, for which all occurrences should be removed 
+    int currsize = arr.size();
+    
+    while(currsize >  arr.size()/2)
+    {
+        int top = pq.top();          // getting max occurrence in current priority queue
+        currsize -= top;             // removing all occurrences 
+        pq.pop();                    // removing element having max occurrence in current priority queue
+        remove_count ++;             // increasing count since we have got one more element for that set 
+        
+    }
+
+  return remove_count;               // returning final count of integers present in set, for which all occurrences should be removed 
+        
+  }
+};
+
+int main() 
+{
+    cout<<"\nEnter number of test cases: ";
+	int t; cin>>t;  //no. of test cases
+
+	/*
+	Constraints:
+    1 <= arr.length <= 105
+    arr's length is even.
+    1 <= arr[i] <= 105
+    */
+
+	while (t--)
+	{
+	    cout<<"\n\nEnter number of elements in array: ";
+	    int n; cin>>n;
+
+	    vector<int> arr (n);
+
+	    cout<<"\nEnter integers in array ( Input ): ";
+	    for (int i=0; i<n; i++)
+	        cin>>arr[i];
+
+	    Solution *ob;
+        cout<<"\nMinimum elements in set required to remove atleast half of the elements are :";
+	    cout <<  ob-> minSetSize(arr);
+	    
+	}
+	return 0;
+}
+
+/*
+SAMPLE INPUT
+2				         // number of test cases 
+6				         // size of the array for test case:1
+7 3 4 3 3 4  	         // integers array / vector for test case:1
+10				         // size of the array for test case:2
+5 5 5 4 4 5 1 2 2 4      // integers array / vector for test case:2
+ 
+SAMPLE OUTPUT (excluding interactive instructions)
+1   ( for test case-1 only by removing occurrences of 1 element(i.e 3) array will be of half in size )
+2   ( for test case-2 we need to remove occurrences of atleast 2 elements (i.e. {5,4} or {5,2}) to make array of size <= half )
+
+COMPLEXITY ANALYSIS : N= no.of cards in desk 
+Time : O(N) to iterate array/vector and save frequencies or . 
+       O(NLogN) for creating maxheap or inserting elements in priority_que
+       Hence, O(NlogN) overall.
+Space: O(N) for inputs and aux space: O(N) to re-create a queue.
+       Hence, O(N) overall.
+*/