From 3d7e00321286a1d1c5ce4b9bad96ae95a6380a8c Mon Sep 17 00:00:00 2001
From: ShwetKhatri2001 <19dcs001@lnmiit.ac.in>
Date: Tue, 3 Aug 2021 11:53:21 +0530
Subject: [PATCH] added c++ solution for 10th question in heap

---
 .../10.Maximum_Score_From_Removing_Stones.cpp | 88 +++++++++++++++++++
 1 file changed, 88 insertions(+)
 create mode 100644 Heap/C++/10.Maximum_Score_From_Removing_Stones.cpp

diff --git a/Heap/C++/10.Maximum_Score_From_Removing_Stones.cpp b/Heap/C++/10.Maximum_Score_From_Removing_Stones.cpp
new file mode 100644
index 0000000..14a391e
--- /dev/null
+++ b/Heap/C++/10.Maximum_Score_From_Removing_Stones.cpp
@@ -0,0 +1,88 @@
+/*
+-> Problem Statement
+- You are given three piles a,b and c of stones, Each turn you choose two different nonempty piles, take one stone from each and add 1 point to your score.
+- You need to return maximum score you can get.
+
+-> Approach Explanation:
+- We need to find maximum moves we can perform where each move is taking two stones from two nonempty piles.
+- Two maximize these moves we need to keep removing stones from piles having maximum and second maximum stones at each move. 
+- To find max and second max values out of a,b and c everytime we will use maxheap(priority_queue).
+- We can keep poping maximum elements from priority_queue and push elements having 1 less value > 0 which will again priortize elements at that stage. 
+- We follow above step till priority_queue contains >=2 elements and increse count for moves everytime.
+*/
+
+#include <iostream>
+#include <map>
+#include <queue>
+using namespace std;
+
+class Solution {
+public:
+    int maximumScore(int a, int b, int c) {
+        
+        priority_queue<int> maxheap;         // pushing initial values of a,b and c to maxheap
+        maxheap.push(a);
+        maxheap.push(b);
+        maxheap.push(c);
+        
+        int score = 0;
+        while (maxheap.size() >= 2) {        // performing required step untill there are two nonempty piles of stones
+            
+            int max = maxheap.top();         // poping maximum and second maximum elements from maxheap
+            maxheap.pop();
+            
+            int secmax = maxheap.top();
+            maxheap.pop();
+            
+            if (max>0 && secmax>0) {
+                maxheap.push(max-1);
+                maxheap.push(secmax-1);      // pushing element having 1 less value to priortize again 
+                score++;                     // increasing score by 1 only if we can push again having value > 0 which indicates stone can be removed from teo piles
+            }
+        }
+        return score;
+        
+    }
+    
+};
+
+int main() 
+{
+    cout<<"\nEnter number of test cases: ";
+	int t; cin>>t;  //no. of test cases
+
+	// Constraint: 1 <= a, b, c <= 10^5
+
+	while (t--)
+	{
+
+
+	    cout<<"\n\nEnter number of stones present in piles a,b and c  : ";
+	    int a; cin>>a;
+        int b; cin>>b;
+        int c; cin>>c;
+
+	    Solution *ob;
+        cout<<"\nK Maximum score you can get is :";
+	    cout <<  ob-> maximumScore(a,b,c);
+	    
+	}
+	return 0;
+}
+
+/*
+SAMPLE INPUT
+3				    // number of test cases 
+5 7 2				// values of a,b and c for test case:1
+1 9 3 	            // values of a,b and c for test case:2
+1000 96 345         // values of a,b and c for test case:1
+
+SAMPLE OUTPUT (excluding interactive instructions)
+7                   // first 3 moves will make (a,b,c) => (2,4,2) and next 4 moves will make them all 0
+4                   // first 2 moves will make (a,b,c) => (1,7,1) and next two will make (0,5,0) then stopped as nonempty piles <2
+441                 
+ 
+COMPLEXITY ANALYSIS : 
+Time : O(max(a,b,c)) since forming max heap is independent of values of a, b and c. 
+Space: O(1), since max heap only stores 3 elements at a time.
+*/
\ No newline at end of file