14. Maximum Diamonds

The problem can be found at the following link: Question Link

My Approach

This is a simple greedy problem where we have to maximize the number of diamonds in bags at each iteration (from 0 to k). We retrieve the maximum number of diamonds from a bag and then put back half of the diamonds into the same bag.

To find the maximum number of diamonds at any given moment, I utilize a priority queue. Firstly, I push all elements of the a array into the priority queue. Then, for k iterations, I execute a loop that retrieves the maximum element from the priority queue (which is at the top), removes it, and replaces it with the halved value.

Time and Auxiliary Space Complexity

• Time Complexity: O(n + k log n), where n is the number of bags (length of array a) and k is the number of iterations. Inserting all elements into the priority queue takes O(n) time, and each iteration takes O(log n) time for heap operations. Therefore, the overall time complexity is O(n + k log n).

• Auxiliary Space Complexity: O(n) since we are using a priority queue to store the elements.

Code (C++)

class Solution {
public:
    long long maxDiamonds(int a[], int n, int k) {
        priority_queue<int> pq;
        for(int i = 0; i < n; ++i)
            pq.push(a[i]);
            
        long long out = 0, pqTop;
        while(k--){
            pqTop = pq.top();
            pq.pop();
            out += pqTop;
            pq.push(pqTop / 2);
        }
        return out;
    }
};

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