Minimum Cost to Connect Sticks

class Solution {
public:
    int minCostToConnectSticks(vector<int>& sticks) {
        int cumulativeCost = 0;
        
        priority_queue<int, vector<int>, greater<int>> minHeap;
        
        // Insert sticks into the heap
        for (int stick : sticks) {
            minHeap.push(stick);
        }
        
        // Continuously merge two smallest sticks until one remains
        while (minHeap.size() > 1) {
            int smallest = minHeap.top(); 
            minHeap.pop();
            int nextSmallest = minHeap.top(); 
            minHeap.pop();
            
            int mergeCost = smallest + nextSmallest;
            cumulativeCost += mergeCost;
            
            minHeap.push(mergeCost);
        }
        
        return cumulativeCost;
    }
};

This solution addresses the problem of finding the minimum cost to connect multiple sticks where the cost to connect any two sticks is the sum of their lengths. The approach uses a greedy algorithm facilitated by a priority queue (or min-heap) to solve the problem efficiently.

• Begin by initializing a variable cumulativeCost to zero. This will store the total minimum cost of connecting all sticks.
• Create a min-heap to always access the two sticks with the smallest lengths.
  • Populate the min-heap with the initial lengths of all sticks provided in the sticks vector.
• Use a loop to continuously merge the two shortest sticks until only one stick remains:
  • Remove the two smallest lengths from the heap.
  • Sum these two lengths to get the cost of merging them.
  • Add this cost to cumulativeCost.
  • Insert the merged stick (the sum of the two smallest lengths) back into the min-heap.
• After exiting the loop, cumulativeCost contains the minimized total cost for connecting all the sticks.

The algorithm primarily leverages the std::priority_queue from the C++ Standard Library, configured to operate as a min-heap, which ensures that the sticks are merged in the optimal order to minimize total cost. This method of summing the smallest available values ensures that each stick is added at the lowest possible cost, making the solution both efficient and effective. The complexity is generally governed by the operations on the heap, making it suitable for reasonably large datasets.

class Solution {
    public int mergeSticks(int[] sticks) {
        int sumCost = 0;
 
        PriorityQueue<Integer> heap = new PriorityQueue<>();
 
        for (int stick : sticks) {
            heap.add(stick);
        }
 
        while (heap.size() > 1) {
            int first = heap.poll();
            int second = heap.poll();
            
            int mergeCost = first + second;
            sumCost += mergeCost;
            
            heap.add(first + second);
        }
 
        return sumCost;
    }
}

The Java solution provided addresses the problem of finding the minimum cost to connect multiple sticks, where the cost is defined as the sum of the lengths of the sticks being connected. Here's a detailed breakdown of the approach used in the code:

• Create an instance of PriorityQueue to help in efficiently accessing the smallest elements, as the goal is to minimize the connection cost.

• Iterate through the array of sticks, adding each stick's length to the priority queue. This ensures that the sticks with the smallest lengths are prioritized for merging, minimizing the cost at each step.

• Continue merging the sticks until only one stick remains in the queue. In each iteration of the loop:

  • Remove the two smallest sticks using poll(), which retrieves and removes the head of the queue.
  • Calculate the cost of merging these two sticks and add it to the total cost.
  • Insert the resultant stick (sum of the two smallest sticks) back into the priority queue.

• Finally, return the accumulated sum of costs, which represents the minimum cost to merge all the sticks into one.

By using a priority queue, the solution efficiently manages the merging process, ensuring that the least costly operations are performed first, which is crucial for minimizing the overall cost.