Minimum Operations to Exceed Threshold Value II

class Solution {
public:
    int minimumOperations(vector<int>& numbers, int threshold) {
        priority_queue<long, vector<long>, greater<long>> heap(numbers.begin(), numbers.end());
        int operations_count = 0;
    
        while (heap.top() < threshold) {
            long smaller = heap.top();
            heap.pop();
            long larger = heap.top();
            heap.pop();
            heap.push(min(smaller, larger) * 2 + max(smaller, larger));
    
            operations_count++;
        }
        return operations_count;
    }
};

The task involves computing the minimum number of operations required to make every number in a collection exceed a specified threshold value. The implementation is in C++ and employs a min-heap approach to efficiently track and modify the smallest numbers until all elements of the collection are above the threshold.

Here’s a description of how the solution works:

• Initialize a min-heap to efficiently get the smallest numbers in the collection.
• Iterate while the smallest number in the heap is less than the threshold:
  • Extract the two smallest elements.
  • Combine these two elements through a specific operation (double the minimum and add the maximum of the two).
  • Add the resulting value back to the heap.
  • Increment the operations count each time this process is applied.

Ensure that these steps result in a solution:

• The heap grabs the two smallest numbers, ensuring the operation has minimal impact on other values and accelerates progression towards the goal.
• The operation defined increases the number minimum enough to gradually push all values above the threshold.
• The algorithm achieves its goal efficiently by focusing on the smallest numbers, requiring fewer operations.

This method offers a practical solution to the problem, leveraging heap data structures to maintain optimal performance and simplicity in the approach.

class Solution {
    
    public int calcMinOps(int[] values, int threshold) {
        PriorityQueue<Long> heap = new PriorityQueue<Long>(
            Arrays.stream(values)
                .mapToLong(num -> (long) num)
                .boxed()
                .collect(Collectors.toList())
        );
        int operations = 0;
    
        while (heap.peek() < threshold) {
            long first = heap.remove();
            long second = heap.remove();
            heap.add(Math.min(first, second) * 2 + Math.max(first, second));
    
            operations++;
        }
        return operations;
    }
}

In this Java implementation, the goal is to determine the minimal number of operations necessary to make the smallest number in an array exceed a given threshold value by using a min-heap strategy.

• Begin by creating a PriorityQueue<Long> and populate it with values from the input array, converting each integer to a long to ensure there are no overflow issues.
• Initialize an operation count to zero.
• Continue processing the heap as long as the smallest value is below the threshold:
  • Remove the two smallest elements from the heap.
  • Combine these two elements using the formula: Math.min(first, second) * 2 + Math.max(first, second).
  • Add the result back to the heap.
  • Increment the operation count after each combination.
• The operation count, upon heap elements exceeding the threshold, is returned as the result.

This method efficiently combines elements to maximize growth in value, thus minimizing the number of operations required to exceed the threshold.

class Solution:
    def minimumSteps(self, numbers: List[int], target: int) -> int:
        heapq.heapify(numbers)
            
        operations_count = 0
        while numbers[0] < target:
            first = heapq.heappop(numbers)
            second = heapq.heappop(numbers)
            heapq.heappush(numbers, min(first, second) * 2 + max(first, second))
                
            operations_count += 1
            
        return operations_count

This program defines a Solution class with the method minimumSteps, aiming to determine the minimum number of operations required to make the smallest number in a list of integers exceed a given target value. Here's a breakdown of the approach:

• Start by converting the numbers list into a min-heap using heapq.heapify to facilitate easy access to the smallest elements.
• Initiate a count for the operations operations_count set to 0.
• Use a while loop to continuously perform operations as long as the smallest element in the heap is below the target value.
• Within the loop:
  • Extract the two smallest elements using heapq.heappop.
  • Calculate the new element by doubling the smaller one and adding the larger then push this new element back into the heap with heapq.heappush.
  • Increment the operations_count by 1 for each operation.
• The loop exits when the smallest number in the heap is equal to or greater than the target, and at this point, the method returns the total count of operations needed (operations_count).

Note that this approach assumes that the provided list numbers has at least two elements to avoid errors while popping from the heap.