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Leetcode 3065. Minimum Operations to Exceed Threshold Value I

Problem Statement

You are given an integer array arr and two integers threshold and maxOperations.

In one operation, you can remove one element from arr and divide it into any numbers of new elements such that their sum is equal to the removed element. The goal is to perform the minimum number of operations such that the sum of the array exceeds threshold.

Return the minimum number of operations required. If it is impossible to achieve this, return -1.

Clarifying Questions

  1. What are the constraints?
    • What is the maximum length of the array arr?
    • What is the range of values for each element in arr, threshold, and maxOperations?
  2. Edge Cases:
    • What if arr is empty?
    • What if the threshold is less than or equal to the sum of arr already?
    • What happens if maxOperations is 0?

Strategy

  1. Initial Check:
    • Calculate the initial sum of arr.
    • If this sum is already greater than or equal to threshold, return 0 since no operations are necessary.
  2. Priority Queue (Max-Heap):
    • Use a max-heap to always choose the maximum element to split.
    • The reason is that splitting a larger element contributes more towards increasing the total number of elements and thus their sum.
  3. Operations:
    • Perform operations iteratively, replacing the maximum heap element with its divided parts.
    • Track the number of operations performed.
  4. Early termination:
    • If at any point the sum exceeds or meets threshold, return the number of operations performed.
  5. Edge Case Handling:
    • If we exhaust maxOperations and still haven’t met the threshold, return -1.

Code

import java.util.Collections;
import java.util.PriorityQueue;

public class MinOperationsToExceedThreshold {

    public int minOperations(int[] arr, int threshold, int maxOperations) {
        long sum = 0;
        for (int num : arr) {
            sum += num;
        }

        if (sum >= threshold) {
            return 0;
        }

        PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
        for (int num : arr) {
            maxHeap.add(num);
        }

        int operations = 0;
        while (operations < maxOperations && sum < threshold && !maxHeap.isEmpty()) {
            int maxElement = maxHeap.poll();
            // Since we are dividing maxElement into 2 parts, we create two new elements.
            int newElement1 = maxElement / 2;
            int newElement2 = maxElement - newElement1;

            sum = sum - maxElement + newElement1 + newElement2;

            maxHeap.add(newElement1);
            maxHeap.add(newElement2);

            operations++;
        }

        return sum >= threshold ? operations : -1;
    }

    public static void main(String[] args) {
        MinOperationsToExceedThreshold solver = new MinOperationsToExceedThreshold();
        int[] arr = {10, 20, 30};
        int threshold = 60;
        int maxOperations = 3;
        System.out.println(solver.minOperations(arr, threshold, maxOperations)); // Output: 1
    }
}

Time Complexity

The solution is efficient given realistic constraints and handles several edge cases appropriately.

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