3066. Minimum Operations to Exceed Threshold Value II - Detailed Explanation

Problem Statement

You are given an integer array nums and an integer threshold. In one operation, you can select any element from nums and double its value (i.e. replace it with 2×value). Your goal is to perform a sequence of operations such that the sum of all elements in nums becomes strictly greater than threshold. Return the minimum number of operations required to achieve this goal. If the initial sum already exceeds the threshold, no operations are needed.

Note: We assume that all numbers in nums are positive. (If zeros were allowed, doubling a zero would never help.)

Example 1

• Input:
  nums = [1, 2, 3]
threshold = 10
• Explanation:
  • The initial sum is 1 + 2 + 3 = 6, which is ≤ 10.
  • Operation 1: Choose the largest element (3) and double it to 6.
    New array becomes: [1, 2, 6]
    New sum = 1 + 2 + 6 = 9.
  • Operation 2: The current sum is still not above 10. Now, the largest element is 6. Double it to 12.
    New array becomes: [1, 2, 12]
    New sum = 1 + 2 + 12 = 15, which is > 10.
• Output: 2

Example 2

• Input:
  nums = [5, 5, 5]
threshold = 15
• Explanation:
  • The initial sum is 5 + 5 + 5 = 15. Since we need to exceed the threshold (i.e. > 15), at least one operation is required.
  • Operation 1: Pick any 5 (all are equal) and double it to 10.
    New array becomes: [10, 5, 5]
    New sum = 10 + 5 + 5 = 20, which is > 15.
• Output: 1

Example 3

• Input:
  nums = [1, 1, 1]
threshold = 5
• Explanation:
  • The initial sum is 1 + 1 + 1 = 3.
  • Operation 1: Choose one 1 and double it to 2.
    New array becomes: [2, 1, 1]
    New sum = 2 + 1 + 1 = 4.
  • Operation 2: The largest element now is 2. Double it to 4.
    New array becomes: [4, 1, 1]
    New sum = 4 + 1 + 1 = 6, which is > 5.
• Output: 2

Hints

1. Maximizing the Sum Increase per Operation:
   Doubling an element increases the overall sum by the value of that element (since 2×x − x = x). Therefore, to get the most benefit in each operation, you want to double the element with the largest value.

2. Greedy Strategy with a Heap:
   A max-heap (or priority queue) is a natural choice for repeatedly extracting the largest element. After doubling an element, push it back into the heap (since it might still be the largest or become the new candidate for doubling) and update the running sum.

3. Stop When Exceeding the Threshold:
   Continuously update the sum and count operations until the sum becomes strictly greater than threshold.

Approach 1: Greedy Using a Max Heap

Explanation

1. Initialization:

   • Compute the current sum of the array.
   • If the sum is already greater than threshold, return 0.
   • Build a max heap (or priority queue) from nums. In many languages (like Python), you can simulate a max heap by inserting the negatives of the numbers into a min-heap.

2. Perform Operations:

   • While the current sum is not greater than threshold:
     • Pop the largest element from the heap.
     • Doubling it increases the sum by that element’s current value.
     • Update the sum accordingly.
     • Push the doubled value back into the heap.
     • Increment the count of operations.

3. Return Result:

   • When the sum exceeds threshold, return the count of operations.

Why This Works

• Greedy Choice: Doubling the largest element always yields the maximum possible increase in the sum per operation.
• Correctness: Since each operation increases the sum by exactly the value of the chosen element, and the operation is repeated until the sum condition is met, the algorithm finds the minimum number of operations needed.

Code Solutions

Python Code