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Given an array of integers nums and a threshold value target, we need to find the minimum number of elements we need to add up from nums to exceed the target value. If it is not possible to exceed the target, return -1.

Clarifying Questions:

  1. Are the numbers in the array positive, negative, or both?
    • The numbers can be both positive and negative.
  2. Are there any constraints on the size of the array or the values within the array?
    • Assume typical constraints for LeetCode problems: array length can be up to (10^5) and values within the array can be as large as (10^4).
  3. Is the array sorted?
    • No, the array is not necessarily sorted.
  4. Can elements be used more than once?
    • No, each element can be used only once.

Strategy:

  1. Sort the array in descending order to start with the largest elements, as this will give us the largest sums with the fewest elements.
  2. Initialize a current_sum to 0 and a counter to keep track of the number of elements added.
  3. Iterate over the sorted array and keep adding elements to the current_sum and increment the counter until current_sum exceeds the target.
  4. If after iterating through the array the current_sum does not exceed the target, return -1.

Time Complexity:

Code:

from typing import List

def min_operations(nums: List[int], target: int) -> int:
    nums.sort(reverse=True)  # sort the array in descending order
    current_sum = 0
    count = 0
    
    for num in nums:
        current_sum += num
        count += 1
        if current_sum > target:
            return count
    
    return -1  # if we exhaust the array and never exceed the target

# Example usage:
# nums = [1, 2, 3, 4, 5]
# target = 11
# Expected output: 3 (we can pick 5, 4, and 3 to exceed 11)
print(min_operations([1, 2, 3, 4, 5], 11))

Explanation of the Example:

Let’s take the sample array [1, 2, 3, 4, 5] and the target 11.

  1. After sorting in descending order: [5, 4, 3, 2, 1]
  2. Start adding from the start and keep track of the sum:
    • Add 5, current_sum = 5, count = 1
    • Add 4, current_sum = 9, count = 2
    • Add 3, current_sum = 12, count = 3
    • Since current_sum (12) now exceeds the target (11), we return the count, which is 3.

This approach ensures we find the minimal number of operations required to exceed the target efficiently.

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