Solution Approach

We implement a binary search solution to find the minimum number of operations needed. The solution follows the standard binary search template for finding the first value where a condition becomes true.

Feasible Function: The feasible(t) function determines if t operations are sufficient to make all numbers <= 0:

1. For each number v in nums:

   • After t operations, every number is reduced by at least t * y
   • If v > t * y, the remaining amount v - t * y must be eliminated by choosing this index
   • When we choose an index, we get extra reduction of x - y per operation
   • So we need ceil((v - t * y) / (x - y)) operations where this index is chosen

2. Count the total number of times we need to choose specific indices:

   cnt = 0
   for v in nums:
       if v > t * y:
           cnt += ceil((v - t * y) / (x - y))

3. If cnt <= t, then t operations are sufficient (return True), otherwise we need more operations (return False)

Binary Search Template:

left, right = 0, max(nums)
first_true_index = -1

while left <= right:
    mid = (left + right) // 2
    if feasible(mid):
        first_true_index = mid
        right = mid - 1  # Try to find smaller valid value
    else:
        left = mid + 1

return first_true_index

The template tracks first_true_index to store the answer when the feasible condition is met. Since we're looking for the minimum operations where feasible(t) becomes true, we update first_true_index = mid when feasible and continue searching left for a potentially smaller value. The algorithm efficiently narrows down the search space by halving it in each iteration, resulting in O(n * log(max(nums))) time complexity.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Example: nums = [5, 3], x = 3, y = 1

Step 1: Set up binary search with template

• left = 0 (minimum possible operations)
• right = max(nums) = 5 (maximum operations we might need)
• first_true_index = -1 (tracks our answer)

Step 2: Binary search iterations using while left <= right

Iteration 1: left = 0, right = 5

• mid = (0 + 5) // 2 = 2
• Check if feasible(2):
  • For nums[0] = 5: remaining = 5 - 2*1 = 3, need ceil(3/2) = 2 selections
  • For nums[1] = 3: remaining = 3 - 2*1 = 1, need ceil(1/2) = 1 selection
  • Total selections needed: 2 + 1 = 3 > 2 → Not feasible
• feasible(2) = False, so left = mid + 1 = 3

Iteration 2: left = 3, right = 5

• mid = (3 + 5) // 2 = 4
• Check if feasible(4):
  • For nums[0] = 5: remaining = 5 - 4*1 = 1, need ceil(1/2) = 1 selection
  • For nums[1] = 3: remaining = 3 - 4*1 = -1 <= 0, need 0 selections
  • Total selections needed: 1 <= 4 → Feasible!
• feasible(4) = True, so first_true_index = 4, right = mid - 1 = 3

Iteration 3: left = 3, right = 3

• mid = (3 + 3) // 2 = 3
• Check if feasible(3):
  • For nums[0] = 5: remaining = 5 - 3*1 = 2, need ceil(2/2) = 1 selection
  • For nums[1] = 3: remaining = 3 - 3*1 = 0 <= 0, need 0 selections
  • Total selections needed: 1 <= 3 → Feasible!
• feasible(3) = True, so first_true_index = 3, right = mid - 1 = 2

Iteration 4: left = 3, right = 2

• left > right, loop terminates

Step 3: Return result

• first_true_index = 3, so the minimum number of operations is 3

Verification:

• Operation 1: Choose index 0 → [5-3, 3-1] = [2, 2]
• Operation 2: Choose index 0 → [2-3, 2-1] = [-1, 1]
• Operation 3: Choose index 1 → [-1-1, 1-3] = [-2, -2]

All values are now <= 0, confirming that 3 operations are sufficient and minimal.

Time and Space Complexity

Time Complexity: O(n * log(max(nums)))

The algorithm uses binary search on the range [0, max(nums)], which requires O(log(max(nums))) iterations. In each iteration, the check function is called, which iterates through all n elements in the nums array to calculate the required operations. Each element requires O(1) time for the arithmetic operations and ceiling division. Therefore, the overall time complexity is O(n * log(max(nums))).

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space. The variables l, r, mid, t, cnt, and v are all scalar values that don't depend on the input size. The check function doesn't use any additional data structures that scale with input size. Therefore, the space complexity is O(1).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Using Wrong Binary Search Template Variant

A common mistake is using the while left < right with right = mid pattern instead of the standard template.

Incorrect Implementation:

left, right = 0, max(nums)
while left < right:
    mid = (left + right) // 2
    if feasible(mid):
        right = mid
    else:
        left = mid + 1
return left

Correct Implementation (Template-compliant):

left, right = 0, max(nums)
first_true_index = -1

while left <= right:
    mid = (left + right) // 2
    if feasible(mid):
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1

return first_true_index

Why the template matters: The standard template explicitly tracks the answer with first_true_index, uses while left <= right, and right = mid - 1 when feasible. This provides clearer semantics and handles edge cases more predictably.

2. Integer Division vs Ceiling Division Error

Another common mistake is incorrectly calculating the ceiling division when determining how many targeted operations are needed for each element.

Incorrect Implementation:

operations_for_this_element = remaining_value // (x - y)  # Wrong! This is floor division

Why it's wrong: When we have a remaining value that isn't perfectly divisible by (x - y), floor division will underestimate the operations needed. For example, if remaining_value = 5 and (x - y) = 2, we need 3 operations (not 2).

Correct Implementation:

# Method 1: Using math.ceil
import math
operations_for_this_element = math.ceil(remaining_value / (x - y))

# Method 2: Manual ceiling division (avoiding floating point)
operations_for_this_element = (remaining_value + (x - y) - 1) // (x - y)

3. Incorrect Binary Search Boundary

Setting the wrong upper bound for binary search can lead to incorrect results or unnecessary iterations.

Incorrect Implementation:

right = sum(nums)  # Wrong! This could be unnecessarily large

Why it's wrong: The maximum operations needed is when we only use global operations (reducing by y), which would be max(nums) // y + 1. Using sum(nums) as the upper bound works but is inefficient.

Correct Implementation:

right = max(nums)  # Sufficient upper bound
# Or more precisely:
right = (max(nums) + y - 1) // y  # Exact maximum operations needed

4. Edge Case: When x equals y

If x == y, the problem becomes trivial but the division by (x - y) would cause a division by zero error.

Solution:

def minOperations(self, nums: List[int], x: int, y: int) -> int:
    if x == y:
        # All operations have the same effect
        return (max(nums) + x - 1) // x

    # Continue with binary search approach...

5. Overflow in Ceiling Division Formula

When using the manual ceiling division formula (a + b - 1) // b, the addition might cause integer overflow in languages with fixed integer sizes.

Solution for Python: Python handles arbitrary precision integers, so this isn't an issue. However, if implementing in other languages:

# Alternative ceiling division without overflow risk
operations_for_this_element = remaining_value // (x - y)
if remaining_value % (x - y) != 0:
    operations_for_this_element += 1