Problem Description

You have an array of integers nums (0-indexed) and two integer values x and y. You can perform operations on this array to reduce all numbers to zero or below.

In each operation, you:

1. Choose any index i from the array (where 0 <= i < nums.length)
2. Decrease the value at nums[i] by x
3. Decrease the values at all other indices (every position except i) by y

Your goal is to find the minimum number of operations needed to make every integer in nums less than or equal to zero.

For example, if you have nums = [10, 5, 8], x = 4, and y = 1, and you perform one operation choosing index 0:

• nums[0] decreases by 4: 10 - 4 = 6
• nums[1] decreases by 1: 5 - 1 = 4
• nums[2] decreases by 1: 8 - 1 = 7
• Result after operation: [6, 4, 7]

You would continue performing operations until all values become <= 0. The problem asks for the minimum number of such operations required.

Intuition

The key observation is that if we can make all numbers <= 0 in t operations, then we can definitely do it in t+1 operations or any number greater than t. This monotonic property suggests we can use binary search to find the minimum number of operations.

Let's think about what happens after t operations:

• Every number gets decreased by y in each operation (since it's not chosen), so each number decreases by at least t * y
• Some numbers might need additional reduction beyond t * y to become <= 0
• We can provide this extra reduction by choosing those specific indices during our t operations

For a number nums[i] to become <= 0 after t operations:

• If nums[i] <= t * y, it's already handled (since every position gets reduced by y in each operation)
• If nums[i] > t * y, we need extra reduction of nums[i] - t * y
• Each time we choose index i, we get an extra reduction of x - y (since we reduce by x instead of y)
• So we need to choose index i at least ceil((nums[i] - t * y) / (x - y)) times

The crucial insight: if the total number of times we need to choose specific indices (sum of all required selections) is <= t, then t operations are sufficient. Otherwise, we need more operations.

This gives us a way to check if a given number of operations t is feasible, which we can use with binary search to find the minimum valid t.

Learn more about Binary Search patterns.

Solution Approach

We implement a binary search solution to find the minimum number of operations needed.

Binary Search Setup:

• Initialize the search range: l = 0 (minimum possible operations) and r = max(nums) (maximum possible operations needed)
• We search for the minimum value of t (number of operations) that satisfies our condition

Check Function: The check(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 Logic:

while l < r:
    mid = (l + r) >> 1  # equivalent to (l + r) // 2
    if check(mid):
        r = mid  # mid operations are sufficient, try smaller
    else:
        l = mid + 1  # mid operations are not enough, need more

The binary search converges when l == r, giving us the minimum number of operations required. The algorithm efficiently narrows down the search space by halving it in each iteration, resulting in O(n * log(max(nums))) time complexity, where n is the length of the array.

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 range

• l = 0 (minimum possible operations)
• r = max(nums) = 5 (maximum operations we might need)

Step 2: Binary search iterations

Iteration 1: l = 0, r = 5

• mid = (0 + 5) // 2 = 2
• Check if 2 operations are sufficient:
  • For nums[0] = 5:
    • After 2 operations, reduced by at least 2 * 1 = 2
    • Remaining: 5 - 2 = 3
    • Need extra reduction via choosing this index: ceil(3 / (3-1)) = ceil(3/2) = 2 times
  • For nums[1] = 3:
    • After 2 operations, reduced by at least 2 * 1 = 2
    • Remaining: 3 - 2 = 1
    • Need extra reduction: ceil(1 / (3-1)) = ceil(1/2) = 1 time
  • Total selections needed: 2 + 1 = 3
  • Since 3 > 2, we cannot do it in 2 operations
• Update: l = mid + 1 = 3

Iteration 2: l = 3, r = 5

• mid = (3 + 5) // 2 = 4
• Check if 4 operations are sufficient:
  • For nums[0] = 5:
    • After 4 operations, reduced by at least 4 * 1 = 4
    • Remaining: 5 - 4 = 1
    • Need extra reduction: ceil(1 / 2) = 1 time
  • For nums[1] = 3:
    • After 4 operations, reduced by at least 4 * 1 = 4
    • Since 3 <= 4, already handled (no extra reduction needed)
  • Total selections needed: 1 + 0 = 1
  • Since 1 <= 4, we can do it in 4 operations
• Update: r = mid = 4

Iteration 3: l = 3, r = 4

• mid = (3 + 4) // 2 = 3
• Check if 3 operations are sufficient:
  • For nums[0] = 5:
    • After 3 operations, reduced by at least 3 * 1 = 3
    • Remaining: 5 - 3 = 2
    • Need extra reduction: ceil(2 / 2) = 1 time
  • For nums[1] = 3:
    • After 3 operations, reduced by at least 3 * 1 = 3
    • Since 3 <= 3, already handled
  • Total selections needed: 1 + 0 = 1
  • Since 1 <= 3, we can do it in 3 operations
• Update: r = mid = 3

Step 3: Binary search converges

• l = r = 3, so the minimum number of operations is 3

Verification of the answer: Let's verify by simulating 3 operations:

• 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 indeed sufficient and minimal.

Solution Implementation

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. Integer Division vs Ceiling Division Error

One of the most common mistakes 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)

2. 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

3. 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...

4. 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