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Minimum Operations to Make Numbers Non-positive

Number: 2729

Difficulty: Hard

Paid? Yes

Companies: Citadel, Snowflake

Problem Description

Given a 0-indexed integer array nums and two integers x and y, you can perform the following operation:

  • Choose an index i (0 <= i < nums.length).
  • Decrement nums[i] by x.
  • Decrement every other element in nums by y.

Return the minimum number of operations required to make every element in nums non-positive (less than or equal to 0).

Key Insights

  • Every operation decrements every index by y and an extra (x - y) for the chosen index.
  • After T operations, every element will have been reduced by at least y * T.
  • For any element still positive, additional operations on that index (each providing an extra x - y reduction) are needed.
  • If for each index the required extra operations (calculated as ceil(max(nums[i] - y * T, 0) / (x - y))) across all indices is ≤ T, then T operations suffice.
  • Binary search can be employed on T (the number of operations) to efficiently find the minimum T that works.

Space and Time Complexity

Time Complexity: O(n log(max(nums[i]) / y)), where n is the length of nums. Space Complexity: O(1) extra space (ignoring input storage).

Solution

We apply binary search over the number of operations, T. For each candidate T, we simulate the effect:

  • Each element gets reduced by y in every operation.
  • Compute the remaining value as: remaining = nums[i] - y * T.
  • If remaining > 0, compute the extra operations needed on that element: ceil(remaining / (x - y)).
  • Sum these extra operations over all elements; if the total extra operations ≤ T, T is a valid candidate. The trick lies in understanding that every operation contributes a global decrement of y and an additional targeted decrement of (x - y) at one index.

We use simple loops and arithmetic (with a ceiling division trick) inside a binary search loop.

Code Solutions

Below are code implementations in Python, JavaScript, C++, and Java with detailed inline comments.

def min_operations(nums, x, y):
    # Helper function to check if T operations are sufficient.
    def is_possible(T):
        extra_ops = 0
        for num in nums:
            # Compute remaining value after applying y decrement T times.
            remaining = num - y * T
            if remaining > 0:
                # Additional operations needed for current num.
                # Ceiling division: (remaining + (x - y) - 1) // (x - y)
                extra_ops += (remaining + (x - y) - 1) // (x - y)
                if extra_ops > T:
                    return False
        return extra_ops <= T

    # Set up a search range for T.
    left, right = 0, max(nums) // y + 1
    while left < right:
        mid = (left + right) // 2
        if is_possible(mid):
            right = mid
        else:
            left = mid + 1
    return left

# Example usage:
print(min_operations([3, 4, 1, 7, 6], 4, 2))  # Expected output: 3
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