Leetcode #2702: Minimum Operations to Make Numbers Non-positive
In this guide, we solve Leetcode #2702 Minimum Operations to Make Numbers Non-positive in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed integer array nums and two integers x and y. In one operation, you must choose an index i such that 0 <= i < nums.length and perform the following: Decrement nums[i] by x.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Array, Binary Search
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: nums = [3,4,1,7,6], x = 4, y = 2
Output: 3
Explanation: You will need three operations. One of the optimal sequence of operations is:
Operation 1: Choose i = 3. Then, nums = [1,2,-1,3,4].
Operation 2: Choose i = 3. Then, nums = [-1,0,-3,-1,2].
Operation 3: Choose i = 4. Then, nums = [-3,-2,-5,-3,-2].
Now, all the numbers in nums are non-positive. Therefore, we return 3.
Python Solution
class Solution:
def minOperations(self, nums: List[int], x: int, y: int) -> int:
def check(t: int) -> bool:
cnt = 0
for v in nums:
if v > t * y:
cnt += ceil((v - t * y) / (x - y))
return cnt <= t
l, r = 0, max(nums)
while l < r:
mid = (l + r) >> 1
if check(mid):
r = mid
else:
l = mid + 1
return l
Complexity
The time complexity is O(log n) or O(n log n). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.