Leetcode #2541: Minimum Operations to Make Array Equal II
In this guide, we solve Leetcode #2541 Minimum Operations to Make Array Equal II 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 two integer arrays nums1 and nums2 of equal length n and an integer k. You can perform the following operation on nums1: Choose two indexes i and j and increment nums1[i] by k and decrement nums1[j] by k.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, Array, Math
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: nums1 = [4,3,1,4], nums2 = [1,3,7,1], k = 3
Output: 2
Explanation: In 2 operations, we can transform nums1 to nums2.
1st operation: i = 2, j = 0. After applying the operation, nums1 = [1,3,4,4].
2nd operation: i = 2, j = 3. After applying the operation, nums1 = [1,3,7,1].
One can prove that it is impossible to make arrays equal in fewer operations.
Python Solution
class Solution:
def minOperations(self, nums1: List[int], nums2: List[int], k: int) -> int:
ans = x = 0
for a, b in zip(nums1, nums2):
if k == 0:
if a != b:
return -1
continue
if (a - b) % k:
return -1
y = (a - b) // k
ans += abs(y)
x += y
return -1 if x else ans // 2
Complexity
The time complexity is , and the space complexity is , where is the length of the array. The space complexity is , where is the length of the array.
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.