Leetcode #1551: Minimum Operations to Make Array Equal
In this guide, we solve Leetcode #1551 Minimum Operations to Make Array Equal 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 have an array arr of length n where arr[i] = (2 * i) + 1 for all valid values of i (i.e., 0 <= i < n). In one operation, you can select two indices x and y where 0 <= x, y < n and subtract 1 from arr[x] and add 1 to arr[y] (i.e., perform arr[x] -=1 and arr[y] += 1).
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Math
Intuition
There is a mathematical invariant or formula that directly leads to the result.
Using math avoids unnecessary loops and reduces complexity.
Approach
Derive the formula or update rule, then compute the answer directly.
Handle edge cases like overflow or zero carefully.
Steps:
- Identify the math relationship.
- Compute the result with a loop or formula.
- Handle edge cases.
Example
Input: n = 3
Output: 2
Explanation: arr = [1, 3, 5]
First operation choose x = 2 and y = 0, this leads arr to be [2, 3, 4]
In the second operation choose x = 2 and y = 0 again, thus arr = [3, 3, 3].
Python Solution
class Solution:
def minOperations(self, n: int) -> int:
return sum(n - (i << 1 | 1) for i in range(n >> 1))
Complexity
The time complexity is , where is the length of the array. The space complexity is .
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.