Leetcode #775: Global and Local Inversions
In this guide, we solve Leetcode #775 Global and Local Inversions 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 an integer array nums of length n which represents a permutation of all the integers in the range [0, n - 1]. The number of global inversions is the number of the different pairs (i, j) where: 0 <= i < j < n nums[i] > nums[j] The number of local inversions is the number of indices i where: 0 <= i < n - 1 nums[i] > nums[i + 1] Return true if the number of global inversions is equal to the number of local inversions.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, 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: nums = [1,0,2]
Output: true
Explanation: There is 1 global inversion and 1 local inversion.
Python Solution
class Solution:
def isIdealPermutation(self, nums: List[int]) -> bool:
mx = 0
for i in range(2, len(nums)):
if (mx := max(mx, nums[i - 2])) > nums[i]:
return False
return True
Complexity
The time complexity is O(n) or O(1). 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.