How do you solve Global and Local Inversions on LeetCode?

Global and Local Inversions (LeetCode #775) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Global inversions include local inversions, but not all global inversions are local inversions.

What is the brute force solution for Global and Local Inversions?

The brute force approach for Global and Local Inversions is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible pair of indices to count global inversions and local inversions. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Global and Local Inversions?

Global and Local Inversions uses the following concepts: Array, Math. The recommended patterns to study are: Array, Sorting.

What are the interview tips for Global and Local Inversions?

Always clarify the definitions of global and local inversions before starting to solve the problem. Think about edge cases, such as arrays with only one element or already sorted arrays. Practice explaining your thought process clearly, as communication is key during interviews.

What are common mistakes when solving Global and Local Inversions?

Counting inversions incorrectly by not distinguishing between global and local inversions. Overlooking the condition that an element must not be more than one position away to maintain equality.

#775

Global and Local Inversions

Medium

ArrayMathArraySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the properties of the permutation to determine if the number of global inversions equals local inversions without explicitly counting all pairs. If an element is more than two positions ahead of its original position, it creates a global inversion without being a local inversion.

⚙️

Algorithm

2 steps

1Step 1: Loop through the array and check if any element nums[i] is greater than i + 1. If yes, return false.
2Step 2: If all elements satisfy the condition, return true.

solution.py5 lines

1def isIdealPermutation(nums):
2    for i in range(len(nums)):
3        if nums[i] > i + 1:
4            return False
5    return True

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the array. The space complexity is O(1) since we use no additional data structures.

1Global inversions include local inversions, but not all global inversions are local inversions.
2If any element is more than one position away from its original index, it creates a global inversion without being a local inversion.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.