LeetCode #775 — MEDIUM

Global and Local Inversions

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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.

Example 1:

Input: nums = [1,0,2]
Output: true
Explanation: There is 1 global inversion and 1 local inversion.

Example 2:

Input: nums = [1,2,0]
Output: false
Explanation: There are 2 global inversions and 1 local inversion.

Constraints:

n == nums.length
1 <= n <= 10⁵
0 <= nums[i] < n
All the integers of nums are unique.
nums is a permutation of all the numbers in the range [0, n - 1].

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[1,0,2]

Example 2

[1,2,0]

Step 02

Core Insight

What unlocks the optimal approach

Where can the 0 be placed in an ideal permutation? What about the 1?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #775: Global and Local Inversions
class Solution {
    public boolean isIdealPermutation(int[] nums) {
        int mx = 0;
        for (int i = 2; i < nums.length; ++i) {
            mx = Math.max(mx, nums[i - 2]);
            if (mx > nums[i]) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #775: Global and Local Inversions
func isIdealPermutation(nums []int) bool {
	mx := 0
	for i := 2; i < len(nums); i++ {
		mx = max(mx, nums[i-2])
		if mx > nums[i] {
			return false
		}
	}
	return true
}

# Accepted solution for LeetCode #775: Global and Local Inversions
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

// Accepted solution for LeetCode #775: Global and Local Inversions
struct Solution;

impl Solution {
    fn is_ideal_permutation(a: Vec<i32>) -> bool {
        let n = a.len();
        for i in 0..n {
            if (a[i] - i as i32).abs() > 1 {
                return false;
            }
        }
        true
    }
}

#[test]
fn test() {
    let a = vec![1, 0, 2];
    let res = true;
    assert_eq!(Solution::is_ideal_permutation(a), res);
    let a = vec![1, 2, 0];
    let res = false;
    assert_eq!(Solution::is_ideal_permutation(a), res);
}

// Accepted solution for LeetCode #775: Global and Local Inversions
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #775: Global and Local Inversions
// class Solution {
//     public boolean isIdealPermutation(int[] nums) {
//         int mx = 0;
//         for (int i = 2; i < nums.length; ++i) {
//             mx = Math.max(mx, nums[i - 2]);
//             if (mx > nums[i]) {
//                 return false;
//             }
//         }
//         return true;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.