This approach checks the condition if there is any inversion other than local inversions by checking for the maximum value condition constraint. We assume that if there exists any i such that nums[i] > nums[j] and j > i + 1, then it implies that the number of global inversions is greater than the number of local inversions.
Therefore, ensure for all i, nums[i] > nums[i+2] does not hold since, for it, nums[i] should be larger than any of the elements up to at least i + 2.
Time Complexity: O(n) because it requires a single traversal of the array.
Space Complexity: O(1) as no additional data structures are used.
1def isIdealPermutation(nums):
2 max_val = -1
3 for i in range(len(nums) - 2):
4 max_val = max(max_val, nums[i])
5 if max_val > nums[i + 2]:
6 return False
7 return TrueThe Python solution follows the same logic, where we dynamically compute the maximum value seen so far and ensure correctness by checking against values two indices ahead.
In this method, we will check each element if it is in a permissible range, conditionally permitting swaps with one element left or right, as can occur with local inversions. Assume an element is misplaced if it is not at i or i ± 1.
Time Complexity: O(n)
Space Complexity: O(1)
1#include <vector>
2#include <cmath>
3
4class Solution {
5public:
6 bool isIdealPermutation(std::vector<int>& nums) {
7 for (int i = 0; i < nums.size(); i++) {
8 if (std::abs(nums[i] - i) > 1) {
9 return false;
10 }
11 }
12 return true;
13 }
14};This solution uses the absolute value check for misplacement beyond what local inversions would allow, verifying the oversteps in distance.