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.
1#include <vector>
2using namespace std;
3
4class Solution {
5public:
6 bool isIdealPermutation(vector<int>& nums) {
7 int maxVal = -1;
8 for (int i = 0; i < nums.size() - 2; i++) {
9 maxVal = max(maxVal, nums[i]);
10 if (maxVal > nums[i + 2]) return false;
11 }
12 return true;
13 }
14};The maximum value seen from the left must be less than or equal to any value we encounter moved two ahead. We ensure to move through the array tracking if there's any possibility of non-local inversions.
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)
1function isIdealPermutation(nums) {
2 for (let i = 0; i < nums.length; i++) {
3 if (Math.abs(nums[i] - i) > 1) {
4 return false;
5 }
6 }
7 return true;
8}With JavaScript, apply the direct index differential to verify classification as this permissible depth check persists, restricted by logical bounds of 1.