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 <stdbool.h>
2
3bool isIdealPermutation(int* nums, int numsSize){
4 int max = -1;
5 for (int i = 0; i < numsSize - 2; i++) {
6 if (max < nums[i]) {
7 max = nums[i];
8 }
9 if (max > nums[i + 2]) {
10 return false;
11 }
12 }
13 return true;
14}We iterate through the array while keeping track of the maximum number seen so far. The core idea is to ensure that the maximum value we pass is less than or equal to both the current value and the follow-up after skipping one (e.g., nums[i+2]).
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.