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.
1function isIdealPermutation(nums) {
2 let maxVal = -1;
3 for (let i = 0; i < nums.length - 2; i++) {
4 maxVal = Math.max(maxVal, nums[i]);
5 if (maxVal > nums[i + 2]) return false;
6 }
7 return true;
8}JavaScript, as usual, captures the principle of maintaining a running 'max' and ensuring no violation against future elements at an offset.
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 <stdbool.h>
2
3bool isIdealPermutation(int* nums, int numsSize){
4 for (int i = 0; i < numsSize; i++) {
5 if (abs(nums[i] - i) > 1) {
6 return false;
7 }
8 }
9 return true;
10}Iterate the array and verify every element index difference from the original is no greater than 1, suggesting only local inversions exist.