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package g0701_0800.s0775_global_and_local_inversions;

// #Medium #Array #Math #2022_03_26_Time_1_ms_(99.74%)_Space_51.6_MB_(90.31%)

public class Solution {
    /*
     * from the above solution, we can tell that if we can find the minimum of A[j] where j >= i + 2,
     * then we could quickly return false, so two steps:
     * 1. remembering minimum
     * 2. scanning from right to left
     * 

* Time: O(n) */ public boolean isIdealPermutation(int[] nums) { int min = nums.length; for (int i = nums.length - 1; i >= 2; i--) { min = Math.min(min, nums[i]); if (nums[i - 2] > min) { return false; } } return true; } }





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