LeetCode-in-Java

2972. Count the Number of Incremovable Subarrays II

Hard

You are given a 0-indexed array of positive integers nums.

A subarray of nums is called incremovable if nums becomes strictly increasing on removing the subarray. For example, the subarray [3, 4] is an incremovable subarray of [5, 3, 4, 6, 7] because removing this subarray changes the array [5, 3, 4, 6, 7] to [5, 6, 7] which is strictly increasing.

Return the total number of incremovable subarrays of nums.

Note that an empty array is considered strictly increasing.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [1,2,3,4]

Output: 10

Explanation: The 10 incremovable subarrays are: [1], [2], [3], [4], [1,2], [2,3], [3,4], [1,2,3], [2,3,4], and [1,2,3,4], because on removing any one of these subarrays nums becomes strictly increasing. Note that you cannot select an empty subarray.

Example 2:

Input: nums = [6,5,7,8]

Output: 7

Explanation: The 7 incremovable subarrays are: [5], [6], [5,7], [6,5], [5,7,8], [6,5,7] and [6,5,7,8]. It can be shown that there are only 7 incremovable subarrays in nums.

Example 3:

Input: nums = [8,7,6,6]

Output: 3

Explanation: The 3 incremovable subarrays are: [8,7,6], [7,6,6], and [8,7,6,6]. Note that [8,7] is not an incremovable subarray because after removing [8,7] nums becomes [6,6], which is sorted in ascending order but not strictly increasing.

Constraints:

Solution

public class Solution {
    public long incremovableSubarrayCount(int[] nums) {
        long ans = 0;
        int n = nums.length;
        int l = 0;
        int r = n - 1;
        while (l + 1 < n && nums[l] < nums[l + 1]) {
            l++;
        }
        while (r > 0 && nums[r - 1] < nums[r]) {
            r--;
        }
        ans = (l == n - 1) ? 0 : 1 + (n - r);
        for (int i = 0; i <= l; i++) {
            while (r < n && nums[r] <= nums[i]) {
                r++;
            }
            ans += n - r + 1;
        }
        return ans;
    }
}