2972. Count the Number of Incremovable Subarrays II

Problem Statement

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:

1 <= nums.length <= 105
1 <= nums[i] <= 109

Intuition

Approach:

Find the left index till what incresing and 
right till what increasing 

Then try to find subarrays count, by taking array from left and picking 
Corresponding right array

Use upperbound to find the right array increasing

Links

https://leetcode.com/problems/count-the-number-of-incremovable-subarrays-ii/

Video Links

https://youtu.be/VmTxOpyPtIE?t=4315

Approach 1:

NlongN

C++

class Solution {
public:
    long long incremovableSubarrayCount(vector<int>& nums) {
        int n = nums.size();
        int left = 0, right = n-1;
        
        while(left+1 < n and nums[left] < nums[left+1])
            left++;

        while(right-1 >= 0 and nums[right] > nums[right-1])
            right--;

        if(left == n-1)
            return (n*(n+1))/2;
        
        long long ans = n-right+1;
        for(int i=0; i<=left; i++){
            int upidx = upper_bound(nums.begin()+right, nums.end(), nums[i])-nums.begin();
            ans += n-upidx+1;
        }

        return ans;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++