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 ofnums.
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
Video Links
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;
}
};