2972. Count the Number of Incremovable Subarrays II

Description

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

 

Solutions

Solution: Two Pointers

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
const incremovableSubarrayCount = function (nums) {
  const n = nums.length;

  const findStartIndex = () => {
    for (let index = n - 2; index >= 0; index--) {
      if (nums[index] >= nums[index + 1]) {
        return index + 1;
      }
    }

    return 0;
  };

  const startIndex = findStartIndex();

  if (startIndex === 0) return (n * (n + 1)) / 2;

  let right = startIndex;
  let result = n - startIndex + 1;

  for (let index = 0; index < startIndex; index++) {
    if (index && nums[index] <= nums[index - 1]) {
      return result;
    }

    while (right < n && nums[index] >= nums[right]) {
      right += 1;
    }

    result += n - right + 1;
  }

  return result;
};