A subarray is a contiguous subsequence of the array.

Return the length of the shortest subarray to remove.

Example 1:

Input: arr = [1,2,3,10,4,2,3,5]
Output: 3
Explanation: The shortest subarray we can remove is [10,4,2] of length 3. The remaining elements after that will be [1,2,3,3,5] which are sorted.
Another correct solution is to remove the subarray [3,10,4].

Example 2:

Input: arr = [5,4,3,2,1]
Output: 4
Explanation: Since the array is strictly decreasing, we can only keep a single element. Therefore we need to remove a subarray of length 4, either [5,4,3,2] or [4,3,2,1].

Example 3:

Input: arr = [1,2,3]
Output: 0
Explanation: The array is already non-decreasing. We do not need to remove any elements.

Example 4:

Input: arr = [1]
Output: 0

Constraints:

• 1 <= arr.length <= 10^5
• 0 <= arr[i] <= 10^9

Solution: Two Pointers

Find the right most j such that arr[j – 1] > arr[j], if not found which means the entire array is sorted return 0. Then we have a non-descending subarray arr[j~n-1].

We maintain two pointers i, j, such that arr[0~i] is non-descending and arr[i] <= arr[j] which means we can remove arr[i+1~j-1] to get a non-descending array. Number of elements to remove is j – i – 1 .

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int findLengthOfShortestSubarray(vector<int>& arr) {
    const int n = arr.size();
    int j = n - 1;
    while (j > 0 && arr[j - 1] <= arr[j]) --j;
    if (j == 0) return 0;
    int ans = j; // remove arr[0~j-1]
    for (int i = 0; i < n; ++i) {
      if (i > 0 && arr[i - 1] > arr[i]) break;
      while (j < n && arr[i] > arr[j]) ++j;      
      // arr[i] <= arr[j], remove arr[i + 1 ~ j - 1]
      ans = min(ans, j - i - 1);
    }
    return ans;
  }
};

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