1574. Shortest Subarray to be Removed to Make Array Sorted

Problem

Given an integer array arr, remove a subarray (can be empty) from arr such that the remaining elements in arr are non-decreasing.

Return the length of the shortest subarray to remove.

A subarray is a contiguous subsequence of the array.

  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.

  Constraints:

• 1 <= arr.length <= 10^5

• 0 <= arr[i] <= 10^9

Solution (Java)

class Solution {
    public int findLengthOfShortestSubarray(int[] arr) {
        int left = 0;
        while (left < arr.length - 1 && arr[left] <= arr[left + 1]) {
            left++;
        }
        if (left == arr.length - 1) {
            return 0;
        }
        int right = arr.length - 1;
        while (right > left && arr[right] >= arr[right - 1]) {
            right--;
        }
        if (right == 0) {
            return arr.length - 1;
        }
        int result = Math.min(arr.length - left - 1, right);
        int i = 0;
        int j = right;
        while (i <= left && j < arr.length) {
            if (arr[j] >= arr[i]) {
                result = Math.min(result, j - i - 1);
                i++;
            } else {
                j++;
            }
        }
        return result;
    }
}

Explain:

nope.

Complexity:

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