LeetCode 1574: Shortest Subarray to be Removed to Make Array Sorted Solution

Master LeetCode problem 1574 (Shortest Subarray to be Removed to Make Array Sorted), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1574. Shortest Subarray to be Removed to Make Array Sorted

Medium

Array Two Pointers Binary Search Stack Monotonic Stack

Problem Explanation

Explanation

To solve this problem, we need to find the minimum number of elements that need to be removed from the given array arr to make the remaining elements non-decreasing. We can achieve this by identifying the longest non-decreasing prefix and suffix of the array and then determining the minimum length of the subarray to be removed.

Find the length of the longest non-decreasing prefix starting from the beginning of the array.
Find the length of the longest non-decreasing suffix starting from the end of the array.
If the entire array is non-decreasing, return 0.
Otherwise, we need to find the minimum subarray to be removed. This can be done by removing the middle part between the prefix and suffix.

The time complexity of this approach is O(n), where n is the number of elements in the array. The space complexity is O(1) as we are using only a constant amount of extra space.

Solution Code

class Solution {
    public int findLengthOfShortestSubarray(int[] arr) {
        int n = arr.length;
        int left = 0, right = n - 1;

        // Find the length of the longest non-decreasing prefix
        while (left + 1 < n && arr[left] <= arr[left + 1]) {
            left++;
        }

        // If the entire array is non-decreasing
        if (left == n - 1) {
            return 0;
        }

        // Find the length of the longest non-decreasing suffix
        while (right > 0 && arr[right - 1] <= arr[right]) {
            right--;
        }

        int res = Math.min(n - left - 1, right);

        int i = 0, j = right;
        while (i <= left && j < n) {
            if (arr[i] <= arr[j]) {
                res = Math.min(res, j - i - 1);
                i++;
            } else {
                j++;
            }
        }

        return res;
    }
}

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