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1574. Shortest Subarray to be Removed to Make Array Sorted

Description

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 <= 105
  • 0 <= arr[i] <= 109

Solutions

  • class Solution {
    public int findLengthOfShortestSubarray(int[] arr) {
        int n = arr.length;
        int i = 0, j = n - 1;
        while (i + 1 < n && arr[i] <= arr[i + 1]) {
            ++i;
        }
        while (j - 1 >= 0 && arr[j - 1] <= arr[j]) {
            --j;
        }
        if (i >= j) {
            return 0;
        }
        int ans = Math.min(n - i - 1, j);
        for (int l = 0; l <= i; ++l) {
            int r = search(arr, arr[l], j);
            ans = Math.min(ans, r - l - 1);
        }
        return ans;
    }

    private int search(int[] arr, int x, int left) {
        int right = arr.length;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (arr[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

  • class Solution {
public:
    int findLengthOfShortestSubarray(vector<int>& arr) {
        int n = arr.size();
        int i = 0, j = n - 1;
        while (i + 1 < n && arr[i] <= arr[i + 1]) {
            ++i;
        }
        while (j - 1 >= 0 && arr[j - 1] <= arr[j]) {
            --j;
        }
        if (i >= j) {
            return 0;
        }
        int ans = min(n - 1 - i, j);
        for (int l = 0; l <= i; ++l) {
            int r = lower_bound(arr.begin() + j, arr.end(), arr[l]) - arr.begin();
            ans = min(ans, r - l - 1);
        }
        return ans;
    }
};

  • class Solution:
    def findLengthOfShortestSubarray(self, arr: List[int]) -> int:
        n = len(arr)
        i, j = 0, n - 1
        while i + 1 < n and arr[i] <= arr[i + 1]:
            i += 1
        while j - 1 >= 0 and arr[j - 1] <= arr[j]:
            j -= 1
        if i >= j:
            return 0
        ans = min(n - i - 1, j)
        for l in range(i + 1):
            r = bisect_left(arr, arr[l], lo=j)
            ans = min(ans, r - l - 1)
        return ans


  • func findLengthOfShortestSubarray(arr []int) int {
	n := len(arr)
	i, j := 0, n-1
	for i+1 < n && arr[i] <= arr[i+1] {
		i++
	}
	for j-1 >= 0 && arr[j-1] <= arr[j] {
		j--
	}
	if i >= j {
		return 0
	}
	ans := min(n-i-1, j)
	for l := 0; l <= i; l++ {
		r := j + sort.SearchInts(arr[j:], arr[l])
		ans = min(ans, r-l-1)
	}
	return ans
}

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