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

In this guide, you will get 1574. Shortest Subarray to be Removed to Make Array Sorted LeetCode Solution with the best time and space complexity. The solution to Shortest Subarray to be Removed to Make Array Sorted problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Shortest Subarray to be Removed to Make Array Sorted solution in C++
4. Shortest Subarray to be Removed to Make Array Sorted solution in Java
5. Shortest Subarray to be Removed to Make Array Sorted solution in Python
6. Additional Resources

Problem Statement of Shortest Subarray to be Removed to Make Array Sorted

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

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(1)

1574. Shortest Subarray to be Removed to Make Array Sorted LeetCode Solution in C++

class Solution {
 public:
  int findLengthOfShortestSubarray(vector<int>& arr) {
    const int n = arr.size();
    int l = 0;
    int r = n - 1;

    // arr[0..l] is non-decreasing.
    while (l < n - 1 && arr[l + 1] >= arr[l])
      ++l;
    // arr[r..n - 1] is non-decreasing.
    while (r > 0 && arr[r - 1] <= arr[r])
      --r;
    // Remove arr[l + 1..n - 1] or arr[0..r - 1].
    int ans = min(n - 1 - l, r);

    // Since arr[0..l] and arr[r..n - 1] are non-decreasing, we place pointers
    // at the rightmost indices, l and n - 1, and greedily shrink them toward
    // the leftmost indices, 0 and r, respectively. By removing arr[i + 1..j],
    // we ensure that `arr` becomes non-decreasing.
    int i = l;
    int j = n - 1;
    while (i >= 0 && j >= r && j > i) {
      if (arr[i] <= arr[j])
        --j;
      else
        --i;
      ans = min(ans, j - i);
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

1574. Shortest Subarray to be Removed to Make Array Sorted LeetCode Solution in Java

class Solution {
  public int findLengthOfShortestSubarray(int[] arr) {
    final int n = arr.length;
    int l = 0;
    int r = n - 1;

    // arr[0..l] is non-decreasing prefix.
    while (l < n - 1 && arr[l + 1] >= arr[l])
      ++l;
    // arr[r..n - 1] is non-decreasing suffix.
    while (r > 0 && arr[r - 1] <= arr[r])
      --r;
    // Remove arr[l + 1..n - 1] or arr[0..r - 1]
    int ans = Math.min(n - 1 - l, r);

    // Since arr[0..l] and arr[r..n - 1] are non-decreasing, we place pointers
    // at the rightmost indices, l and n - 1, and greedily shrink them toward
    // the leftmost indices, 0 and r, respectively. By removing arr[i + 1..j],
    // we ensure that `arr` becomes non-decreasing.
    int i = l;
    int j = n - 1;
    while (i >= 0 && j >= r && j > i) {
      if (arr[i] <= arr[j])
        --j;
      else
        --i;
      ans = Math.min(ans, j - i);
    }

    return ans;
  }
}
// code provided by PROGIEZ

1574. Shortest Subarray to be Removed to Make Array Sorted LeetCode Solution in Python

class Solution:
  def findLengthOfShortestSubarray(self, arr: list[int]) -> int:
    n = len(arr)
    l = 0
    r = n - 1

    # arr[0..l] is non-decreasing.
    while l < n - 1 and arr[l + 1] >= arr[l]:
      l += 1
    # arr[r..n - 1] is non-decreasing.
    while r > 0 and arr[r - 1] <= arr[r]:
      r -= 1
    # Remove arr[l + 1..n - 1] or arr[0..r - 1].
    ans = min(n - 1 - l, r)

    # Since arr[0..l] and arr[r..n - 1] are non-decreasing, we place pointers
    # at the rightmost indices, l and n - 1, and greedily shrink them toward
    # the leftmost indices, 0 and r, respectively. By removing arr[i + 1..j],
    # we ensure that `arr` becomes non-decreasing.
    i = l
    j = n - 1
    while i >= 0 and j >= r and j > i:
      if arr[i] <= arr[j]:
        j -= 1
      else:
        i -= 1
      ans = min(ans, j - i)

    return ans
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

Happy Coding! Keep following PROGIEZ for more updates and solutions.