LeetCode #1574 — MEDIUM

Shortest Subarray to be Removed to Make Array Sorted

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers · Binary Search · Stack

Example 1

[1,2,3,10,4,2,3,5]

Example 2

[5,4,3,2,1]

Example 3

[1,2,3]

Related Problems

Count the Number of Incremovable Subarrays II (count-the-number-of-incremovable-subarrays-ii)
Count the Number of Incremovable Subarrays I (count-the-number-of-incremovable-subarrays-i)

Step 02

Core Insight

What unlocks the optimal approach

The key is to find the longest non-decreasing subarray starting with the first element or ending with the last element, respectively.
After removing some subarray, the result is the concatenation of a sorted prefix and a sorted suffix, where the last element of the prefix is smaller than the first element of the suffix.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1574: Shortest Subarray to be Removed to Make Array Sorted
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;
    }
}

// Accepted solution for LeetCode #1574: Shortest Subarray to be Removed to Make Array Sorted
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
}

# Accepted solution for LeetCode #1574: Shortest Subarray to be Removed to Make Array Sorted
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

// Accepted solution for LeetCode #1574: Shortest Subarray to be Removed to Make Array Sorted
struct Solution;

impl Solution {
    fn find_length_of_shortest_subarray(arr: Vec<i32>) -> i32 {
        let n = arr.len();
        let mut l = 0;
        while l + 1 < n && arr[l] <= arr[l + 1] {
            l += 1;
        }
        if l == n - 1 {
            return 0;
        }
        let mut r = n - 1;
        while r > 0 && arr[r - 1] <= arr[r] {
            r -= 1;
        }
        let mut res = (n - 1 - l).min(r);
        let mut i = 0;
        let mut j = r;
        while i <= l && j < n {
            if arr[i] <= arr[j] {
                res = res.min(j - i - 1);
                i += 1;
            } else {
                j += 1;
            }
        }
        res as i32
    }
}

#[test]
fn test() {
    let arr = vec![1, 2, 3, 10, 4, 2, 3, 5];
    let res = 3;
    assert_eq!(Solution::find_length_of_shortest_subarray(arr), res);
    let arr = vec![5, 4, 3, 2, 1];
    let res = 4;
    assert_eq!(Solution::find_length_of_shortest_subarray(arr), res);
    let arr = vec![1, 2, 3];
    let res = 0;
    assert_eq!(Solution::find_length_of_shortest_subarray(arr), res);
    let arr = vec![1];
    let res = 0;
    assert_eq!(Solution::find_length_of_shortest_subarray(arr), res);
}

// Accepted solution for LeetCode #1574: Shortest Subarray to be Removed to Make Array Sorted
function findLengthOfShortestSubarray(arr: number[]): number {
    let [l, r, n] = [0, arr.length - 1, arr.length];

    while (r && arr[r - 1] <= arr[r]) r--;
    if (r === 0) return 0;

    let ans = r;
    while (l < r && (!l || arr[l - 1] <= arr[l])) {
        while (r < n && arr[l] > arr[r]) r++;
        ans = Math.min(ans, r - l - 1);
        l++;
    }

    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.