What is the time complexity of Shortest Subarray to be Removed to Make Array Sorted?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Shortest Subarray to be Removed to Make Array Sorted?

This problem uses Array, Two Pointers, Binary Search, Stack, Monotonic Stack. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Shortest Subarray to be Removed to Make Array Sorted?

Insight 1: The remaining array after removing the subarray can be thought of as two non-decreasing sequences joined together: a prefix and a suffix of the original array. Insight 2: We can use two pointers, one at the beginning and one at the end of the array, to find the longest non-decreasing prefix and suffix respectively.

Which companies ask LeetCode 1574. Shortest Subarray to be Removed to Make Array Sorted in interviews?

This problem is frequently asked at DE Shaw, razorpay.

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

Shortest Subarray to be Removed to Make Array Sorted - Complete Solution Guide

Shortest Subarray to be Removed to Make Array Sorted is LeetCode problem 1574, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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].

Detailed Explanation

The problem asks us to find the length of the shortest subarray that can be removed from a given integer array `arr` such that the remaining elements are in non-decreasing order (sorted in ascending order or with equal values). The removed subarray can be empty, meaning no elements need to be removed if the array is already sorted. The goal is to minimize the length of the removed subarray.

Solution Approach

The solution uses a two-pointer approach to find the longest non-decreasing prefix and suffix of the array. Then, it iterates through the prefix and suffix to find the shortest subarray to remove to make the remaining array non-decreasing. It first calculates the length of the shortest subarray to remove by removing either the entire prefix or the entire suffix. Then, it iterates through the prefix and suffix, comparing elements to find the shortest subarray to remove such that the remaining elements form a non-decreasing sequence.

Step-by-Step Algorithm

Step 1: Find the longest non-decreasing prefix of the array using a `left` pointer. Increment `left` as long as `arr[left] <= arr[left + 1]`.
Step 2: Find the longest non-decreasing suffix of the array using a `right` pointer. Decrement `right` as long as `arr[right - 1] <= arr[right]`.
Step 3: Calculate the initial minimum length of the subarray to remove as the minimum of `n - 1 - left` (removing the prefix) and `right` (removing the suffix).
Step 4: Use two pointers `i` and `j`, initialized to 0 and `right` respectively, to iterate through the prefix and suffix. Compare `arr[i]` and `arr[j]`. If `arr[i] <= arr[j]`, it means we can combine the prefix up to index `i` and the suffix starting from index `j` to form a non-decreasing sequence. Update the minimum length of the subarray to remove as `min(ans, j - i - 1)`. Increment `i`. Otherwise, increment `j`.
Step 5: Return the minimum length of the subarray to remove.

Key Insights

Insight 1: The remaining array after removing the subarray can be thought of as two non-decreasing sequences joined together: a prefix and a suffix of the original array.
Insight 2: We can use two pointers, one at the beginning and one at the end of the array, to find the longest non-decreasing prefix and suffix respectively.
Insight 3: Once we have the longest prefix and suffix, we can iterate through them to find the optimal removal length by considering the cases where elements from the prefix and suffix combine to form a larger non-decreasing sequence.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers, Binary Search, Stack, Monotonic Stack.

Companies

Asked at: DE Shaw, razorpay.

Shortest Subarray to be Removed to Make Array Sorted | LeetCode 1574

Medium

Array Two Pointers Binary Search Stack Monotonic Stack

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

Solution Code

Available in:

class Solution {
    public int findLengthOfShortestSubarray(int[] arr) {
        int n = arr.length;
        if (n <= 1) {
            return 0;
        }

        int left = 0;
        while (left + 1 < n && arr[left] <= arr[left + 1]) {
            left++;
        }

        if (left == n - 1) {
            return 0;
        }

        int right = n - 1;
        while (right > 0 && arr[right - 1] <= arr[right]) {
            right--;
        }

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

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

        return ans;
    }
}

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Problem Info

Description

Problem 1574: Shortest Subarray to be Removed to Make Array Sorted - Detailed explanation and solution approaches.

Topics

Array Two Pointers Binary Search Stack Monotonic Stack

Companies

DE Shawrazorpay

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1574

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Find the longest non-decreasing prefix of the array using a `left` pointer. Increment `left` as long as `arr[left] <= arr[left + 1]`.
2
Step 2: Find the longest non-decreasing suffix of the array using a `right` pointer. Decrement `right` as long as `arr[right - 1] <= arr[right]`.
3
Step 3: Calculate the initial minimum length of the subarray to remove as the minimum of `n - 1 - left` (removing the prefix) and `right` (removing the suffix).
4
Step 4: Use two pointers `i` and `j`, initialized to 0 and `right` respectively, to iterate through the prefix and suffix. Compare `arr[i]` and `arr[j]`. If `arr[i] <= arr[j]`, it means we can combine the prefix up to index `i` and the suffix starting from index `j` to form a non-decreasing sequence. Update the minimum length of the subarray to remove as `min(ans, j - i - 1)`. Increment `i`. Otherwise, increment `j`.
5
Step 5: Return the minimum length of the subarray to remove.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because we use a two-pointer approach, iterating through the array a limited number of times directly proportional to the array's length. The space complexity is O(1) because we only use a few integer variables, regardless of the input array's size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Binary Search Approach

Iterate through all possible prefixes. For each prefix, use binary search to find the smallest suffix that can be combined with the prefix to form a sorted array. This involves a binary search for each prefix.

Time: O(n log n)Space: O(1)

Trade-offs: While this approach is valid, the two-pointer approach is more efficient as it avoids the O(log n) cost of binary search for each prefix. The binary search approach might be preferred if memory usage is extremely restricted as it's slightly simpler to implement in a limited memory environment, however the provided solution already has optimal space complexity.

✨

Key Insights

1Insight 1: The remaining array after removing the subarray can be thought of as two non-decreasing sequences joined together: a prefix and a suffix of the original array.
2Insight 2: We can use two pointers, one at the beginning and one at the end of the array, to find the longest non-decreasing prefix and suffix respectively.
3Insight 3: Once we have the longest prefix and suffix, we can iterate through them to find the optimal removal length by considering the cases where elements from the prefix and suffix combine to form a larger non-decreasing sequence.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Edge Cases

Edge case 1: Empty array or array with a single element. The solution correctly returns 0 in this case because no removal is needed.
Edge case 2: The array is already sorted. The solution correctly returns 0 because no removal is needed.
Edge case 3: The array is strictly decreasing. The solution correctly handles this by removing all elements except one, resulting in a subarray of length n-1 to be removed.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the array is already sorted, leading to incorrect results.

✗Common mistake 2: Incorrectly calculating the length of the subarray to remove. Make sure to carefully consider the indices when calculating the length of the removed subarray.

✗Common mistake 3: Not considering the case where removing the entire prefix or suffix results in the shortest removal length.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data cleaning where you want to remove a small chunk of corrupted or out-of-order data from a larger dataset to maintain the integrity of the dataset's sorted order.

🏦Real-world scenario 2: Optimizing time series data. You might identify a short period where sensor readings are unreliable or noisy and removing those would yield a cleaner sorted signal for analysis.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 13, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Shortest Subarray to be Removed to Make Array Sorted"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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