What is the time complexity of Count the Number of Incremovable Subarrays II?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Count the Number of Incremovable Subarrays II?

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

What are the key insights for solving Count the Number of Incremovable Subarrays II?

Insight 1: The problem can be solved by identifying the longest strictly increasing prefix and suffix of the array. The incremovable subarrays are essentially those whose removal allows for the prefix and suffix to be concatenated into a strictly increasing sequence. Insight 2: Iterating through all possible prefixes and then finding a suitable suffix using two pointers is a key step. The two-pointer technique helps determine the shortest suffix that can be combined with the current prefix to form a strictly increasing array.

Which companies ask LeetCode 2972. Count the Number of Incremovable Subarrays II in interviews?

This problem is frequently asked at DE Shaw, IBM.

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Count the Number of Incremovable Subarrays II - LeetCode 2972 Solution

Count the Number of Incremovable Subarrays II - Complete Solution Guide

Count the Number of Incremovable Subarrays II is LeetCode problem 2972, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 0-indexed array of positive integers nums . A subarray of nums is called incremovable if nums becomes strictly increasing on removing the subarray. For example, the subarray [3, 4] is an incremovable subarray of [5, 3, 4, 6, 7] because removing this subarray changes the array [5, 3, 4, 6, 7] to [5, 6, 7] which is strictly increasing. Return the total number of incremovable subarrays of nums . Note that an empty array is considered strictly increasing. A subarray is a contiguous n

Detailed Explanation

The problem asks us to count the number of 'incremovable' subarrays within a given array of positive integers. A subarray is considered 'incremovable' if, upon removing it from the original array, the remaining elements form a strictly increasing sequence. An empty array is also considered strictly increasing. We need to find all possible contiguous subarrays that satisfy this condition and return the total count.

Solution Approach

The solution first identifies the longest strictly increasing prefix and suffix of the given array. Then, it iterates through each possible prefix ending at index `i` (up to the end of the longest increasing prefix). For each such prefix, it finds the smallest suffix starting at index `j` such that every element in the suffix is greater than the last element of the prefix (or the prefix is empty). All suffixes starting at index `j` are valid, and the number of such suffixes is added to the total count. The algorithm leverages the fact that once a valid starting index `j` for the suffix is found for a prefix, all larger values of `j` will also result in valid suffixes since the suffix is increasing.

Step-by-Step Algorithm

Step 1: Find the end index `l` of the longest strictly increasing prefix of the array.
Step 2: Find the start index `r` of the longest strictly increasing suffix of the array.
Step 3: Initialize a counter `count` to store the number of incremovable subarrays.
Step 4: Consider the case where the remaining part is just a valid suffix. Add `n - r + 1` to `count` to account for all suffixes starting at or after index `r` (including an empty suffix).
Step 5: Iterate through all possible prefixes ending at index `i` from 0 to `l`.
Step 6: For each prefix, find the smallest index `j` (starting from `r`) such that `nums[j] > nums[i]`.
Step 7: Add `n - j + 1` to `count` because all suffixes starting from index `j` onwards are valid.
Step 8: Return the final `count`.

Key Insights

Insight 1: The problem can be solved by identifying the longest strictly increasing prefix and suffix of the array. The incremovable subarrays are essentially those whose removal allows for the prefix and suffix to be concatenated into a strictly increasing sequence.
Insight 2: Iterating through all possible prefixes and then finding a suitable suffix using two pointers is a key step. The two-pointer technique helps determine the shortest suffix that can be combined with the current prefix to form a strictly increasing array.
Insight 3: An empty prefix is also a possibility, meaning the array can be made strictly increasing just by removing a prefix from the original array, leaving only the suffix.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

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

Companies

Asked at: DE Shaw, IBM.

Count the Number of Incremovable Subarrays II | LeetCode 2972

Hard

Array Two Pointers Binary Search

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public long incremovableSubarrayCount(int[] nums) {
        int n = nums.length;
        
        // Find the end index 'l' of the longest strictly increasing prefix
        int l = 0;
        while (l + 1 < n && nums[l] < nums[l + 1]) {
            l++;
        }
        
        // If the entire array is strictly increasing, any subarray is incremovable
        if (l == n - 1) {
            return (long)n * (n + 1) / 2;
        }
        
        // Find the start index 'r' of the longest strictly increasing suffix
        int r = n - 1;
        while (r > 0 && nums[r - 1] < nums[r]) {
            r--;
        }
        
        // An incremovable subarray's removal leaves a concatenation of a valid prefix
        // and a valid suffix of the original array.
        // A valid prefix is nums[0...i] where i <= l.
        // A valid suffix is nums[j...n-1] where j >= r.

        // Case 1: The remaining part is just a valid suffix (the prefix is empty).
        // This corresponds to removing a prefix of the original array.
        // Valid suffixes are nums[k...n-1] for k >= r, plus the empty suffix.
        // This gives (n - r) + 1 possibilities.
        long count = n - r + 1;
        
        // Case 2: The remaining part has a non-empty prefix.
        // For each valid prefix nums[0...i] (where 0 <= i <= l),
        // count the number of valid suffixes that can follow it.
        int j = r;
        for (int i = 0; i <= l; i++) {
            // For the prefix ending at nums[i], we need to find the first element
            // in the suffix part (from index r onwards) that is strictly
            // greater than nums[i].
            while (j < n && nums[j] <= nums[i]) {
                j++;
            }
            
            // All suffixes starting from index j onwards are valid.
            // These are nums[j...n-1], nums[j+1...n-1], ..., and the empty suffix.
            // The number of such suffixes is (n - j) + 1.
            count += n - j + 1;
            
        }
        
        return count;
    }
}

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Problem Info

Description

Problem 2972: Count the Number of Incremovable Subarrays II - Detailed explanation and solution approaches.

Topics

Array Two Pointers Binary Search

Companies

DE ShawIBM

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:2972

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

8 steps

1
Step 1: Find the end index `l` of the longest strictly increasing prefix of the array.
2
Step 2: Find the start index `r` of the longest strictly increasing suffix of the array.
3
Step 3: Initialize a counter `count` to store the number of incremovable subarrays.
4
Step 4: Consider the case where the remaining part is just a valid suffix. Add `n - r + 1` to `count` to account for all suffixes starting at or after index `r` (including an empty suffix).
5
Step 5: Iterate through all possible prefixes ending at index `i` from 0 to `l`.
6
Step 6: For each prefix, find the smallest index `j` (starting from `r`) such that `nums[j] > nums[i]`.
7
Step 7: Add `n - j + 1` to `count` because all suffixes starting from index `j` onwards are valid.
8
Step 8: Return the final `count`.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The outer loop iterates at most `n` times (for finding prefixes). The inner loop (finding the suffix starting point `j`) can also iterate up to `n` times in the worst case. Therefore, the overall time complexity is O(n*n) = O(n^2). The algorithm uses only a few constant space variables, so the space complexity is O(1).

📖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

Instead of linearly searching for the suitable suffix starting point `j`, binary search can be used. For each prefix, we can use binary search to find the first element in the suffix that is strictly greater than the last element of the prefix.

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

Trade-offs: Binary search might improve performance compared to a linear search, especially for larger arrays. However, the constant factor overhead of binary search might make it less efficient for smaller arrays. This trade-off depends heavily on the input data and the implementation.

✨

Key Insights

1Insight 1: The problem can be solved by identifying the longest strictly increasing prefix and suffix of the array. The incremovable subarrays are essentially those whose removal allows for the prefix and suffix to be concatenated into a strictly increasing sequence.
2Insight 2: Iterating through all possible prefixes and then finding a suitable suffix using two pointers is a key step. The two-pointer technique helps determine the shortest suffix that can be combined with the current prefix to form a strictly increasing array.
3Insight 3: An empty prefix is also a possibility, meaning the array can be made strictly increasing just by removing a prefix from the original array, leaving only the suffix.

📚

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!

Edge Cases

Edge case 1: When the entire array is strictly increasing. In this case, any subarray is incremovable, and the solution correctly calculates the number of subarrays as n*(n+1)/2.
Edge case 2: When the array is decreasing. The prefix and suffix will consist of only the first and last element respectively. The algorithm will still iterate correctly and return the proper count.
Edge case 3: An empty array is considered strictly increasing after removing the subarray. This case is implicitly handled by accounting for an empty suffix.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the entire array is strictly increasing, leading to an incorrect count.

✗Common mistake 2: Incorrectly identifying the longest strictly increasing prefix and suffix, leading to incorrect bounds for the loops.

✗Common mistake 3: Failing to include the empty suffix, resulting in undercounting the number of incremovable subarrays.

✗Common mistake 4: Assuming that the subarray removed *must* include an element. Removing a prefix to leave just the suffix is also a valid scenario.

✗Common mistake 5: Overlooking the fact that the suffix must be *strictly* increasing. The while loop condition `nums[j] <= nums[i]` is crucial for ensuring a strictly increasing sequence after removing the subarray.

✅

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 analysis where you need to identify sections of data that, when removed, result in a monotonically increasing trend for better forecasting or anomaly detection.

🏦Real-world scenario 2: Resource allocation where you want to remove a subset of tasks such that the remaining tasks can be executed in a strictly increasing order of priority or efficiency.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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: September 1, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 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.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Count the Number of Incremovable Subarrays II"?

For this hard-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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