What is the time complexity of Subarray With Elements Greater Than Varying Threshold?

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

What data structures are used to solve Subarray With Elements Greater Than Varying Threshold?

This problem uses Array, Stack, Union Find, Monotonic Stack. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Subarray With Elements Greater Than Varying Threshold?

Insight 1: Monotonic stack is crucial for efficiently finding the next smaller element to the left and right for each element in the array. This helps determine the maximum possible size 'k' for which the element could be the minimum in that subarray. Insight 2: The problem can be solved in O(n) time using the monotonic stack approach because each element is pushed onto and popped off the stack at most once.

Which companies ask LeetCode 2334. Subarray With Elements Greater Than Varying Threshold in interviews?

This problem is frequently asked at instabase.

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Subarray With Elements Greater Than Varying Threshold - LeetCode 2334 Solution

Subarray With Elements Greater Than Varying Threshold - Complete Solution Guide

Subarray With Elements Greater Than Varying Threshold is LeetCode problem 2334, 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 an integer array nums and an integer threshold . Find any subarray of nums of length k such that every element in the subarray is greater than threshold / k . Return the size of any such subarray . If there is no such subarray, return -1 . A subarray is a contiguous non-empty sequence of elements within an array. Example 1: Input: nums = [1,3,4,3,1], threshold = 6 Output: 3 Explanation: The subarray [3,4,3] has a size of 3, and every element is greater than 6 / 3 = 2. Note that thi

Detailed Explanation

The problem asks us to find the size of any subarray within a given array `nums` that satisfies a specific condition. The condition is that every element in the subarray must be greater than `threshold / k`, where `k` is the size (length) of the subarray, and `threshold` is a given integer. If such a subarray exists, we return its size. If no such subarray exists, we return -1.

Solution Approach

The solution utilizes monotonic stacks to efficiently determine the leftmost and rightmost boundaries for each element `nums[i]` within which `nums[i]` is the minimum value. Once we have these boundaries, we can calculate the maximum possible size `k` of a subarray where `nums[i]` could be the minimum. We then check if `nums[i] > threshold / k`. If this condition holds, we have found a valid subarray size and return `k`.

Step-by-Step Algorithm

Step 1: Initialize two arrays, `left` and `right`, to store the indices of the nearest smaller elements to the left and right of each element in `nums`, respectively. Initialize them with default values of -1 for `left` and `n` (length of nums) for `right`.
Step 2: Use a monotonic increasing stack to find the `left` array. Iterate through `nums` from left to right. For each element `nums[i]`, pop elements from the stack until the stack is empty or the element at the top of the stack is smaller than `nums[i]`. The index at the top of the stack (if not empty) is the index of the nearest smaller element to the left of `nums[i]`. Store this index in `left[i]`.
Step 3: Push the current index `i` onto the stack.
Step 4: Use another monotonic increasing stack (or clear the first one) to find the `right` array. Iterate through `nums` from right to left. For each element `nums[i]`, pop elements from the stack until the stack is empty or the element at the top of the stack is smaller than `nums[i]`. The index at the top of the stack (if not empty) is the index of the nearest smaller element to the right of `nums[i]`. Store this index in `right[i]`.
Step 5: Push the current index `i` onto the stack.
Step 6: Iterate through `nums`. For each element `nums[i]`, calculate the potential subarray size `k` as `right[i] - left[i] - 1`. This represents the length of the subarray where `nums[i]` is the smallest element.
Step 7: Check if `k > 0` and if `nums[i] > threshold / k`. If both conditions are true, return `k` because we found a valid subarray size.
Step 8: If the loop completes without finding a valid subarray size, return -1.

Key Insights

Insight 1: Monotonic stack is crucial for efficiently finding the next smaller element to the left and right for each element in the array. This helps determine the maximum possible size 'k' for which the element could be the minimum in that subarray.
Insight 2: The problem can be solved in O(n) time using the monotonic stack approach because each element is pushed onto and popped off the stack at most once.
Insight 3: It's important to compare `nums[i]` to `threshold / k` using floating-point division to handle potential precision issues, especially when `nums[i]` and `threshold` are integers.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Stack, Union Find, Monotonic Stack.

Companies

Asked at: instabase.

Subarray With Elements Greater Than Varying Threshold | LeetCode 2334

Hard

Array Stack Union Find Monotonic Stack

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

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public int validSubarraySize(int[] nums, int threshold) {
        int n = nums.length;
        
        int[] left = new int[n];
        for (int i = 0; i < n; i++) {
            left[i] = -1;
        }
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < n; i++) {
            while (!stack.isEmpty() && nums[stack.peek()] >= nums[i]) {
                stack.pop();
            }
            if (!stack.isEmpty()) {
                left[i] = stack.peek();
            }
            stack.push(i);
        }
        
        int[] right = new int[n];
        for (int i = 0; i < n; i++) {
            right[i] = n;
        }
        stack = new Stack<>();
        for (int i = n - 1; i >= 0; i--) {
            while (!stack.isEmpty() && nums[stack.peek()] >= nums[i]) {
                stack.pop();
            }
            if (!stack.isEmpty()) {
                right[i] = stack.peek();
            }
            stack.push(i);
        }
        
        for (int i = 0; i < n; i++) {
            int k = right[i] - left[i] - 1;
            if (k > 0 && (double)nums[i] > (double)threshold / k) {
                return k;
            }
        }
        
        return -1;
    }
}

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Problem Info

Description

Problem 2334: Subarray With Elements Greater Than Varying Threshold - Detailed explanation and solution approaches.

Topics

Array Stack Union Find Monotonic Stack

Companies

instabase

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2334

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: Initialize two arrays, `left` and `right`, to store the indices of the nearest smaller elements to the left and right of each element in `nums`, respectively. Initialize them with default values of -1 for `left` and `n` (length of nums) for `right`.
2
Step 2: Use a monotonic increasing stack to find the `left` array. Iterate through `nums` from left to right. For each element `nums[i]`, pop elements from the stack until the stack is empty or the element at the top of the stack is smaller than `nums[i]`. The index at the top of the stack (if not empty) is the index of the nearest smaller element to the left of `nums[i]`. Store this index in `left[i]`.
3
Step 3: Push the current index `i` onto the stack.
4
Step 4: Use another monotonic increasing stack (or clear the first one) to find the `right` array. Iterate through `nums` from right to left. For each element `nums[i]`, pop elements from the stack until the stack is empty or the element at the top of the stack is smaller than `nums[i]`. The index at the top of the stack (if not empty) is the index of the nearest smaller element to the right of `nums[i]`. Store this index in `right[i]`.
5
Step 5: Push the current index `i` onto the stack.
6
Step 6: Iterate through `nums`. For each element `nums[i]`, calculate the potential subarray size `k` as `right[i] - left[i] - 1`. This represents the length of the subarray where `nums[i]` is the smallest element.
7
Step 7: Check if `k > 0` and if `nums[i] > threshold / k`. If both conditions are true, return `k` because we found a valid subarray size.
8
Step 8: If the loop completes without finding a valid subarray size, return -1.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The monotonic stack operations guarantee that we process each element a constant number of times. The space complexity stems from the need to store the left and right boundaries for each element.

📖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

Brute Force

Iterate through all possible subarrays and check if each subarray satisfies the given condition. For each subarray of size `k`, iterate through its elements and check if each element is greater than `threshold / k`.

Time: O(n^2), where n is the length of the input array `nums`. This is because there are O(n^2) possible subarrays, and checking each subarray takes O(k) time, where k is at most n.Space: O(1), as it doesn't require any extra space.

Trade-offs: The brute force approach is simple to understand but inefficient for large input arrays. It is only suitable for small values of n. The monotonic stack approach provides a significantly better time complexity of O(n).

✨

Key Insights

1Insight 1: Monotonic stack is crucial for efficiently finding the next smaller element to the left and right for each element in the array. This helps determine the maximum possible size 'k' for which the element could be the minimum in that subarray.
2Insight 2: The problem can be solved in O(n) time using the monotonic stack approach because each element is pushed onto and popped off the stack at most once.
3Insight 3: It's important to compare `nums[i]` to `threshold / k` using floating-point division to handle potential precision issues, especially when `nums[i]` and `threshold` are integers.

📚

Key Terms Explained

For Students

Array

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

Stack

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

Edge Cases

Edge case 1: Empty input array. While the problem states `1 <= nums.length`, consider an empty array or a case where no subarray satisfies the condition. The code handles the case where no valid subarray is found by returning -1.
Edge case 2: threshold is very large. The solution handles this by performing floating point division in the comparison `nums[i] > threshold / k` to avoid integer overflow issues.
Edge case 3: threshold is 0. This won't cause an exception in the given solutions, but the condition for a valid subarray becomes nums[i] > 0, which essentially requires finding a subarray with all positive numbers. The code correctly handles this.
Edge case 4: All numbers in `nums` are the same. The monotonic stack will still correctly find the left and right boundaries.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the subarray size `k`. The correct calculation is `right[i] - left[i] - 1`. Off-by-one errors are common.

✗Common mistake 2: Integer division. Using integer division to compare `nums[i]` and `threshold / k` may lead to incorrect results. It is important to use floating-point division by casting one of the operands to a `double` to ensure accurate comparison.

✗Common mistake 3: Not handling the edge cases of empty input or when no valid subarray exists. The solution should return -1 in these cases.

✅

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: Stock span problem. Finding the span of a stock price in a given range, where the span is the number of consecutive days before the current day when the stock price was less than or equal to the current day's price. The monotonic stack approach is used to efficiently solve the stock span problem.

🏦Real-world scenario 2: Finding the largest rectangular area in a histogram. The histogram can be represented as an array of heights, and the largest rectangular area can be found by determining the left and right boundaries for each bar in the histogram, similar to the approach used in this problem.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: November 22, 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.

What problems are best solved with a stack?

Stacks excel at: matching problems (parentheses validation), next greater/smaller element, expression evaluation, backtracking, and undo operations. The key pattern is LIFO (Last In, First Out) - when you need to access the most recently added element.

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 "Subarray With Elements Greater Than Varying Threshold"?

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 stack techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?