What is the time complexity of Longest Non-decreasing Subarray From Two Arrays?

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

What data structures are used to solve Longest Non-decreasing Subarray From Two Arrays?

This problem uses Array, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

Which companies ask LeetCode 2771. Longest Non-decreasing Subarray From Two Arrays in interviews?

This problem is frequently asked at Citadel.

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Longest Non-decreasing Subarray From Two Arrays - LeetCode 2771 Solution

Longest Non-decreasing Subarray From Two Arrays - Complete Solution Guide

Longest Non-decreasing Subarray From Two Arrays is LeetCode problem 2771, 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

You are given two 0-indexed integer arrays nums1 and nums2 of length n . Let's define another 0-indexed integer array, nums3 , of length n . For each index i in the range [0, n - 1] , you can assign either nums1[i] or nums2[i] to nums3[i] . Your task is to maximize the length of the longest non-decreasing subarray in nums3 by choosing its values optimally. Return an integer representing the length of the longest non-decreasing subarray in nums3 . Note: A subarray is a contiguous non-empty sequen

Detailed Explanation

The problem asks us to find the length of the longest non-decreasing subarray that can be formed by constructing a third array `nums3`. For each index `i`, the element `nums3[i]` can be either `nums1[i]` or `nums2[i]`. The goal is to choose the values for `nums3` optimally to maximize the length of its longest non-decreasing subarray. A non-decreasing subarray means that each element in the subarray is greater than or equal to the previous one. The inputs are two integer arrays `nums1` and `nums2` of equal length `n`. The output is a single integer representing the maximum achievable length of the non-decreasing subarray in `nums3`. The length of the input arrays `n` is between 1 and 10^5, and each element in the arrays is between 1 and 10^9.

Solution Approach

The solution uses dynamic programming to find the maximum length of the non-decreasing subarray. We maintain two variables, `dp1` and `dp2`, which store the length of the longest non-decreasing subarray ending at the current index using `nums1[i]` and `nums2[i]` respectively. We iterate through the arrays and update `dp1` and `dp2` at each index based on whether `nums1[i]` and `nums2[i]` can extend the previous non-decreasing subarrays formed by either `nums1[i-1]` or `nums2[i-1]`. The maximum of `dp1`, `dp2`, and the current maximum length `ans` is updated at each step.

Step-by-Step Algorithm

Step 1: Initialize `n` as the length of the input arrays, `dp1` and `dp2` to 1 (since a single element is a non-decreasing subarray of length 1), and `ans` to 1.
Step 2: Iterate from `i = 1` to `n - 1`.
Step 3: Calculate `temp1`, the length of the longest non-decreasing subarray ending at index `i` using `nums1[i]`. Initialize `temp1` to 1.
Step 4: If `nums1[i] >= nums1[i-1]`, then we can extend the previous non-decreasing subarray ending with `nums1[i-1]`. Update `temp1 = max(temp1, dp1 + 1)`.
Step 5: If `nums1[i] >= nums2[i-1]`, then we can extend the previous non-decreasing subarray ending with `nums2[i-1]`. Update `temp1 = max(temp1, dp2 + 1)`.
Step 6: Calculate `temp2`, the length of the longest non-decreasing subarray ending at index `i` using `nums2[i]`. Initialize `temp2` to 1.
Step 7: If `nums2[i] >= nums1[i-1]`, then we can extend the previous non-decreasing subarray ending with `nums1[i-1]`. Update `temp2 = max(temp2, dp1 + 1)`.
Step 8: If `nums2[i] >= nums2[i-1]`, then we can extend the previous non-decreasing subarray ending with `nums2[i-1]`. Update `temp2 = max(temp2, dp2 + 1)`.
Step 9: Update `dp1 = temp1` and `dp2 = temp2`.
Step 10: Update `ans = max(ans, max(dp1, dp2))`.
Step 11: After the loop completes, return `ans`.

Key Insights

Insight 1: Dynamic Programming is suitable because the optimal solution for a larger subarray can be built upon the optimal solutions of smaller subarrays. We can keep track of the length of the longest non-decreasing subarray ending at each index.
Insight 2: We need to maintain two DP values for each index, one representing the longest non-decreasing subarray ending with `nums1[i]` and the other ending with `nums2[i]`.
Insight 3: The space complexity can be optimized to O(1) since we only need to consider the results from the previous index to compute the current result.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Citadel.

Longest Non-decreasing Subarray From Two Arrays | LeetCode 2771

Medium

Array Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public int maxNonDecreasingLength(int[] nums1, int[] nums2) {
        int n = nums1.length;
        int dp1 = 1;
        int dp2 = 1;
        int ans = 1;

        for (int i = 1; i < n; i++) {
            int temp1 = 1;
            if (nums1[i] >= nums1[i - 1]) {
                temp1 = Math.max(temp1, dp1 + 1);
            }
            if (nums1[i] >= nums2[i - 1]) {
                temp1 = Math.max(temp1, dp2 + 1);
            }

            int temp2 = 1;
            if (nums2[i] >= nums1[i - 1]) {
                temp2 = Math.max(temp2, dp1 + 1);
            }
            if (nums2[i] >= nums2[i - 1]) {
                temp2 = Math.max(temp2, dp2 + 1);
            }

            dp1 = temp1;
            dp2 = temp2;

            ans = Math.max(ans, Math.max(dp1, dp2));
        }

        return ans;
    }
}

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Problem Info

Description

Problem 2771: Longest Non-decreasing Subarray From Two Arrays - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

Citadel

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2771

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

11 steps

1
Step 1: Initialize `n` as the length of the input arrays, `dp1` and `dp2` to 1 (since a single element is a non-decreasing subarray of length 1), and `ans` to 1.
2
Step 2: Iterate from `i = 1` to `n - 1`.
3
Step 3: Calculate `temp1`, the length of the longest non-decreasing subarray ending at index `i` using `nums1[i]`. Initialize `temp1` to 1.
4
Step 4: If `nums1[i] >= nums1[i-1]`, then we can extend the previous non-decreasing subarray ending with `nums1[i-1]`. Update `temp1 = max(temp1, dp1 + 1)`.
5
Step 5: If `nums1[i] >= nums2[i-1]`, then we can extend the previous non-decreasing subarray ending with `nums2[i-1]`. Update `temp1 = max(temp1, dp2 + 1)`.
6
Step 6: Calculate `temp2`, the length of the longest non-decreasing subarray ending at index `i` using `nums2[i]`. Initialize `temp2` to 1.
7
Step 7: If `nums2[i] >= nums1[i-1]`, then we can extend the previous non-decreasing subarray ending with `nums1[i-1]`. Update `temp2 = max(temp2, dp1 + 1)`.
8
Step 8: If `nums2[i] >= nums2[i-1]`, then we can extend the previous non-decreasing subarray ending with `nums2[i-1]`. Update `temp2 = max(temp2, dp2 + 1)`.
9
Step 9: Update `dp1 = temp1` and `dp2 = temp2`.
10
Step 10: Update `ans = max(ans, max(dp1, dp2))`.
11
Step 11: After the loop completes, return `ans`.

💡

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 algorithm iterates through the input array once. Inside the loop, only constant time operations like comparison and assignment are performed. Therefore, the time complexity is linear. Only a fixed number of variables are used. Thus, space complexity is constant.

📖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

Recursion with Memoization

A recursive approach can be used to explore all possible combinations of elements from nums1 and nums2 to construct nums3 and find the longest non-decreasing subarray. Memoization can be used to avoid redundant calculations.

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

Trade-offs: Recursion with memoization has the same time complexity as the iterative DP approach but uses O(n) space due to the memoization table and call stack. The iterative approach is generally preferred due to its lower space complexity and reduced overhead.

✨

Key Insights

1Insight 1: Dynamic Programming is suitable because the optimal solution for a larger subarray can be built upon the optimal solutions of smaller subarrays. We can keep track of the length of the longest non-decreasing subarray ending at each index.
2Insight 2: We need to maintain two DP values for each index, one representing the longest non-decreasing subarray ending with `nums1[i]` and the other ending with `nums2[i]`.
3Insight 3: The space complexity can be optimized to O(1) since we only need to consider the results from the previous index to compute the current result.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: When the input arrays have only one element (n = 1), the algorithm correctly returns 1, as a single element forms a non-decreasing subarray of length 1.
Edge case 2: When all elements in both arrays are non-decreasing, the algorithm correctly computes the length of the entire array as the longest non-decreasing subarray. For example, nums1 = [1, 2, 3], nums2 = [1, 2, 3] will return 3.
Edge case 3: When all elements in both arrays are strictly decreasing, the algorithm correctly returns 1 because no non-decreasing subarray longer than 1 exists. For example, nums1 = [3, 2, 1], nums2 = [3, 2, 1] will return 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering both `nums1[i]` and `nums2[i]` as potential candidates for extending the non-decreasing subarray. The algorithm must explore both options to find the optimal solution.

✗Common mistake 2: Incorrectly updating `dp1` and `dp2`. It's important to use temporary variables to store the new values of `dp1` and `dp2` before updating them to avoid using the updated values in the same iteration.

✗Common mistake 3: Not initializing `ans` to 1, causing an incorrect result when the longest non-decreasing subarray has length 1.

✅

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: Optimizing resource allocation. Consider two possible production schedules (nums1 and nums2) for each day. The goal is to choose a schedule for each day to maximize the duration of a period where production is non-decreasing, representing stable growth.

🏦Real-world scenario 2: Analyzing stock prices. Given two sources of stock price data (nums1 and nums2) for a particular stock, determine the longest period where the stock price is non-decreasing, by choosing the most favorable data source for each day. This can help identify periods of sustained growth or stability.

💡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: August 11, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Longest Non-decreasing Subarray From Two Arrays"?

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

🤖 Need help with this problem?