What is the time complexity of Check if There is a Valid Partition For The Array?

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

What data structures are used to solve Check if There is a Valid Partition For The Array?

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

What are the key insights for solving Check if There is a Valid Partition For The Array?

Insight 1: Dynamic programming is an effective approach to solve this problem since the solution to a subproblem (prefix of the array) can be used to solve larger subproblems. Insight 2: The state of the DP can be represented by whether a partition is valid up to a given index. Since the conditions only depend on the previous 2 or 3 elements, we only need to keep track of the validity of the last 3 states to achieve O(1) space complexity.

Which companies ask LeetCode 2369. Check if There is a Valid Partition For The Array in interviews?

This problem is frequently asked at Coinbase.

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Check if There is a Valid Partition For The Array - LeetCode 2369 Solution

Check if There is a Valid Partition For The Array - Complete Solution Guide

Check if There is a Valid Partition For The Array is LeetCode problem 2369, 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 a 0-indexed integer array nums . You have to partition the array into one or more contiguous subarrays. We call a partition of the array valid if each of the obtained subarrays satisfies one of the following conditions: The subarray consists of exactly 2, equal elements. For example, the subarray [2,2] is good. The subarray consists of exactly 3, equal elements. For example, the subarray [4,4,4] is good. The subarray consists of exactly 3 consecutive increasing elements, that is, t

Detailed Explanation

The problem requires you to determine if a given 0-indexed integer array `nums` can be partitioned into one or more contiguous subarrays, such that each subarray satisfies one of the following conditions: 1. The subarray contains exactly two equal elements (e.g., `[2, 2]`). 2. The subarray contains exactly three equal elements (e.g., `[4, 4, 4]`). 3. The subarray contains exactly three consecutive increasing elements (e.g., `[3, 4, 5]`). The input is an integer array `nums`, and the output is a boolean value indicating whether a valid partition exists.

Solution Approach

The solution uses dynamic programming to determine if a valid partition exists. `dp[i]` represents whether the prefix of the array up to index `i-1` can be validly partitioned. The algorithm iterates through the array, and for each index `i`, it checks if the last 2 or 3 elements can form a valid subarray based on the problem's conditions. It updates the `dp` state based on the validity of smaller subproblems (prefixes). Instead of storing a full `dp` array, the algorithm utilizes only three variables (`dp_prev_three`, `dp_prev_two`, and `dp_prev_one`) to store the last three `dp` states, thus achieving O(1) space complexity.

Step-by-Step Algorithm

Step 1: Initialize `dp_prev_three` to `True`, representing that an empty array is validly partitioned.
Step 2: Initialize `dp_prev_two` to `False`, representing that a single-element array is not validly partitioned.
Step 3: Initialize `dp_prev_one` to `True` if the first two elements are equal, and `False` otherwise, representing the validity of the first two elements being partitioned.
Step 4: Iterate from index 3 to the end of the array (inclusive).
Step 5: For each index `i`, calculate `current_dp` based on the following conditions: * If the last two elements are equal (`nums[i-2] == nums[i-1]`), then `current_dp` is `True` if `dp_prev_two` is `True`. * If the last three elements are all equal (`nums[i-3] == nums[i-2] == nums[i-1]`) or the last three elements are consecutive increasing numbers (`nums[i-3] + 1 == nums[i-2] and nums[i-2] + 1 == nums[i-1]`), then `current_dp` is `True` if `dp_prev_three` is `True`. * `current_dp` will be `True` if either of these conditions is true, otherwise `False`.
Step 6: Update `dp_prev_three`, `dp_prev_two`, and `dp_prev_one` to shift the DP window forward (i.e., `dp_prev_three = dp_prev_two`, `dp_prev_two = dp_prev_one`, `dp_prev_one = current_dp`).
Step 7: After the loop finishes, return the value of `dp_prev_one`, which represents whether the entire array can be validly partitioned.

Key Insights

Insight 1: Dynamic programming is an effective approach to solve this problem since the solution to a subproblem (prefix of the array) can be used to solve larger subproblems.
Insight 2: The state of the DP can be represented by whether a partition is valid up to a given index. Since the conditions only depend on the previous 2 or 3 elements, we only need to keep track of the validity of the last 3 states to achieve O(1) space complexity.
Insight 3: The base cases are crucial. An empty array is considered valid. A single-element array is invalid. A two-element array is valid if the elements are equal.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Coinbase.

Check if There is a Valid Partition For The Array | LeetCode 2369

Medium

Array Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public boolean validPartition(int[] nums) {
        int n = nums.length;
        
        // dp[i] is true if prefix nums[0...i-1] can be partitioned.
        // We need dp[i-2] and dp[i-3] to compute dp[i].
        // We can use O(1) space by storing only the last 3 dp values.
        
        // dp_prev_three, dp_prev_two, dp_prev_one store dp[i-3], dp[i-2], dp[i-1]
        
        // Initialize for loop starting at i=3
        // dp[0] = True, dp[1] = False, dp[2] = (nums[0] == nums[1])
        if (n < 2) {
            return false;
        }
            
        boolean dp_prev_three = true;
        boolean dp_prev_two = false;
        boolean dp_prev_one = (nums[0] == nums[1]);

        for (int i = 3; i <= n; i++) {
            boolean current_dp = false;
            
            // Case 1: Subarray of 2 equal elements ending at i-1
            // Depends on dp[i-2]
            if (nums[i-2] == nums[i-1]) {
                current_dp = current_dp || dp_prev_two;
            }
            
            // Case 2 & 3: Subarray of 3 elements ending at i-1
            // Depends on dp[i-3]
            if ((nums[i-3] == nums[i-2] && nums[i-2] == nums[i-1]) || 
               (nums[i-3] + 1 == nums[i-2] && nums[i-2] + 1 == nums[i-1])) {
                current_dp = current_dp || dp_prev_three;
            }
            
            boolean temp = dp_prev_two;
            dp_prev_two = dp_prev_one;
            dp_prev_one = current_dp;
            dp_prev_three = temp;
            
        }
        
        return dp_prev_one;
    }
}

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Problem Info

Description

Problem 2369: Check if There is a Valid Partition For The Array - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

Coinbase

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2369

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

7 steps

1
Step 1: Initialize `dp_prev_three` to `True`, representing that an empty array is validly partitioned.
2
Step 2: Initialize `dp_prev_two` to `False`, representing that a single-element array is not validly partitioned.
3
Step 3: Initialize `dp_prev_one` to `True` if the first two elements are equal, and `False` otherwise, representing the validity of the first two elements being partitioned.
4
Step 4: Iterate from index 3 to the end of the array (inclusive).
5
Step 5: For each index `i`, calculate `current_dp` based on the following conditions: * If the last two elements are equal (`nums[i-2] == nums[i-1]`), then `current_dp` is `True` if `dp_prev_two` is `True`. * If the last three elements are all equal (`nums[i-3] == nums[i-2] == nums[i-1]`) or the last three elements are consecutive increasing numbers (`nums[i-3] + 1 == nums[i-2] and nums[i-2] + 1 == nums[i-1]`), then `current_dp` is `True` if `dp_prev_three` is `True`. * `current_dp` will be `True` if either of these conditions is true, otherwise `False`.
6
Step 6: Update `dp_prev_three`, `dp_prev_two`, and `dp_prev_one` to shift the DP window forward (i.e., `dp_prev_three = dp_prev_two`, `dp_prev_two = dp_prev_one`, `dp_prev_one = current_dp`).
7
Step 7: After the loop finishes, return the value of `dp_prev_one`, which represents whether the entire array can be validly partitioned.

💡

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 linear because we process each element of the array only once. The space complexity is constant because the number of variables used does not depend on the size of the input.

📖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

Tabulation DP with array

Instead of using O(1) space, we can use an array `dp` of size `n+1`. `dp[i]` will store whether the prefix of the array up to index `i-1` can be validly partitioned. The logic for updating `dp[i]` is the same as the main approach, but instead of updating three variables, we update `dp[i]` using `dp[i-2]` and `dp[i-3]`.

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

Trade-offs: This approach is conceptually simpler to understand but uses O(n) space, whereas the main solution uses O(1) space. Use the main solution when space is a constraint.

✨

Key Insights

1Insight 1: Dynamic programming is an effective approach to solve this problem since the solution to a subproblem (prefix of the array) can be used to solve larger subproblems.
2Insight 2: The state of the DP can be represented by whether a partition is valid up to a given index. Since the conditions only depend on the previous 2 or 3 elements, we only need to keep track of the validity of the last 3 states to achieve O(1) space complexity.
3Insight 3: The base cases are crucial. An empty array is considered valid. A single-element array is invalid. A two-element array is valid if the elements are equal.

📚

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: Empty array or array with a single element. The code explicitly handles the case where the array has fewer than 2 elements, returning `false` because it's impossible to form a valid partition.
Edge case 2: The beginning of the array. The initialization ensures that `dp_prev_three`, `dp_prev_two`, and `dp_prev_one` are correctly set for the first few elements of the array. `dp_prev_three` being `True` allows an empty array (before processing any elements) to be considered valid. `dp_prev_two` being `False` allows the case when first element in array is invalid.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly initializing the `dp` variables. The initialization is critical for handling base cases and ensuring that the DP logic works correctly.

✗Common mistake 2: Not correctly handling the consecutive increasing numbers case. It is important to check that both consecutive differences are equal to 1, not just one of them. `(nums[i-3] + 1 == nums[i-2] and nums[i-2] + 1 == nums[i-1])`.

✗Common mistake 3: Incorrect index access for subarrays of length 2 and 3. Off-by-one errors can lead to incorrect results.

✅

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 compression. Breaking a sequence of data into repeating patterns or increasing sequences can be used for efficient data storage.

🏦Real-world scenario 2: Financial data analysis. Identifying sequences of similar or consecutively increasing stock prices can be used for trend analysis and investment strategies.

💡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: June 30, 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 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 29

❓

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 "Check if There is a Valid Partition For The Array"?

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?