What is the time complexity of Convert an Array Into a 2D Array With Conditions?

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

What data structures are used to solve Convert an Array Into a 2D Array With Conditions?

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

What are the key insights for solving Convert an Array Into a 2D Array With Conditions?

Insight 1: The maximum frequency of any number in `nums` determines the minimum number of rows required in the 2D array. If a number appears `k` times, you need at least `k` rows to accommodate all instances of that number without duplicates in any row. Insight 2: A greedy approach of filling rows sequentially as you iterate through `nums` works well. We can keep track of how many times each number has already been placed in a row using a hash map.

Which companies ask LeetCode 2610. Convert an Array Into a 2D Array With Conditions in interviews?

This problem is frequently asked at Gojek.

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Convert an Array Into a 2D Array With Conditions - LeetCode 2610 Solution

Convert an Array Into a 2D Array With Conditions - Complete Solution Guide

Convert an Array Into a 2D Array With Conditions is LeetCode problem 2610, 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 an integer array nums . You need to create a 2D array from nums satisfying the following conditions: The 2D array should contain only the elements of the array nums . Each row in the 2D array contains distinct integers. The number of rows in the 2D array should be minimal . Return the resulting array . If there are multiple answers, return any of them. Note that the 2D array can have a different number of elements on each row. Example 1: Input: nums = [1,3,4,1,2,3,1] Output: [[1,3,

Detailed Explanation

The problem requires converting a given integer array `nums` into a 2D array. The 2D array must satisfy the following conditions: 1. It should only contain elements from the original `nums` array. 2. Each row in the 2D array must contain distinct integers (no duplicates within a row). 3. The number of rows in the 2D array should be minimized. This means we need to pack as many distinct numbers as possible into each row before starting a new row. The goal is to return one valid 2D array that meets these criteria. The order of rows or elements within a row does not matter as long as the conditions are met.

Solution Approach

The provided solution uses a greedy approach to construct the 2D array row by row. It iterates through the input array `nums`. For each number, it checks how many times that number has already appeared in the 2D array (which implicitly determines the row index where we should place it next). If the number's count equals the current number of rows, a new row is created. The number is then added to the appropriate row, and the count for that number is incremented. This ensures that no row contains duplicate numbers and that the number of rows is minimized.

Step-by-Step Algorithm

Step 1: Initialize a hash map (counts) to store the frequency of each number that has been added to the 2D array so far. Initialize an empty list (ans) to represent the 2D array.
Step 2: Iterate through the input array nums, element by element.
Step 3: For each element `num`, check the count of `num` in the hash map `counts`. This value represents the row index where we should try to insert `num` next.
Step 4: If the count of `num` is equal to the number of rows currently in `ans`, it means all existing rows already contain `num`. Therefore, create a new row and append it to `ans`.
Step 5: Append `num` to the row at index `counts[num]` in `ans`.
Step 6: Increment the count of `num` in the hash map `counts`.
Step 7: After processing all elements in nums, return the resulting 2D array `ans`.

Key Insights

Insight 1: The maximum frequency of any number in `nums` determines the minimum number of rows required in the 2D array. If a number appears `k` times, you need at least `k` rows to accommodate all instances of that number without duplicates in any row.
Insight 2: A greedy approach of filling rows sequentially as you iterate through `nums` works well. We can keep track of how many times each number has already been placed in a row using a hash map.
Insight 3: No sorting or complex calculations are required. The problem focuses on efficiently managing frequencies and distributing elements across rows.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table.

Companies

Asked at: Gojek.

Convert an Array Into a 2D Array With Conditions | LeetCode 2610

Medium

Array Hash Table

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public List<List<Integer>> findMatrix(int[] nums) {
        Map<Integer, Integer> counts = new HashMap<>();
        List<List<Integer>> ans = new ArrayList<>();
        for (int num : nums) {
            int row = counts.getOrDefault(num, 0);
            if (row == ans.size()) {
                ans.add(new ArrayList<>());
            }
            ans.get(row).add(num);
            counts.put(num, counts.getOrDefault(num, 0) + 1);
        }
        return ans;
    }
}

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Problem Info

Description

Problem 2610: Convert an Array Into a 2D Array With Conditions - Detailed explanation and solution approaches.

Topics

Array Hash Table

Companies

Gojek

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2610

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 a hash map (counts) to store the frequency of each number that has been added to the 2D array so far. Initialize an empty list (ans) to represent the 2D array.
2
Step 2: Iterate through the input array nums, element by element.
3
Step 3: For each element `num`, check the count of `num` in the hash map `counts`. This value represents the row index where we should try to insert `num` next.
4
Step 4: If the count of `num` is equal to the number of rows currently in `ans`, it means all existing rows already contain `num`. Therefore, create a new row and append it to `ans`.
5
Step 5: Append `num` to the row at index `counts[num]` in `ans`.
6
Step 6: Increment the count of `num` in the hash map `counts`.
7
Step 7: After processing all elements in nums, return the resulting 2D array `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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the single loop that iterates through the input array `nums`. Inside the loop, the operations on the hash map (get and put) take O(1) on average. The space complexity is determined by the size of the hash map and the 2D array, both of which can grow up to the size of the input array in the worst-case scenario.

📖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

Sorting and Iteration

Sort the array first. Then iterate through the sorted array and create a 2D array. For each number, find the next available row (row with no occurrence of this number).

Time: O(n log n) due to the sorting step.Space: O(n) for the 2D array.

Trade-offs: Sorting is slower than the hashmap approach, making this less efficient overall. This approach is useful if memory is highly constrained since the hashmap will take extra memory that sorting in place would not.

Frequency Counting with Array/List Manipulation

Instead of a hash map, use an array to count frequencies (assuming the range of numbers is limited). This approach may require array resizing during building. This can be useful when the numbers are in a small and known range.

Time: O(n)Space: O(n) in the worst case where each number appears a different number of times. The array for frequency counting will take O(n) in space.

Trade-offs: If memory is limited and nums[i] is less than or equal to nums.length then this approach may be better because you are not allocating and deallocating memory for hash tables and keys during program execution. This approach is beneficial when the range of numbers is very small (and constrained by problem's constraint). Note, you can only do this if you have an upper bound on the values that are passed.

✨

Key Insights

1Insight 1: The maximum frequency of any number in `nums` determines the minimum number of rows required in the 2D array. If a number appears `k` times, you need at least `k` rows to accommodate all instances of that number without duplicates in any row.
2Insight 2: A greedy approach of filling rows sequentially as you iterate through `nums` works well. We can keep track of how many times each number has already been placed in a row using a hash map.
3Insight 3: No sorting or complex calculations are required. The problem focuses on efficiently managing frequencies and distributing elements across rows.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: Empty input array: The code correctly handles an empty input array by returning an empty 2D array. The loop will simply not execute.
Edge case 2: Input array with all distinct elements: The code creates a single row in the 2D array containing all the distinct elements. The frequency count for each element will remain at 0, so they all get added to row 0.
Edge case 3: Input array with all the same elements: The code creates `k` rows, where `k` is the number of times the element appears. Each row will contain only a single instance of that element. This ensures that each row contains distinct integers, as required by the problem.
Edge case 4: Input array with a mix of repeating and distinct elements: The algorithm dynamically adjusts the number of rows as it encounters repeating elements, efficiently placing each element into the first available row.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the minimum number of rows required. Failing to realize that the maximum frequency of any number dictates the minimum number of rows leads to an incorrect solution.

✗Common mistake 2: Adding duplicate numbers in the same row. Forgetting to check if a number already exists in a row before adding it violates the problem's constraints.

✗Common mistake 3: Allocating insufficient memory for the 2D array in C. Dynamic memory allocation must be done carefully in C, taking into account the maximum possible size of each row and the overall number of rows.

✅

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: Task scheduling. Imagine scheduling tasks to different machines. Each task requires a specific resource. To avoid resource conflicts, you want to schedule tasks requiring the same resource on different machines (rows). The algorithm ensures that tasks are distributed across the minimal number of machines.

🏦Real-world scenario 2: Data partitioning. Consider distributing data across multiple storage nodes. You want to minimize the number of nodes used while ensuring that each node contains distinct data entries. This algorithm provides an efficient way to partition the data across nodes.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 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

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

@DevExpert92Jul 14

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

@AlgoMasterJul 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 are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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 "Convert an Array Into a 2D Array With Conditions"?

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

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