What is the time complexity of Divide Array in Sets of K Consecutive Numbers?

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

What data structures are used to solve Divide Array in Sets of K Consecutive Numbers?

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

What are the key insights for solving Divide Array in Sets of K Consecutive Numbers?

Insight 1: Sorting the input array (or at least identifying the unique elements in sorted order) is crucial to efficiently check for consecutive numbers. Insight 2: Using a hash map (or counter) to store the frequency of each number allows us to quickly determine if we have enough instances of consecutive numbers to form the desired sets.

Which companies ask LeetCode 1296. Divide Array in Sets of K Consecutive Numbers in interviews?

This problem is frequently asked at Waymo.

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Divide Array in Sets of K Consecutive Numbers - LeetCode 1296 Solution

Divide Array in Sets of K Consecutive Numbers - Complete Solution Guide

Divide Array in Sets of K Consecutive Numbers is LeetCode problem 1296, 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

Given an array of integers nums and a positive integer k , check whether it is possible to divide this array into sets of k consecutive numbers. Return true if it is possible . Otherwise, return false . Example 1: Input: nums = [1,2,3,3,4,4,5,6], k = 4 Output: true Explanation: Array can be divided into [1,2,3,4] and [3,4,5,6]. Example 2: Input: nums = [3,2,1,2,3,4,3,4,5,9,10,11], k = 3 Output: true Explanation: Array can be divided into [1,2,3] , [2,3,4] , [3,4,5] and [9,10,11]. Example 3: Inpu

Detailed Explanation

The problem asks us to determine if a given array of integers `nums` can be divided into sets of `k` consecutive numbers. In other words, we need to check if it's possible to group the numbers in `nums` into sub-arrays, where each sub-array contains `k` consecutive integers. The output should be `true` if such a division is possible, and `false` otherwise. The array `nums` can contain duplicates and is unsorted. Also, `k` will always be a positive integer.

Solution Approach

The provided solution uses a greedy approach. First, it checks if the length of the input array `nums` is divisible by `k`. If not, it's impossible to divide the array into sets of size `k`, so we return `false`. Then, it counts the occurrences of each number in `nums` using a hash map (or `Counter` in Python). Next, it sorts the unique numbers in `nums`. Finally, it iterates through the sorted numbers. For each number, it checks if there are enough consecutive numbers following it (i.e., num+1, num+2, ..., num+k-1) with the same frequency (or at least the needed frequency) to form a valid group. If at any point, it's not possible to form a group of `k` consecutive numbers starting from a number, it returns `false`. Otherwise, it returns `true`.

Step-by-Step Algorithm

Step 1: Check if the length of `nums` is divisible by `k`. If not, return `false`.
Step 2: Create a hash map `counts` to store the frequency of each number in `nums`.
Step 3: Extract the distinct elements of nums and sort them in ascending order. Store these in a container like sorted_keys.
Step 4: Iterate through `sorted_keys`.
Step 5: For each number `start_num` in `sorted_keys`, check if its count in `counts` is greater than 0. If not, continue to the next number.
Step 6: If the count of `start_num` is greater than 0, determine the number of groups `num_groups` that can be formed starting from `start_num`. This is equal to `counts[start_num]`.
Step 7: Iterate from `i = 0` to `k-1`. In each iteration, check if `counts[start_num + i]` is greater than or equal to `num_groups`. If not, return `false`.
Step 8: If `counts[start_num + i]` is greater than or equal to `num_groups`, subtract `num_groups` from `counts[start_num + i]` to indicate that these numbers have been used to form the groups.
Step 9: If all numbers have been processed without returning `false`, return `true`.

Key Insights

Insight 1: Sorting the input array (or at least identifying the unique elements in sorted order) is crucial to efficiently check for consecutive numbers.
Insight 2: Using a hash map (or counter) to store the frequency of each number allows us to quickly determine if we have enough instances of consecutive numbers to form the desired sets.
Insight 3: The greedy approach of starting with the smallest number and forming consecutive groups works because if a valid division exists, starting with the smallest will always lead to a solution.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Greedy, Sorting.

Companies

Asked at: Waymo.

Divide Array in Sets of K Consecutive Numbers | LeetCode 1296

Medium

Array Hash Table Greedy Sorting

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;

class Solution {
    public boolean isPossibleDivide(int[] nums, int k) {
        if (nums.length % k != 0) {
            return false;
        }
        
        Map<Integer, Integer> counts = new HashMap<>();
        for (int num : nums) {
            counts.put(num, counts.getOrDefault(num, 0) + 1);
        }
        
        Arrays.sort(nums);
        int[] sorted_keys_arr = Arrays.stream(nums).distinct().toArray();

        
        for (int start_num : sorted_keys_arr) {
             if (counts.getOrDefault(start_num,0) > 0) {
                int num_groups = counts.get(start_num);
                for (int i = 0; i < k; i++) {
                    int num = start_num + i;
                    if (counts.getOrDefault(num,0) < num_groups) {
                        return false;
                    }
                    counts.put(num, counts.get(num) - num_groups);
                }
            }
        }
        
        return true;
    }
}

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Problem Info

Description

Problem 1296: Divide Array in Sets of K Consecutive Numbers - Detailed explanation and solution approaches.

Topics

Array Hash Table Greedy Sorting

Companies

Waymo

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1296

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

9 steps

1
Step 1: Check if the length of `nums` is divisible by `k`. If not, return `false`.
2
Step 2: Create a hash map `counts` to store the frequency of each number in `nums`.
3
Step 3: Extract the distinct elements of nums and sort them in ascending order. Store these in a container like sorted_keys.
4
Step 4: Iterate through `sorted_keys`.
5
Step 5: For each number `start_num` in `sorted_keys`, check if its count in `counts` is greater than 0. If not, continue to the next number.
6
Step 6: If the count of `start_num` is greater than 0, determine the number of groups `num_groups` that can be formed starting from `start_num`. This is equal to `counts[start_num]`.
7
Step 7: Iterate from `i = 0` to `k-1`. In each iteration, check if `counts[start_num + i]` is greater than or equal to `num_groups`. If not, return `false`.
8
Step 8: If `counts[start_num + i]` is greater than or equal to `num_groups`, subtract `num_groups` from `counts[start_num + i]` to indicate that these numbers have been used to form the groups.
9
Step 9: If all numbers have been processed without returning `false`, return `true`.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the sorting step. The hash map operations (insertion, retrieval) take O(1) on average, and the loops iterate at most n times. The space complexity is determined by the size of the hash map, which can grow up to O(n) in the worst case (all numbers are unique).

📖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

Using a priority queue (heap)

Instead of sorting, you could use a priority queue to maintain the smallest number. Then, repeatedly extract the smallest element and try to form a sequence starting from that number. This approach avoids sorting the entire array upfront.

Time: O(n log n), where n is the number of elements in nums, because heap operations have logarithmic complexity.Space: O(n), where n is the size of nums to store the counts and the heap.

Trade-offs: The priority queue approach is conceptually similar but might be more complex to implement. Sorting the unique keys is usually faster in practice.

✨

Key Insights

1Insight 1: Sorting the input array (or at least identifying the unique elements in sorted order) is crucial to efficiently check for consecutive numbers.
2Insight 2: Using a hash map (or counter) to store the frequency of each number allows us to quickly determine if we have enough instances of consecutive numbers to form the desired sets.
3Insight 3: The greedy approach of starting with the smallest number and forming consecutive groups works because if a valid division exists, starting with the smallest will always lead to a solution.

📚

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!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `nums` is empty. The code correctly handles this implicitly as the length check in step 1 will catch it and since an empty array is technically divisible by any number k.
Edge case 2: `k = 1`. The code correctly handles this since any array can be divided into sets of size 1.
Edge case 3: `nums` contains only one unique number and `k > 1`. The code will return false if there isn't at least k copies of that one number.
Edge case 4: Very large input numbers. The code can handle large numbers (within the integer limits) because it uses hash maps and performs arithmetic operations on integers.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking if `len(nums)` is divisible by `k` before proceeding. This can lead to incorrect results and unnecessary computation.

✗Common mistake 2: Incorrectly updating the counts after forming a group. It's crucial to subtract the number of groups formed from the counts of all the numbers in the group.

✗Common mistake 3: Not handling the case where the count of a number is less than the required number of groups. This can lead to incorrect results.

✗Common mistake 4: Forgetting to extract only distinct elements before sorting when they are used to form the consecutive sets.

✅

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: Resource allocation. Imagine you have a set of tasks with different priorities (represented by numbers) and you need to group them into batches of consecutive priority levels for processing. This algorithm can help determine if such a grouping is possible.

🏦Real-world scenario 2: Card game analysis. In a card game, you might want to determine if a player has a hand that can be divided into sets of consecutive cards of a certain length. This problem directly models that scenario.

💡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.

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: July 7, 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 13

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

@InterviewProJul 6

❓

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 "Divide Array in Sets of K Consecutive Numbers"?

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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