How do you solve Hand of Straights on LeetCode?

Hand of Straights (LeetCode #846) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: The total number of cards must be divisible by groupSize to form complete groups.

What is the time complexity of Hand of Straights?

The optimal solution for Hand of Straights runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the frequency counts.

What is the brute force solution for Hand of Straights?

The brute force approach for Hand of Straights is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries to generate all possible combinations of groups of cards and checks if they can be rearranged into valid groups. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Hand of Straights?

Hand of Straights uses the following concepts: Array, Hash Table, Greedy, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Hand of Straights?

Always consider edge cases, such as when the hand length is not divisible by groupSize. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice writing out the frequency map logic, as it is a common pattern in many problems.

What are common mistakes when solving Hand of Straights?

Failing to check if the total number of cards is divisible by groupSize. Not properly managing counts in the frequency map, leading to incorrect group formations.

#846

Hand of Straights

Medium

ArrayHash TableGreedySortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

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

The optimal approach uses a frequency map to count occurrences of each card and then attempts to form groups by checking if the required consecutive cards are available. This is efficient because we avoid unnecessary searches and directly manage counts.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each card using a HashMap.
2Step 2: Sort the unique keys of the HashMap.
3Step 3: Iterate through the sorted keys and for each key, try to form a group of consecutive cards by checking and decrementing their counts in the HashMap.

solution.py14 lines

1# Full working Python code
2
3def canArrange(hand, groupSize):
4    from collections import Counter
5    if len(hand) % groupSize != 0:
6        return False
7    count = Counter(hand)
8    for card in sorted(count.keys()):
9        if count[card] > 0:
10            for i in range(groupSize):
11                if count[card + i] < count[card]:
12                    return False
13                count[card + i] -= count[card]
14    return True

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the frequency counts.

1The total number of cards must be divisible by groupSize to form complete groups.
2Using a frequency map allows us to efficiently manage counts and check for consecutive numbers.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.