Hand Of Straights Leetcode Problem 846 [Python Solution]

Hand Of Straights, even though it does not sound like it, is a Leetcode problem and we’ll break down the problem, examine the constraints.

Also, we will discuss a brute force approach, and then delve into an efficient Python solution.

By the end, you’ll have a clear understanding of how to solve this problem and its time and space complexity.

Problem Overview

Problem Statement: Alice has some number of cards, and she wants to rearrange the cards into groups, where each group consists of a specified number of consecutive cards.

Given an integer array hand, where hand[i] is the value written on the i-th card, and an integer groupSize, your task is to determine if Alice can rearrange the cards to form such groups.

Return true if it’s possible, and false otherwise.

Example 1:

Input: hand = [1,2,3,6,2,3,4,7,8], groupSize = 3
Output: true

Explanation: Alice’s hand can be rearranged as [1,2,3],[2,3,4],[6,7,8].

Example 2:

Input: hand = [1,2,3,4,5], groupSize = 4
Output: false

Explanation: Alice’s hand cannot be rearranged into groups of size 4.

Constraints:

1 ≤ hand.length ≤ 10^4
0 ≤ hand[i] ≤ 10^9
1 ≤ groupSize ≤ hand.length

Now, let’s dive into a solution for this problem.

Understanding the Constraints

Before we jump into the solution, let’s discuss the constraints given in the problem:

The length of the hand array can be as large as 10,000 elements, indicating that we need an efficient algorithm to solve the problem.
The values on the cards can be in the range from 0 to 10^9, so the input values can be quite large.
The groupSize will be a positive integer, and it should be less than or equal to the length of the hand array.

These constraints tell us that we need to come up with an efficient algorithm that can handle large inputs while maintaining a reasonable time complexity.

Hand of Straights LeetCode Problem Solution

1. Bruteforce Approach

Let’s start with a brute force approach to tackle the problem.

The brute force approach would involve checking all possible combinations of groups to see if we can rearrange the cards according to the rules.

Here’s a Python implementation of this approach:

def isNStraightHand(hand, groupSize):
    if len(hand) % groupSize:
        return False

    # Create a dictionary to count the occurrences of each card value
    count = {}
    for n in hand:
        count[n] = 1 + count.get(n, 0)

    # Convert unique card values into a min-heap
    minHeap = list(count.keys())
    heapq.heapify(minHeap)

    while minHeap:
        first = minHeap[0]  # Get the minimum value from the min-heap
        for i in range(first, first + groupSize):
            if i not in count:
                return False
            count[i] -= 1
            if count[i] == 0:
                if i != minHeap[0]:  # Pop the minimum value from the min-heap
                    return False
                heapq.heappop(minHeap)

    return True

Now, let’s break down the key steps in this brute force approach:

We first check if the length of the hand array is divisible by groupSize.

If it’s not, we return False since we won’t be able to create valid groups.

We create a dictionary count to count the occurrences of each card value in the hand array.

This helps us keep track of how many cards of each value we have.

We convert the unique card values into a min-heap minHeap.

Using a min-heap ensures that we always select the smallest available value when forming groups.

We iterate through the minHeap while it’s not empty.

For each value, we check if we can create a group of consecutive cards starting from that value.

If not, we return False.

If we can form a group, we decrement the counts in the count dictionary.

If the count becomes zero, we remove that value from the minHeap.

If we successfully iterate through the entire minHeap, it means we were able to form valid groups, and we return True.

Time and Space Complexity

Now, let’s analyze the time and space complexity of our solution:

Time Complexity: The main loop runs until the minHeap is empty, and in each iteration, we perform constant time operations.

However, creating the minHeap has a time complexity of O(n), and since the loop can execute up to n/groupSize times, the overall time complexity is O(n) + O(n/groupSize), which simplifies to O(n).

Space Complexity: We use additional space for the count dictionary, which can store unique card values.

In the worst case, this dictionary can have n unique values, resulting in a space complexity of O(n).

Additionally, the minHeap stores unique values from count, so it also has a space complexity of O(n).

This brute force approach provides a reasonable solution for the problem, but it can be optimized further.

In the following section, we’ll discuss a more efficient approach.

2. An Efficient Approach with Min-Heap

In the brute force approach, we created a min-heap from unique card values and repeatedly popped the minimum value to form groups.

This approach works, but we can simplify it further.

Let’s start by understanding the core idea:

We know that we need to create groups of consecutive card values.
The minimum value of a group will determine its starting point.
If we can form a group starting from a minimum value, we should be able to create other groups as well.

Here’s the updated Python implementation for this efficient approach:

View on GitHub ➹

def isNStraightHand(self, hand: List[int], groupSize: int) -> bool:
        if len(hand) % groupSize:
            return False

        count = collections.Counter(hand)  # Count occurrences of each card value
        minHeap = list(count.keys())  # Unique card values

        heapq.heapify(minHeap)  # Convert to a min-heap

        while minHeap:
            first = minHeap[0]  # Get the minimum value from the min-heap
            for i in range(first, first + groupSize):
                if i not in count:
                    return False
                count[i] -= 1
                if count[i] == 0:
                    heapq.heappop(minHeap)

        return True

In this updated solution, we’ve improved the code by using the collections.Counter function to count the occurrences of each card value, and we use the minHeap to store unique card values.

The rest of the algorithm remains the same, where we iterate through the minHeap to form groups.

This efficient approach maintains the same time and space complexity as the brute force approach, but it simplifies the code and enhances readability.

Reasoning Behind Our Approach

The reasoning behind our approach is straightforward.

We leverage a min-heap to efficiently track the minimum value to start each group.

By counting the occurrences of card values using a dictionary, we can ensure that we have the necessary cards to create groups of consecutive values.

If we can create a group starting from the minimum value, we continue to form other groups until the min-heap is empty.

The code is designed to handle large input sizes efficiently, making it a suitable solution for the problem.

Conclusion

In this guide, we tackled the Hand Of Straights problem on LeetCode.

We explored the problem statement, examined the constraints, discussed a brute force approach, and presented an efficient Python solution.

By understanding the problem and the reasoning behind our approach, you should be well-equipped to solve similar problems that require grouping and counting elements efficiently.

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Solving coding problems is a valuable skill, and continuous practice will help you become a better problem solver.

You can find the original problem on LeetCode at the following link.

Happy coding!