Hand of Straights

class Solution {
public:
    bool canArrange(vector<int>& cards, int size) {
        if (cards.size() % size != 0) {
            return false;
        }

        // Creating a map to count frequencies of each card
        unordered_map<int, int> frequency;
        for (int card : cards) {
            frequency[card]++;
        }

        for (int card : cards) {
            int firstCard = card;
            // Locate the beginning of the straight
            while (frequency[firstCard - 1]) {
                firstCard--;
            }

            // Check for consecutive sequences from the located start
            while (firstCard <= card) {
                while (frequency[firstCard]) {
                    // Attempting to collect a sequence of length 'size'
                    for (int current = firstCard; current < firstCard + size; current++) {
                        if (!frequency[current]) {
                            return false;
                        }
                        frequency[current]--;
                    }
                }
                firstCard++;
            }
        }
        
        return true;
    }
};

This C++ solution is designed to check if an array of integers (cards) can be arranged into groups where each group has a specific size of strictly consecutive integers.

Follow these steps for understanding the method executed in the code:

1. Check if the total number of cards is divisible by the group size, size. If not, the method returns false immediately.

2. Use an unordered map, frequency, to count the occurrences of each card. This prepares for grouping by tracking how many times each integer appears in the input vector.

3. Iterate through the cards. For each card, identify the smallest consecutive integer that could start a valid group by decrementing firstCard if the previous number exists in frequency.

4. From the established starting point, check if a full group of consecutive cards of the required size can be formed. This is done by attempting to count down from frequency for each card in the expected sequence. If a card required for a sequence is missing (frequency[current] becomes zero), return false.

5. If all potential starts lead to valid sequences where consecutive groups are successfully formed, the function returns true, indicating that the entire set of cards can indeed be divided into the desired groups.

This approach leverages both counting and mapping to efficiently determine the possibility of forming the required sequences, ensuring that cards are grouped only when appropriate consecutive groups can be made.

class Solution {

    public boolean canDivideIntoSequences(int[] cards, int size) {
        if (cards.length % size != 0) {
            return false;
        }

        // Store each card count in a HashMap
        HashMap<Integer, Integer> counts = new HashMap<>();
        for (int card : cards) {
            int currentCount = counts.getOrDefault(card, 0);
            counts.put(card, currentCount + 1);
        }

        for (int card : cards) {
            int minCard = card;
            // Find the minimum card that starts a potential group
            while (counts.getOrDefault(minCard - 1, 0) > 0) {
                minCard--;
            }

            // Try to construct the groups starting from minCard
            while (minCard <= card) {
                while (counts.getOrDefault(minCard, 0) > 0) {
                    // Attempt to form a straight group of 'size' consecutive cards
                    for (
                        int sequenceCard = minCard;
                        sequenceCard < minCard + size;
                        sequenceCard++
                    ) {
                        if (counts.getOrDefault(sequenceCard, 0) == 0) {
                            return false;
                        }
                        counts.put(sequenceCard, counts.get(sequenceCard) - 1);
                    }
                }
                minCard++;
            }
        }

        return true;
    }
}

The "Hand of Straights" problem involves determining whether a given array of integer cards can be partitioned into groups of consecutive sequences, where each group must have a specified number of cards (size). This solution implements the necessary functionality in Java.

• Use an early exit condition to check if the total number of cards is completely divisible by the specified group size. If it isn't, return false immediately.

• Initialize a HashMap to count the occurrences of each card value.

• Iterate over each card in the provided array, updating the count for each card in the HashMap.

• For constructing the groups, find the smallest card value that can possibly start a new sequence. This is determined by checking for the lowest card in sequence that has not yet been completely used up in a previous grouping.

• Attempt to construct sequences, or straights, starting from this minimum card value and extending by the length determined by size.

• For each sequence, verify that there are enough cards available to complete the sequence. If any card in the sequence from minCard to minCard + size - 1 is missing or has been exhausted in the counts, the entire method returns false.

• If all sequences are successfully constructed with the available cards, the function concludes by returning true.

This solution fundamentally relies on mapping and also implementing a logical check for consecutive sequences within a mapped structure. Ensure consistent handling of card counts, and orderly checking of sequential group possibilities is essential for validating the card distribution according to the problem requirements.

class Solution:
    def canFormGroups(self, cards: List[int], size: int) -> bool:
        if len(cards) % size != 0:
            return False

        # Dictionary to count occurrences of each card
        count_map = Counter(cards)

        for card in sorted(cards):
            # Identify the smallest card not fully used in a group
            while count_map[card - 1] > 0:
                card -= 1

            # Begin at the lowest card and try forming groups
            while card in count_map and count_map[card] > 0:
                # Validate and decrement counts for the group
                for c in range(card, card + size):
                    if count_map[c] == 0:
                        return False
                    count_map[c] -= 1
                card += 1

        return True

This Python solution checks if it's possible to rearrange a list of integers (cards) into groups of consecutive numbers, each of size specified by size. If the total number of cards isn't a multiple of the group size, it immediately returns False since grouping is impossible.

Here's how the solution works:

• First, it checks if the total number of cards is divisible by the group size. If not, it returns False.
• It utilizes a Counter from the collections library to store the frequency of each card.
• The main logic iterates over the sorted list of cards. For each card, it attempts to find the lowest possible starting number of a consecutive sequence that hasn't been completely used in forming previous groups.
• For each potential starting number, it looks ahead 'size' cards to see if a valid group can be formed. If any card within this range is missing from count_map, it returns False as it is not possible to form a group.
• If a valid group is formed from a starting card, it decrements the count of each card in that group from the count_map.
• The code iterates until all possible groups are successfully formed or it comes across a group that cannot be formed.

The effectiveness of this solution relies on the ability to sort the cards and verify consecutive sequences, ensuring that all cards can be grouped without any remaining. If it successfully groups all cards, it returns True. The use of a counter and sorting ensures that the groups are formed with the smallest numbers available first.