Alice has some number of cards and she wants to rearrange the cards into groups so that each group is of size groupSize, and consists of groupSize consecutive cards.
Given an integer array hand where hand[i] is the value written on the ith card and an integer groupSize, return true if she can rearrange the cards, or 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 can not be rearranged into groups of 4.
Constraints:
1 <= hand.length <= 1040 <= hand[i] <= 1091 <= groupSize <= hand.length
Note: This question is the same as 1296: https://leetcode.com/problems/divide-array-in-sets-of-k-consecutive-numbers/
## Solution(Python)
```Python
import heapq
class Solution:
def isNStraightHand(self, hand: List[int], groupSize: int) -> bool:
return self.most_optimal(hand, groupSize)
# Time Complexity: O(n logn + NK)
# Space Complexity: O(n)
def map(self, hands: List[int], groupSize: int) -> bool:
# Attempt
# # 1,2,3,,6,2,3,4,7,8
# # groupSize = 3
# # if n% g != 0 False
# # b = n//g
# # b bcukets
# # hashmap = [1,2,2,1,1]
# # smallest_cur = min(hashmap)
# # [[],[],[]]
# # smallest_cur = min(hashmap)
# #
# # [] 1 cur+=1
# # 2 , 3
# # if hashmap[smallest_cur] == 0
# # reryurn Fa;se
# # hashmap[smallest_cur] -=1
# # return True
# #
n = len(hands)
if n% groupSize!= 0:
return False
# no_of_buckets = n // groupSize
# MAX = max(hand)
# MIN = min(hand)
# counts = [0] * ( MAX -MIN +1)
cardCount = Counter(hands)
min_heap = list(cardCount.keys())
heapq.heapify(min_heap)
while min_heap:
current_card = min_heap[0]
for i in range(groupSize):
if cardCount[current_card + i] == 0:
return False
cardCount[current_card + i] -= 1
if cardCount[current_card + i] <= 0:
heapq.heappop(min_heap)
return True
def optimal(self, hands: List[int], groupSize: int) -> bool: # hand = [1,2,3,6,2,3,4,7,8], groupSize = 3
cardCount = Counter(hands) # {1:1, 2:2, 3:2, 6:1, 4:1,7:1,8:1}
# Sorted list of card values
sortedCards = sorted(cardCount.keys()) # 1 2 3 4 6 7 8
# Queue to keep track of the number of new groups
# starting with each card value
groupStartQueue = deque() # []
lastCard = -1
currentOpenGroups = 0
for currentCard in sortedCards: # 1 2 3 4 6 7 8 -> 6
# Check if there are any discrepancies in the sequence
# or more groups are required than available cards
if (
currentOpenGroups > 0 and currentCard > lastCard + 1 #
) or currentOpenGroups > cardCount[currentCard]: #
return False
groupStartQueue.append(cardCount[currentCard] - currentOpenGroups) # 0, 0, 1
lastCard = currentCard # 6
currentOpenGroups = cardCount[currentCard] # 1
# Maintain the queue size to be equal to groupSize
if len(groupStartQueue) == groupSize: #
currentOpenGroups -= groupStartQueue.popleft() # 0
return currentOpenGroups == 0
def most_optimal(self, hand: List[int], groupSize: int) -> bool:
if len(hand) % groupSize != 0:
return False
# Counter to store the count of each card value
card_count = Counter(hand)
for card in hand:
start_card = card
# Find the start of the potential straight sequence
while card_count[start_card - 1]:
start_card -= 1
# Process the sequence starting from start_card
while start_card <= card:
while card_count[start_card]:
# Check if we can form a consecutive sequence
# of groupSize cards
for next_card in range(start_card, start_card + groupSize):
if not card_count[next_card]:
return False
card_count[next_card] -= 1
start_card += 1
return True
```