846. Hand of Straights
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 <= 104
0 <= hand[i] <= 109
1 <= groupSize <= hand.length
Note: This question is the same as 1296: https://leetcode.com/problems/divide-array-in-sets-of-k-consecutive-numbers/
Solution:
class Solution {
public boolean isNStraightHand(int[] hand, int groupSize) {
if (hand.length % groupSize != 0){
return false;
}
if (groupSize == 1){
return true;
}
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for (int i = 0; i < hand.length; i++){
map.put(hand[i], map.getOrDefault(hand[i],0) + 1);
}
PriorityQueue<Integer> pq = new PriorityQueue<Integer>((a,b) -> a-b);
for (Map.Entry<Integer, Integer> entry : map.entrySet()){ // n
pq.offer(entry.getKey()); // logn
}
while(!pq.isEmpty()){
int min = pq.peek();
for (int i = min; i < min + groupSize; i++){
if (map.getOrDefault(i, -1) == -1){
return false;
}
map.put(i, map.get(i) - 1);
if (map.get(i) == 0){
map.remove(i);
if (pq.peek() == i){
pq.poll();
}else{
return false;
}
}
}
}
return true;
}
}
// TC: O(groupSize*nlogn)
// SC: O(n)