Description#
Given an array of integers nums and a positive integer k, check whether it is possible to divide this array into sets of k consecutive numbers.
Return true if it is possible. Otherwise, return false.
Example 1:
Input: nums = [1,2,3,3,4,4,5,6], k = 4
Output: true
Explanation: Array can be divided into [1,2,3,4] and [3,4,5,6].
Example 2:
Input: nums = [3,2,1,2,3,4,3,4,5,9,10,11], k = 3
Output: true
Explanation: Array can be divided into [1,2,3] , [2,3,4] , [3,4,5] and [9,10,11].
Example 3:
Input: nums = [1,2,3,4], k = 3
Output: false
Explanation: Each array should be divided in subarrays of size 3.
Constraints:
1 <= k <= nums.length <= 1051 <= nums[i] <= 109
Note: This question is the same as 846:
https://leetcode.com/problems/hand-of-straights/Solutions#
Solution 1#
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| class Solution:
def isPossibleDivide(self, nums: List[int], k: int) -> bool:
cnt = Counter(nums)
for v in sorted(nums):
if cnt[v]:
for x in range(v, v + k):
if cnt[x] == 0:
return False
cnt[x] -= 1
if cnt[x] == 0:
cnt.pop(x)
return True
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| class Solution {
public boolean isPossibleDivide(int[] nums, int k) {
Map<Integer, Integer> cnt = new HashMap<>();
for (int v : nums) {
cnt.put(v, cnt.getOrDefault(v, 0) + 1);
}
Arrays.sort(nums);
for (int v : nums) {
if (cnt.containsKey(v)) {
for (int x = v; x < v + k; ++x) {
if (!cnt.containsKey(x)) {
return false;
}
cnt.put(x, cnt.get(x) - 1);
if (cnt.get(x) == 0) {
cnt.remove(x);
}
}
}
}
return true;
}
}
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| class Solution {
public:
bool isPossibleDivide(vector<int>& nums, int k) {
unordered_map<int, int> cnt;
for (int& v : nums) ++cnt[v];
sort(nums.begin(), nums.end());
for (int& v : nums) {
if (cnt.count(v)) {
for (int x = v; x < v + k; ++x) {
if (!cnt.count(x)) {
return false;
}
if (--cnt[x] == 0) {
cnt.erase(x);
}
}
}
}
return true;
}
};
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| func isPossibleDivide(nums []int, k int) bool {
cnt := map[int]int{}
for _, v := range nums {
cnt[v]++
}
sort.Ints(nums)
for _, v := range nums {
if _, ok := cnt[v]; ok {
for x := v; x < v+k; x++ {
if _, ok := cnt[x]; !ok {
return false
}
cnt[x]--
if cnt[x] == 0 {
delete(cnt, x)
}
}
}
}
return true
}
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Solution 2#
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| from sortedcontainers import SortedDict
class Solution:
def isPossibleDivide(self, nums: List[int], k: int) -> bool:
if len(nums) % k != 0:
return False
sd = SortedDict()
for h in nums:
if h in sd:
sd[h] += 1
else:
sd[h] = 1
while sd:
v = sd.peekitem(0)[0]
for i in range(v, v + k):
if i not in sd:
return False
if sd[i] == 1:
sd.pop(i)
else:
sd[i] -= 1
return True
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| class Solution {
public boolean isPossibleDivide(int[] nums, int k) {
if (nums.length % k != 0) {
return false;
}
TreeMap<Integer, Integer> tm = new TreeMap<>();
for (int h : nums) {
tm.put(h, tm.getOrDefault(h, 0) + 1);
}
while (!tm.isEmpty()) {
int v = tm.firstKey();
for (int i = v; i < v + k; ++i) {
if (!tm.containsKey(i)) {
return false;
}
if (tm.get(i) == 1) {
tm.remove(i);
} else {
tm.put(i, tm.get(i) - 1);
}
}
}
return true;
}
}
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| class Solution {
public:
bool isPossibleDivide(vector<int>& nums, int k) {
if (nums.size() % k != 0) return false;
map<int, int> mp;
for (int& h : nums) mp[h] += 1;
while (!mp.empty()) {
int v = mp.begin()->first;
for (int i = v; i < v + k; ++i) {
if (!mp.count(i)) return false;
if (mp[i] == 1)
mp.erase(i);
else
mp[i] -= 1;
}
}
return true;
}
};
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| func isPossibleDivide(nums []int, k int) bool {
if len(nums)%k != 0 {
return false
}
m := treemap.NewWithIntComparator()
for _, h := range nums {
if v, ok := m.Get(h); ok {
m.Put(h, v.(int)+1)
} else {
m.Put(h, 1)
}
}
for !m.Empty() {
v, _ := m.Min()
for i := v.(int); i < v.(int)+k; i++ {
if _, ok := m.Get(i); !ok {
return false
}
if v, _ := m.Get(i); v.(int) == 1 {
m.Remove(i)
} else {
m.Put(i, v.(int)-1)
}
}
}
return true
}
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