# 1296. Divide Array in Sets of K Consecutive Numbers

## 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 <= 105
• 1 <= nums[i] <= 109

Note: This question is the same as 846: https://leetcode.com/problems/hand-of-straights/

## Solutions

• 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;
}
}

• 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;
}
};

• 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


• 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
}