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# 1296. Divide Array in Sets of K Consecutive Numbers

https\://leetcode.com/problems/divide-array-in-sets-of-k-consecutive-numbers/

Given an array of integers `nums` and a positive integer `k`, find whether it's possible to divide this array into sets of `k` consecutive numbers\
Return `True` if its 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 = [3,3,2,2,1,1], k = 3
Output: true
```

**Example 4:**

```
Input: nums = [1,2,3,4], k = 3
Output: false
Explanation: Each array should be divided in subarrays of size 3.
```

**Constraints:**

* `1 <= nums.length <= 10^5`
* `1 <= nums[i] <= 10^9`
* `1 <= k <= nums.length`

数组能否分成长度都为K的集合，每个集合内部满足差为1的等差序列。排好序后依次对每个元素找它后面K-1个是否存在，不存在说明不能分，存在就把它从数组中移除；继续遍历。

```cpp
class Solution {
public:
    bool isPossibleDivide(vector<int>& nums, int k) {
        const int N = nums.size();
        if (N % k != 0) return false;
        multiset<int> s(nums.begin(), nums.end());
        while (!s.empty()) {
            const auto num = *(s.begin());
            s.erase(s.begin());
            for (int i = 1; i < k; ++i) {
                if (s.empty()) return false;
                auto it = s.find(num + i);
                if (it == s.end()) return false;
                s.erase(it);
            }
        }
        return true;
    }
};
```