Formatted question description: https://leetcode.ca/all/1296.html

# 1296. Divide Array in Sets of K Consecutive Numbers (Medium)

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

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

Related Topics:
Array, Greedy

Similar Questions:

## Solution 1.

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

// Time: O(NlogM + MK) where M is the number of unique numbers in A.
// Space: O(N)
class Solution {
public:
bool isPossibleDivide(vector<int>& A, int k) {
if (A.size() % k) return false;
map<int, int> m;
for (int n : A) ++m[n];
for (auto &[n, cnt] : m) {
if (cnt == 0) continue;
for (int i = 1; i < k; ++i) {
if (m.count(n + i) == 0 || m[n + i] < cnt) return false;
m[n + i] -= cnt;
}
}
return true;
}
};


Java

class Solution {
public boolean isPossibleDivide(int[] nums, int k) {
int length = nums.length;
if (length % k != 0)
return false;
Arrays.sort(nums);
List<Integer> list = new ArrayList<Integer>();
for (int num : nums)
while (!list.isEmpty()) {
int min = list.remove(0);
for (int i = min + 1; i <= min + k - 1; i++) {
if (!list.contains(i))
return false;
list.remove(new Integer(i));
}
}
return true;
}
}