1296. 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

// Greedy
bool isPossibleDivide(vector<int>& nums, int k) { // time: O(n*log(n)); space: O(n)
    if (nums.size() % k) return false;
    map<int, int> cnt;
    int freq = 0;
    for (const int& num : nums) ++cnt[num];
    for (auto it = cnt.begin(); it != cnt.end(); ++it) {
        if (!it->second) continue;
        freq = it->second;
        it->second -= freq;
        for (int i = 1; i < k; ++i) {
            if (cnt[it->first + i] < freq) return false;
            else cnt[it->first + i] -= freq;
        }
    }
    return true;
}

// Greedy
bool isPossibleDivide(vector<int>& nums, int k) { // time: O(nlogn); space: O(n)
    if (nums.size() % k) return false;
    sort(nums.begin(), nums.end());
    unordered_map<int, int> cnt;
    for (const int& num : nums) ++cnt[num];
    for (const int& num : nums) {
        if (!cnt[num]) continue;
        --cnt[num];
        bool found = false;
        for (int i = 1; i < k; ++i) {
            if (!cnt[num + i]--) break;
            if (i == k - 1) found = true;
        }
        if (!found) return false;
    }
    return true;
}