README_EN.md

1296. Divide Array in Sets of K Consecutive Numbers

中文文档

Description

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 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 = [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 <= k <= nums.length <= 105
• 1 <= nums[i] <= 109

 

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

Solutions

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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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