1296. Divide Array in Sets of K Consecutive Numbers
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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
class Solution { public boolean isPossibleDivide(int[] nums, int k) { boolean res = false; int l = nums.length; if(l % k != 0) return res; Map<Integer, Integer> map = new HashMap(); PriorityQueue<Integer> q = new PriorityQueue(); for(int i: nums){ map.put(i, map.getOrDefault(i, 0)+1); } for(int i: map.keySet()) q.offer(i); while(!q.isEmpty()){ int cur = q.poll(); if(map.get(cur) == 0) continue; int times = map.get(cur); for(int i = 0; i < k; i++){ if(!map.containsKey(cur + i) || map.get(cur + i) < times) return false; map.put(cur + i, map.get(cur + i) - times); } l -= k * times; } return l == 0; } }
首先判断有没有k倍的元素,然后用map记录每个数出现的数量,然后用pq记下出现的数字,并用全局变量l记录数组长度。
先拿出一个key,如果对应的value==0就跳过,否则就用times出现的数量。
从0到k-1循环,如果没有cur+i或者cur+i剩余数量少于times说明必不可能成功,return false,更新map中每个key的value
更新剩余长度l
判断l==0?
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