[LeetCode] 1296. Divide Array in Sets of K Consecutive Numbers

轉自LeetCode

Given an array of integers nums and a positive integer k, find whether it is 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

Solution

659. Split Array into Consecutive Subsequences 的衍生題

這次有要求每個 subsequence 的長度

可以用同一個思路來解，並且可以簡化

只要檢查到，沒辦法形成一個新的 subsequence，直接回傳 false

其餘的思路一樣

kotlin

	class Solution {
	fun isPossibleDivide(nums: IntArray, k: Int): Boolean {
	return if(nums.size % k != 0) {
	false
	} else {
	val freqMap = mutableMapOf<Int,Int>()
	for(n in nums) {
	freqMap[n] = freqMap[n]?.let{it+1} ?: 1
	}
	nums.sort()
	for(n in nums) {
	if(freqMap[n]!! == 0) {
	continue
	}
	for(i in 1 until k) {
	if(freqMap[n+i] == null || freqMap[n+i]!! <= 0) {
	return false
	} else {
	freqMap[n+i] = freqMap[n+i]!! - 1
	}
	}
	freqMap[n] = freqMap[n]!! - 1
	}
	true
	}
	}
	}

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