1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers

1296. Divide Array in Sets of K Consecutive Numbers

题目地址

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

题目描述

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

Exactly Same as 846. Hand of Straights

代码

Approach #1

1. Count number of different cards to a map

2. Loop from the smallest card number.

3. Everytime we meet a new card i, we cut off i - i + k - 1 from the counter.

Time O(M log M + MK), where M is the number of different cards.

class Solution {
    public boolean isPossibleDivide(int[] nums, int k) {
        if (nums.length % k != 0) return false;
        Map<Integer, Integer> map = new TreeMap();
        for (int num: nums) {
            map.put(num, map.getOrDefault(num, 0) + 1);
        }

        for (int num: map.keySet()) {
            int cnt = map.get(num);
            if (cnt == 0)   continue;

            for (int i = num; i < k + num; i++) {
                if (!map.containsKey(i) || map.get(i) <= 0) {
                    return false;
                }
                map.put(i, map.get(i) - cnt);
            }
        }

        return true;
    }
}

Approach #2

1. Count number of different cards to a map c

2. Cur represent current open straight groups.

3. In a deque start, we record the number of opened a straight group.

4. Loop from the smallest card number.

 public boolean isPossibleDivide(int[] A, int k) {
    Map<Integer, Integer> c = new TreeMap<>();
    for (int i : A)     c.put(i, c.getOrDefault(i, 0) + 1);

    Queue<Integer> start = new LinkedList<>();
    int last_checked = -1, opened = 0;
    for (int i : c.keySet()) {
        if (opened > 0 && i > last_checked + 1 || opened > c.get(i)) return false;
        start.add(c.get(i) - opened);
        last_checked = i; opened = c.get(i);
        if (start.size() == k) opened -= start.remove();
    }
    return opened == 0;
}