Last Updated on 2023-01-04 by Clay
題目
You are given a 0-indexed integer array tasks, where tasks[i] represents the difficulty level of a task. In each round, you can complete either 2 or 3 tasks of the same difficulty level.
Return the minimum rounds required to complete all the tasks, or -1 if it is not possible to complete all the tasks.
Example 1:
Input: tasks = [2,2,3,3,2,4,4,4,4,4] Output: 4 Explanation: To complete all the tasks, a possible plan is: - In the first round, you complete 3 tasks of difficulty level 2. - In the second round, you complete 2 tasks of difficulty level 3. - In the third round, you complete 3 tasks of difficulty level 4. - In the fourth round, you complete 2 tasks of difficulty level 4. It can be shown that all the tasks cannot be completed in fewer than 4 rounds, so the answer is 4.
Example 2:
Input: tasks = [2,3,3] Output: -1 Explanation: There is only 1 task of difficulty level 2, but in each round, you can only complete either 2 or 3 tasks of the same difficulty level. Hence, you cannot complete all the tasks, and the answer is -1.
Constraints:
1 <= tasks.length <= 1051 <= tasks[i] <= 109
題目給定一串數值陣列，陣列中的每個元素都代表一個指定難度的任務。我們每個回合只能解同一種類型的任務，每次可以解 2 個或是 3 個同種任務。
我們要計算的是，最少要花多少回合，才能把所有任務解決掉呢？如果有無法解決的情況，則回傳 -1。
解題思路
哈希表（Hash Table）
- 我們可以使用哈希表來儲存每種種類的任務有多少個。
- 針對每種特定種類的任務，我們先判斷是否能被 3 整除。這時候會分成兩種情況
- 被 3 整除：代表最少的解決回合數即為『任務數除以 3』
- 不被 3 整除：無論是餘 1 還是餘 2，最少的解決回合數都是『任務數除以 3 加 1』
- 把所有任務的最少解決回合數加總，即為我們要的答案；如果中途有任何一種特定任務的數量為 1，則代表我們無論如何都解決不了，所以回傳
-1
哈希表 複雜度
| Time Complexity | O(n) |
| Space Complexity | O(n) |
C++ 範例程式碼
class Solution {
public:
int minimumRounds(vector<int>& tasks) {
// Init
int rounds = 0;
unordered_map<int, int> mp;
for (int& task: tasks) {
++mp[task];
}
// Count `rounds`
for (auto& it: mp) {
if (it.second == 1) {
return -1;
}
rounds += it.second % 3 ? it.second / 3 + 1 : it.second / 3;
}
return rounds;
}
};Python 範例程式碼
class Solution:
def minimumRounds(self, tasks: List[int]) -> int:
# Init
rounds = 0
mp = {}
for task in tasks:
mp[task] = mp.get(task, 0) + 1
# Count `rounds`
for key in mp:
if mp[key] == 1:
return -1
rounds += mp[key] // 3 + 1 if mp[key] % 3 else mp[key] // 3
return rounds