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
Code
第一次看到這題的想法是直接用 hash table 記錄下次數，然後再計算一次做 3 個 job 最多需要幾次。
特別注意在做除法時要將 variable cast 成 float type 才會得到浮點數的計算結果。（float div = (float)count / (float)3;）
class Solution {
public:
int minimumRounds(vector<int>& tasks) {
unordered_map<int, int> myMap;
for(int i = 0; i < tasks.size(); i++) {
if(myMap.find(tasks[i]) == myMap.end()) {
myMap[tasks[i]] = 1;
} else {
myMap[tasks[i]] += 1;
}
}
int rounds = 0;
for(auto i: myMap) {
int count = i.second;
if(count < 2) return -1;
else {
float div = (float)count / (float)3;
rounds += ceil(div);
}
}
return rounds;
}
};Key Insight
(freq + 2) / 3 就足以判斷需要多少 round，而不需要像上面那樣計算浮點數再取 ceiling。
class Solution {
public:
int minimumRounds(vector<int>& tasks) {
unordered_map<int, int> count;
for(int t: tasks) {
++count[t];
}
int rounds = 0;
for(auto &i: count) {
int freq = i.second;
if(freq < 2) return -1;
rounds += (freq + 2) / 3;
}
return rounds;
}
};