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1953. Maximum Number of Weeks for Which You Can Work
Description
There are n
projects numbered from 0
to n  1
. You are given an integer array milestones
where each milestones[i]
denotes the number of milestones the i^{th}
project has.
You can work on the projects following these two rules:
 Every week, you will finish exactly one milestone of one project. You must work every week.
 You cannot work on two milestones from the same project for two consecutive weeks.
Once all the milestones of all the projects are finished, or if the only milestones that you can work on will cause you to violate the above rules, you will stop working. Note that you may not be able to finish every project's milestones due to these constraints.
Return the maximum number of weeks you would be able to work on the projects without violating the rules mentioned above.
Example 1:
Input: milestones = [1,2,3] Output: 6 Explanation: One possible scenario is:  During the 1^{st} week, you will work on a milestone of project 0.  During the 2^{nd} week, you will work on a milestone of project 2.  During the 3^{rd} week, you will work on a milestone of project 1.  During the 4^{th} week, you will work on a milestone of project 2.  During the 5^{th} week, you will work on a milestone of project 1.  During the 6^{th} week, you will work on a milestone of project 2. The total number of weeks is 6.
Example 2:
Input: milestones = [5,2,1] Output: 7 Explanation: One possible scenario is:  During the 1^{st} week, you will work on a milestone of project 0.  During the 2^{nd} week, you will work on a milestone of project 1.  During the 3^{rd} week, you will work on a milestone of project 0.  During the 4^{th} week, you will work on a milestone of project 1.  During the 5^{th} week, you will work on a milestone of project 0.  During the 6^{th} week, you will work on a milestone of project 2.  During the 7^{th} week, you will work on a milestone of project 0. The total number of weeks is 7. Note that you cannot work on the last milestone of project 0 on 8^{th} week because it would violate the rules. Thus, one milestone in project 0 will remain unfinished.
Constraints:
n == milestones.length
1 <= n <= 10^{5}
1 <= milestones[i] <= 10^{9}
Solutions

class Solution { public long numberOfWeeks(int[] milestones) { int mx = 0; long s = 0; for (int e : milestones) { s += e; mx = Math.max(mx, e); } long rest = s  mx; return mx > rest + 1 ? rest * 2 + 1 : s; } }

class Solution { public: long long numberOfWeeks(vector<int>& milestones) { int mx = *max_element(milestones.begin(), milestones.end()); long long s = accumulate(milestones.begin(), milestones.end(), 0LL); long long rest = s  mx; return mx > rest + 1 ? rest * 2 + 1 : s; } };

class Solution: def numberOfWeeks(self, milestones: List[int]) > int: mx, s = max(milestones), sum(milestones) rest = s  mx return rest * 2 + 1 if mx > rest + 1 else s

func numberOfWeeks(milestones []int) int64 { mx := slices.Max(milestones) s := 0 for _, x := range milestones { s += x } rest := s  mx if mx > rest+1 { return int64(rest*2 + 1) } return int64(s) }

function numberOfWeeks(milestones: number[]): number { const mx = Math.max(...milestones); const s = milestones.reduce((a, b) => a + b, 0); const rest = s  mx; return mx > rest + 1 ? rest * 2 + 1 : s; }