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Formatted question description: https://leetcode.ca/all/2141.html
2141. Maximum Running Time of N Computers
 Difficulty: Hard.
 Related Topics: Array, Binary Search, Greedy, Sorting.
 Similar Questions: Minimum Moves to Equal Array Elements, Sell DiminishingValued Colored Balls, Maximum Number of Tasks You Can Assign, Minimum Time to Complete Trips, Minimum Amount of Time to Fill Cups.
Problem
You have n
computers. You are given the integer n
and a 0indexed integer array batteries
where the ith
battery can run a computer for batteries[i]
minutes. You are interested in running all n
computers simultaneously using the given batteries.
Initially, you can insert at most one battery into each computer. After that and at any integer time moment, you can remove a battery from a computer and insert another battery any number of times. The inserted battery can be a totally new battery or a battery from another computer. You may assume that the removing and inserting processes take no time.
Note that the batteries cannot be recharged.
Return the **maximum number of minutes you can run all the n
computers simultaneously.**
Example 1:
Input: n = 2, batteries = [3,3,3]
Output: 4
Explanation:
Initially, insert battery 0 into the first computer and battery 1 into the second computer.
After two minutes, remove battery 1 from the second computer and insert battery 2 instead. Note that battery 1 can still run for one minute.
At the end of the third minute, battery 0 is drained, and you need to remove it from the first computer and insert battery 1 instead.
By the end of the fourth minute, battery 1 is also drained, and the first computer is no longer running.
We can run the two computers simultaneously for at most 4 minutes, so we return 4.
Example 2:
Input: n = 2, batteries = [1,1,1,1]
Output: 2
Explanation:
Initially, insert battery 0 into the first computer and battery 2 into the second computer.
After one minute, battery 0 and battery 2 are drained so you need to remove them and insert battery 1 into the first computer and battery 3 into the second computer.
After another minute, battery 1 and battery 3 are also drained so the first and second computers are no longer running.
We can run the two computers simultaneously for at most 2 minutes, so we return 2.
Constraints:

1 <= n <= batteries.length <= 105

1 <= batteries[i] <= 109
Solution

class Solution { private boolean isPossibeToRun(int n, int[] batteries, long avgTime) { long duration = 0; for (long ele : batteries) { duration += Math.min(ele, avgTime); } return avgTime * n <= duration; } public long maxRunTime(int n, int[] batteries) { long startTime = 0; long sum = 0; long ans = 0; for (long ele : batteries) { sum += ele; } long endTime = sum; while (startTime <= endTime) { long avgTime = (startTime + endTime) / 2; if (isPossibeToRun(n, batteries, avgTime)) { ans = avgTime; startTime = avgTime + 1; } else { endTime = avgTime  1; } } return ans; } }

# 2141. Maximum Running Time of N Computers # https://leetcode.com/problems/maximumrunningtimeofncomputers class Solution: def maxRunTime(self, n: int, batteries: List[int]) > int: def good(k): target = n * k have = 0 for x in batteries: have += min(x, k) return have >= target total = sum(batteries) left, right = 0, total // n res = 1 while left <= right: mid = left + (right  left) // 2 if good(mid): res = mid left = mid + 1 else: right = mid  1 return res

impl Solution { #[allow(dead_code)] pub fn max_run_time(n: i32, batteries: Vec<i32>) > i64 { // First sort the batteries let mut batteries = batteries; let m = batteries.len() as i32; batteries.sort(); let mut extra_sum: i64 = 0; for i in 0..(m  n) as usize { extra_sum += batteries[i] as i64; } let mut i = (m  n) as usize; let mut cur_height = batteries[i]; let mut ret = cur_height as i64; while extra_sum != 0 { if i + 1 == m as usize { assert!(cur_height == *batteries.last().unwrap()); ret += extra_sum / n as i64; break; } if batteries[i] == batteries[i + 1] { i += 1; continue; } let diff = extra_sum / (i  (m  n) as usize + 1) as i64; if (cur_height as i64 + diff) <= batteries[i + 1] as i64 { ret = cur_height as i64 + diff; break; } else { extra_sum = (batteries[i + 1]  batteries[i]) as i64 * (i  (m  n) as usize + 1) as i64; ret = batteries[i + 1] as i64; } i += 1; if i != m as usize { cur_height = batteries[i]; } } ret } }

class Solution { public: long long maxRunTime(int n, vector<int>& batteries) { long long l = 0, r = 0; for (int x : batteries) { r += x; } while (l < r) { long long mid = (l + r + 1) >> 1; long long s = 0; for (int x : batteries) { s += min(1LL * x, mid); } if (s >= n * mid) { l = mid; } else { r = mid  1; } } return l; } };

func maxRunTime(n int, batteries []int) int64 { l, r := 0, 0 for _, x := range batteries { r += x } for l < r { mid := (l + r + 1) >> 1 s := 0 for _, x := range batteries { s += min(x, mid) } if s >= n*mid { l = mid } else { r = mid  1 } } return int64(l) } func min(a, b int) int { if a < b { return a } return b }

function maxRunTime(n: number, batteries: number[]): number { let l = 0n; let r = 0n; for (const x of batteries) { r += BigInt(x); } while (l < r) { const mid = (l + r + 1n) >> 1n; let s = 0n; for (const x of batteries) { s += BigInt(Math.min(x, Number(mid))); } if (s >= mid * BigInt(n)) { l = mid; } else { r = mid  1n; } } return Number(l); }
Explain:
nope.
Complexity:
 Time complexity : O(n).
 Space complexity : O(n).