LeetCode #2141 — HARD

Maximum Running Time of N Computers

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You have n computers. You are given the integer n and a 0-indexed integer array batteries where the i^th 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

Step 01

Brute Force Baseline

Problem summary: You have n computers. You are given the integer n and a 0-indexed 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Greedy

Example 1

2
[3,3,3]

Example 2

2
[1,1,1,1]

Core Insight

What unlocks the optimal approach

For a given running time, can you determine if it is possible to run all n computers simultaneously?
Try to use Binary Search to find the maximal running time

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2141: Maximum Running Time of N Computers
class Solution {
    public long maxRunTime(int n, int[] batteries) {
        long l = 0, r = 0;
        for (int x : batteries) {
            r += x;
        }
        while (l < r) {
            long mid = (l + r + 1) >> 1;
            long s = 0;
            for (int x : batteries) {
                s += Math.min(mid, x);
            }
            if (s >= n * mid) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }
}

// Accepted solution for LeetCode #2141: Maximum Running Time of N Computers
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)
}

# Accepted solution for LeetCode #2141: Maximum Running Time of N Computers
class Solution:
    def maxRunTime(self, n: int, batteries: List[int]) -> int:
        l, r = 0, sum(batteries)
        while l < r:
            mid = (l + r + 1) >> 1
            if sum(min(x, mid) for x in batteries) >= n * mid:
                l = mid
            else:
                r = mid - 1
        return l

// Accepted solution for LeetCode #2141: Maximum Running Time of N Computers
impl Solution {
    pub fn max_run_time(n: i32, batteries: Vec<i32>) -> i64 {
        let n = n as i64;
        let mut l: i64 = 0;
        let mut r: i64 = batteries.iter().map(|&x| x as i64).sum();

        while l < r {
            let mid = (l + r + 1) >> 1;
            let mut s: i64 = 0;

            for &x in &batteries {
                let v = x as i64;
                s += if v < mid { v } else { mid };
            }

            if s >= n * mid {
                l = mid;
            } else {
                r = mid - 1;
            }
        }

        l
    }
}

// Accepted solution for LeetCode #2141: Maximum Running Time of N Computers
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);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log M)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.