LeetCode #2594 — MEDIUM

Minimum Time to Repair Cars

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an integer array ranks representing the ranks of some mechanics. ranks_i is the rank of the i^th mechanic. A mechanic with a rank r can repair n cars in r * n² minutes.

You are also given an integer cars representing the total number of cars waiting in the garage to be repaired.

Return the minimum time taken to repair all the cars.

Note: All the mechanics can repair the cars simultaneously.

Example 1:

Input: ranks = [4,2,3,1], cars = 10
Output: 16
Explanation: 
- The first mechanic will repair two cars. The time required is 4 * 2 * 2 = 16 minutes.
- The second mechanic will repair two cars. The time required is 2 * 2 * 2 = 8 minutes.
- The third mechanic will repair two cars. The time required is 3 * 2 * 2 = 12 minutes.
- The fourth mechanic will repair four cars. The time required is 1 * 4 * 4 = 16 minutes.
It can be proved that the cars cannot be repaired in less than 16 minutes.

Example 2:

Input: ranks = [5,1,8], cars = 6
Output: 16
Explanation: 
- The first mechanic will repair one car. The time required is 5 * 1 * 1 = 5 minutes.
- The second mechanic will repair four cars. The time required is 1 * 4 * 4 = 16 minutes.
- The third mechanic will repair one car. The time required is 8 * 1 * 1 = 8 minutes.
It can be proved that the cars cannot be repaired in less than 16 minutes.

Constraints:

1 <= ranks.length <= 10⁵
1 <= ranks[i] <= 100
1 <= cars <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given an integer array ranks representing the ranks of some mechanics. ranksi is the rank of the ith mechanic. A mechanic with a rank r can repair n cars in r * n2 minutes. You are also given an integer cars representing the total number of cars waiting in the garage to be repaired. Return the minimum time taken to repair all the cars. Note: All the mechanics can repair the cars simultaneously.

Baseline thinking

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

Pattern signal: Array · Binary Search

Example 1

[4,2,3,1]
10

Example 2

[5,1,8]
6

Core Insight

What unlocks the optimal approach

For a predefined fixed time, can all the cars be repaired?
Try using binary search on the answer.

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

Upper-end input sizes

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 #2594: Minimum Time to Repair Cars
class Solution {
    public long repairCars(int[] ranks, int cars) {
        long left = 0, right = 1L * ranks[0] * cars * cars;
        while (left < right) {
            long mid = (left + right) >> 1;
            long cnt = 0;
            for (int r : ranks) {
                cnt += Math.sqrt(mid / r);
            }
            if (cnt >= cars) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

// Accepted solution for LeetCode #2594: Minimum Time to Repair Cars
func repairCars(ranks []int, cars int) int64 {
	return int64(sort.Search(ranks[0]*cars*cars, func(t int) bool {
		cnt := 0
		for _, r := range ranks {
			cnt += int(math.Sqrt(float64(t / r)))
		}
		return cnt >= cars
	}))
}

# Accepted solution for LeetCode #2594: Minimum Time to Repair Cars
class Solution:
    def repairCars(self, ranks: List[int], cars: int) -> int:
        def check(t: int) -> bool:
            return sum(int(sqrt(t // r)) for r in ranks) >= cars

        return bisect_left(range(ranks[0] * cars * cars), True, key=check)

// Accepted solution for LeetCode #2594: Minimum Time to Repair Cars
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2594: Minimum Time to Repair Cars
// class Solution {
//     public long repairCars(int[] ranks, int cars) {
//         long left = 0, right = 1L * ranks[0] * cars * cars;
//         while (left < right) {
//             long mid = (left + right) >> 1;
//             long cnt = 0;
//             for (int r : ranks) {
//                 cnt += Math.sqrt(mid / r);
//             }
//             if (cnt >= cars) {
//                 right = mid;
//             } else {
//                 left = mid + 1;
//             }
//         }
//         return left;
//     }
// }

// Accepted solution for LeetCode #2594: Minimum Time to Repair Cars
function repairCars(ranks: number[], cars: number): number {
    let left = 0;
    let right = ranks[0] * cars * cars;
    while (left < right) {
        const mid = left + Math.floor((right - left) / 2);
        let cnt = 0;
        for (const r of ranks) {
            cnt += Math.floor(Math.sqrt(mid / r));
        }
        if (cnt >= cars) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

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.