ou 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 c

repairCars .js

/**
 * @param {number[]} ranks
 * @param {number} cars
 * @return {number}
 */
var repairCars = function (ranks, cars) {
    // Initialize the binary search range:
    // 'left' represents the minimum possible time (1 minute).
    // 'right' is the maximum possible time (if the slowest mechanic repairs all cars).
    let left = 1;
    let right = Math.min(...ranks) * cars * cars;

    /**
     * Helper function to determine if it's possible to repair all cars
     * within the given time `mid`.
     * @param {number} mid - The current time being tested.
     * @return {boolean} - True if all cars can be repaired within `mid` time, otherwise false.
     */
    function canRepairInTime(mid) {
        let totalCars = 0; // Total number of cars repaired across all mechanics.
        
        // Calculate the number of cars each mechanic can repair within `mid` time.
        for (let rank of ranks) {
            let count = Math.floor(Math.sqrt(mid / rank)); // Cars repaired by this mechanic.
            totalCars += count;
            
            // If the total cars repaired meet or exceed the required `cars`, return true early.
            if (totalCars >= cars) return true;
        }
        
        // If the total cars repaired are less than required, return false.
        return false;
    }

    // Binary search to find the minimum time.
    while (left < right) {
        let mid = Math.floor((left + right) / 2); // Calculate the midpoint of the current range.
        
        // Check if it's feasible to repair all cars in `mid` time.
        if (canRepairInTime(mid)) {
            right = mid; // If feasible, reduce the upper limit (try to find a smaller time).
        } else {
            left = mid + 1; // If not feasible, increase the lower limit.
        }
    }

    // The minimum time required to repair all cars is stored in 'left'.
    return left;
};