Minimum Time to repair cars

Problem TC: O(nlog(k)), where k is the max time in the range between 1 to rn^2 class Solution { public long repairCars(int[] ranks, int cars) { long low = 1;// atleat one lowute will be needed long high = 0; for(int i : ranks){ high = Math.max(i,high);//mechanic with the largest rank will take greatest time to process all the cars } high = high cars * cars; long val = 0; while(low=cars; } }

Mar 16, 2025 - 14:37

Problem
TC: O(nlog(k)), where k is the max time in the range between 1 to r*n^2

class Solution {
    public long repairCars(int[] ranks, int cars) {
        long low = 1;// atleat one lowute will be needed 
        long high = 0;
        for(int i : ranks){
            high = Math.max(i,high);//mechanic with the largest rank will take greatest time to process all the cars
        }
        high = high *cars * cars;
        long val = 0;
        while(low<=high){
            long mid = (long)(low+high)/2l;

            if(check(mid,ranks,cars)){
                val = mid;
                high = mid-1;
            }
            else low = mid+1;
        }
        return val;
    }

    public boolean check(long time, int ranks[], int cars){
        //now we have to find out how much cars a mechanic can fix in given time 'time'
        //we know r*n^2 = time i.e rankOfMechanic*Math.pow(noOfCars,2) will give the time each mechanic will take to process n cars
        //we have to find out no. of cars for each mechanic for the give time
        //if total car processed in time 'time' is greater than or equal to 'cars' then the time is good enough time 
        //and we can look for time smaller that the time 'time'
        //n = Math.sqrt((time/r))

        int totalCar= 0;
        for(int rank : ranks){
            totalCar+= Math.sqrt(time/rank);
        }
        return totalCar>=cars;
    }
}