# Minimum Time Required

### Problem Statement :

```You are planning production for an order. You have a number of machines that each have a fixed number of days to produce an item. Given that all the machines operate simultaneously, determine the minimum number of days to produce the required order.

For example, you have to produce goal = 10 items. You have three machines that take machines = [2, 3, 2 ] days to produce an item. The following is a schedule of items produced:

Day Production  Count
2   2               2
3   1               3
4   2               5
6   3               8
8   2              10
It takes 8 days to produce 10  items using these machines.

Function Description

Complete the minimumTime function in the editor below. It should return an integer representing the minimum number of days required to complete the order.

minimumTime has the following parameter(s):

machines: an array of integers representing days to produce one item per machine
goal: an integer, the number of items required to complete the order

Input Format

The first line consist of two integers n  and goal, the size of machines  and the target production.
The next line contains n space-separated integers, machines[ i ].

Constraints

1  <=  n  <= 10^5
1  <=  goal  <=  10^9
1  <=   machines[ i ]  <=  10^9

Output Format

Return the minimum time required to produce goal items considering all machines work simultaneously.```

### Solution :

```                            ```Solution in C :

In   C :

#include <stdio.h>

int main() {
int n;
long long goal, ans;
scanf("%d%lld", &n, &goal);
long long machines[n];
for (int i = 0; i < n; i++) {
scanf("%lld", &machines[i]);
}
long long low = 1;
long long high = 1e18;
while (low <= high) {
long long mid = (low + high) / 2;
long long done = 0;
for (int i = 0; i < n; i++) {
done += mid / machines[i];
if (done >= goal)
break;
}
if (done >= goal) {
high = mid - 1;
ans = mid;
} else
low = mid + 1;
}
printf("%lld\n", ans);
return 0;
}```
```

```                        ```Solution in C++ :

In   C ++ :

#include<bits/stdc++.h>
using namespace std;

int main()
{
int n;
long long goal,ans;
scanf("%d%lld",&n,&goal);
long long machines[n];
for(int i=0;i<n;i++)
{
scanf("%lld",&machines[i]);
}
long long low=1;
long long high=1e18;
while(low<=high)
{
long long mid=(low+high)/2;
long long done=0;
for(int i=0;i<n;i++)
{
done+=mid/machines[i];
if(done>=goal)
break;
}
if(done>=goal)
{
high=mid-1;
ans=mid;
}
else
low=mid+1;
}
printf("%lld\n",ans);
return 0;
}```
```

```                        ```Solution in Java :

In    Java  :

static long minTime(long[] machines, long goal) {
Arrays.sort(machines);
long max = machines[machines.length - 1];
long minDays = 0;
long maxDays = max*goal;
long result = -1;
while (minDays < maxDays) {
long mid = (minDays + maxDays) / 2;
long unit = 0;
for (long machine : machines) {
unit += mid / machine;
}
if (unit < goal) {
minDays = mid+1;
} else {
result = mid;
maxDays = mid;
}
}
return result;
}```
```

```                        ```Solution in Python :

In  Python3  :

def minTime(machines, goal):

machines.sort()

low_rate = machines
lower_bound = (goal // (len(machines) / low_rate))
high_rate = machines[-1]
upper_bound = (goal // (len(machines) / high_rate)) + 1

while lower_bound < upper_bound:

num_days = (lower_bound + upper_bound) // 2
total = getNumItems(machines, goal, num_days)
if total >= goal:
upper_bound = num_days
else:
lower_bound = num_days + 1

return int(lower_bound)

def getNumItems(machines, goal, num_days):

total = 0

for machine in machines:
total += (num_days // machine)

```

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