# Find minimum time to finish all jobs with given constraints

Given an array of jobs with different time requirements. There are K identical assignees available and we are also given how much time an assignee takes to do one unit of job. Find the minimum time to finish all jobs with following constraints.

• An assignee can be assigned only contiguous jobs. For example, an assignee cannot be assigned jobs 1 and 3, but not 2.
• Two assignees cannot share (or co-assigned) a job, i.e., a job cannot be partially assigned to one assignee and partially to other.

Input :

```K:     Number of assignees available.
T:     Time taken by an assignee to finish one unit
of job
job[]: An array that represents time requirements of
different jobs.```

Example:

```Input:  k = 2, T = 5, job[] = {4, 5, 10}
Output: 50
The minimum time required to finish all jobs is 50.
There are 2 assignees available. We get this time by
assigning {4, 5} to first assignee and {10} to second
assignee.

Input:  k = 4, T = 5, job[] = {10, 7, 8, 12, 6, 8}
Output: 75
We get this time by assigning {10} {7, 8} {12} and {7, 8}
```

We strongly recommend you to minimize your browser and try this yourself first.
The idea is to use Binary Search. Think if we have a function (say isPossible()) that tells us if it’s possible to finish all jobs within a given time and number of available assignees. We can solve this problem by doing a binary search for the answer. If middle point of binary search is not possible, then search in second half, else search in first half. Lower bound for Binary Search for minimum time can be set as 0. The upper bound can be obtained by adding all given job times.

Now how to implement isPossible()? This function can be implemented using Greedy Approach. Since we want to know if it is possible to finish all jobs within a given time, we traverse through all jobs and keep assigning jobs to current assignee one by one while a job can be assigned within the given time limit. When time taken by current assignee exceeds the given time, create a new assignee and start assigning jobs to it. If number of assignees become more than k, then return false, else return true.

```// C++ program to find minimum time to finish all jobs with
// given number of assignees
#include<bits/stdc++.h>
using namespace std;

// Utility function to get maximum element in job[0..n-1]
int getMax(int arr[], int n)
{
int result = arr[0];
for (int i=1; i<n; i++)
if (arr[i] > result)
result = arr[i];
return result;
}

// Returns true if it is possible to finish jobs[] within
// given time 'time'
bool isPossible(int time, int K, int job[], int n)
{
// cnt is count of current assignees required for jobs
int cnt = 1;

int curr_time = 0; //  time assigned to current assignee

for (int i = 0; i < n;)
{
// If time assigned to current assignee exceeds max,
// increment count of assignees.
if (curr_time + job[i] > time) {
curr_time = 0;
cnt++;
}
else { // Else add time of job to current time and move
// to next job.
curr_time += job[i];
i++;
}
}

// Returns true if count is smaller than k
return (cnt <= K);
}

// Returns minimum time required to finish given array of jobs
// k --> number of assignees
// T --> Time required by every assignee to finish 1 unit
// m --> Number of jobs
int findMinTime(int K, int T, int job[], int n)
{
// Set start and end for binary search
// end provides an upper limit on time
int end = 0, start = 0;
for (int i = 0; i < n; ++i)
end += job[i];

int ans = end; // Initialize answer

// Find the job that takes maximum time
int job_max = getMax(job, n);

// Do binary search for minimum feasible time
while (start <= end)
{
int mid = (start + end) / 2;

// If it is possible to finish jobs in mid time
if (mid >= job_max && isPossible(mid, K, job, n))
{
ans = min(ans, mid);  // Update answer
end = mid - 1;
}
else
start = mid + 1;
}

return (ans * T);
}

// Driver program
int main()
{
int job[] =  {10, 7, 8, 12, 6, 8};
int n = sizeof(job)/sizeof(job[0]);
int k = 4, T = 5;
cout << findMinTime(k, T, job, n) << endl;
return 0;
}
```

Output:

`75`

Thanks to Gaurav Ahirwar for suggesting above solution.

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