This document describes an algorithm to find the minimum time required to finish all jobs given a set of constraints. It takes as input the number of available assignees (K), the time taken by each assignee to finish one unit of work (T), and an array of job times (job[]). It uses a binary search approach to find the minimum feasible time. It ensures that assignees are only assigned contiguous jobs and that jobs are not shared between assignees. The algorithm returns the minimum completion time multiplied by T. It provides two examples applying the algorithm.

1. •Yastee A. Shah (16it148)
By:
Design and
Analyze
Algorithm
(IT-341)

2. TOPIC :
FIND MINIMUM TIME TO
FINISH ALL JOBS WITH
GIVEN CONSTRAINTS..!
2

3. Variables used :
3
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.

4. 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.
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5. Code:
5
// 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;
}

6. …..(2)
6
// Returns minimum time required to finish given array of jobs
// k --> number of assignees
// T --> Time required by every assignee to finish 1 unit
// n --> 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);

7. …..(3)
7
// 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);
}

8. …..(4)
8
// 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;
}

9. …..(5)
9
// 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

10. …..(6)
10
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);
}

11. Example :
11
(1)
Input: k = 2, T = 5, job[] = {4, 5, 10}
Output: 50 The minimum time required to finish all the jobs is
50. There are 2 assignees available. We get this time by
assigning {4, 5} to first assignee and {10} to second
assignee.
(2)
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
{6, 8}

12. Thank you
For Watching..☺
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