FIND MINIMUM TIME TO FINISH ALL JOBS WITH GIVEN CONSTRAINTS.pptx

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

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

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.

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.
4

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;
}

…..(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);

…..(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);
}

…..(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;
}

…..(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

…..(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);
}

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}

Thank you
For Watching..☺
12

FIND MINIMUM TIME TO FINISH ALL JOBS WITH GIVEN CONSTRAINTS.pptx

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