# Completion time of a given process in round robin

We are given n-processes with their completion times in form of an array. We need to find the time instant when a given process p ends if the scheduling process is round robin and time slice is 1-sec.

Examples :

```Input : arr[] = {3, 2, 4, 2}, p = 1
Output : Completion time = 6
Explanation : Snap of process for every second is as:
Time    |   Process Array
0       |   {3, 2, 4, 2}
1       |   {2, 2, 4, 2}
2       |   {2, 1, 4, 2}
3       |   {2, 1, 3, 2}
4       |   {2, 1, 3, 1}
5       |   {1, 1, 3, 1}
6       |   {1, 0, 3, 1}

Input : arr[] = {2, 4, 1, 3}, p = 2
Output :Completion time = 3
Explanation : Snap of process for every second is as:
Time    |   Process Array
0       |   {2, 4, 1, 3}
1       |   {1, 4, 1, 3}
2       |   {1, 3, 1, 3}
3       |   {1, 3, 0, 3}
```

Brute Force :The basic approach for solving this problem is to apply round robin algorithm with time slice 1. But the time complexity of that approach will be O(ΣAi) i.e. summation of all process’s time, which is quite high.

Efficient Approach: The idea is based on below observations.
1) All processes with CPU time less than arr[p] would complete before arr[p]. We simply need to add time of these processes.
2) We also need to add time of arr[p].
3) For every process x with CPU time more than arr[p], two cases arise :
…..(i) If x is on left of arr[p] (scheduled before arr[p]), then this process takes arr[p] time of CPU before p finishes.
…..(ii) If x is on right of arr[p] (scheduled after arr[p]), then this process takes arr[p]-1 time of CPU before p finishes.

Algorithm :

```time_req = 0;

// Add time for process on left of p
// (Scheduled before p in a round of
// 1 unit time slice)
for (int i=0; i<p; i++)
{
if (arr[i] < arr[p])
time_req += arr[i];
else
time_req += arr[p];
}

// step 2 : Add time of process p
time_req += arr[p];

// Add time for process on right
// of p (Scheduled after p in
// a round of 1 unit time slice)
for (int i=p+1; i<n; i++)
{
if (arr[i] < arr[p])
time_req += arr[i];
else
time_req += arr[p]-1;
}
```

## C++

 `// Program to find end time of a process  ` `// p in round robin scheduling with unit ` `// time slice. ` `#include ` `using` `namespace` `std; ` ` `  `// Returns completion time of p. ` `int` `completionTime(``int` `arr[], ``int` `n, ``int` `p) { ` ` `  `  ``// Initialize result ` `  ``int` `time_req = 0; ` ` `  `  ``// Step 1 : Add time of processes on left ` `  ``//  of p (Scheduled before p) ` `  ``for` `(``int` `i = 0; i < p; i++) { ` `    ``if` `(arr[i] < arr[p]) ` `      ``time_req += arr[i]; ` `    ``else` `      ``time_req += arr[p]; ` `  ``} ` ` `  `  ``// Step 2 : Add time of p ` `  ``time_req += arr[p]; ` ` `  `  ``// Step 3 : Add time of processes on right ` `  ``//  of p (Scheduled after p) ` `  ``for` `(``int` `i = p + 1; i < n; i++) { ` `    ``if` `(arr[i] < arr[p]) ` `      ``time_req += arr[i]; ` `    ``else` `      ``time_req += arr[p] - 1; ` `  ``} ` ` `  `  ``return` `time_req; ` `} ` ` `  `// driver program ` `int` `main() { ` `  ``int` `arr[] = {3, 5, 2, 7, 6, 1}; ` `  ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `  ``int` `p = 2; ` `  ``cout << ``"Completion time = "`  `       ``<< completionTime(arr, n, p); ` `  ``return` `0; ` `} `

## Java

 `// Program to find end time of a process  ` `// p in round robin scheduling with unit ` `// time slice. ` `class` `GFG ` `{ ` `    ``// Returns completion time of p. ` `    ``static` `int` `completionTime(``int` `arr[], ``int` `n, ``int` `p) { ` `         `  `        ``// Initialize result ` `        ``int` `time_req = ``0``; ` `         `  `        ``// Step 1 : Add time of processes on left ` `        ``// of p (Scheduled before p) ` `        ``for` `(``int` `i = ``0``; i < p; i++) { ` `            ``if` `(arr[i] < arr[p]) ` `                ``time_req += arr[i]; ` `            ``else` `                ``time_req += arr[p]; ` `        ``} ` `         `  `        ``// Step 2 : Add time of p ` `        ``time_req += arr[p]; ` `         `  `        ``// Step 3 : Add time of processes on right ` `        ``// of p (Scheduled after p) ` `        ``for` `(``int` `i = p + ``1``; i < n; i++) { ` `            ``if` `(arr[i] < arr[p]) ` `                ``time_req += arr[i]; ` `            ``else` `                ``time_req += arr[p] - ``1``; ` `        ``} ` `         `  `        ``return` `time_req; ` `    ``} ` `         `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``int` `arr[] = {``3``, ``5``, ``2``, ``7``, ``6``, ``1``}; ` `        ``int` `n =arr.length;; ` `        ``int` `p = ``2``; ` `         `  `        ``System.out.print(``"Completion time = "``+ ` `            ``completionTime(arr, n, p)); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python

 `# Program to find end time of a process  ` `# p in round robin scheduling with unit ` `# time slice. ` ` `  `# Returns completion time of p. ` `def` `completionTime(arr, n, p) : ` ` `  `    ``# Initialize result ` `    ``time_req ``=` `0` `     `  `    ``# Step 1 : Add time of processes on ` `    ``# left of p (Scheduled before p) ` `    ``for` `i ``in` `range``(``0``, p):  ` `        ``if` `(arr[i] < arr[p]): ` `            ``time_req ``+``=` `arr[i] ` `        ``else``: ` `            ``time_req ``+``=` `arr[p] ` `     `  `     `  `    ``# Step 2 : Add time of p ` `    ``time_req ``+``=` `arr[p] ` `     `  `    ``# Step 3 : Add time of processes on  ` `    ``# right of p (Scheduled after p) ` `    ``for` `i ``in` `range``(p ``+` `1``, n):  ` `        ``if` `(arr[i] < arr[p]): ` `            ``time_req ``+``=` `arr[i] ` `        ``else``: ` `            ``time_req ``+``=` `arr[p] ``-` `1` `     `  `    ``return` `time_req ` `     `  ` `  `# driver program ` `arr ``=` `[``3``, ``5``, ``2``, ``7``, ``6``, ``1``] ` `n ``=` `len``(arr)  ` `p ``=` `2` `print``(``"Completion time ="``,  ` `        ``completionTime(arr, n, p)) ` ` `  ` `  `# This code is contributed by  ` `# Smitha Dinesh Semwal `

## C#

 `// C# program to find end time of a process  ` `// p in round robin scheduling with unit ` `// time slice. ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Returns completion time of p. ` `    ``static` `int` `completionTime(``int` `[]arr, ` `                            ``int` `n, ``int` `p) ` `    ``{ ` `         `  `        ``// Initialize result ` `        ``int` `time_req = 0; ` `         `  `        ``// Step 1 : Add time of processes ` `        ``// on left of p (Scheduled before p) ` `        ``for` `(``int` `i = 0; i < p; i++) { ` `            ``if` `(arr[i] < arr[p]) ` `                ``time_req += arr[i]; ` `            ``else` `                ``time_req += arr[p]; ` `        ``} ` `         `  `        ``// Step 2 : Add time of p ` `        ``time_req += arr[p]; ` `         `  `        ``// Step 3 : Add time of processes on ` `        ``// right of p (Scheduled after p) ` `        ``for` `(``int` `i = p + 1; i < n; i++) { ` `            ``if` `(arr[i] < arr[p]) ` `                ``time_req += arr[i]; ` `            ``else` `                ``time_req += arr[p] - 1; ` `        ``} ` `         `  `        ``return` `time_req; ` `    ``} ` `         `  `    ``// Driver code ` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``int` `[]arr = {3, 5, 2, 7, 6, 1}; ` `        ``int` `n =arr.Length;; ` `        ``int` `p = 2; ` `         `  `        ``Console.WriteLine(``"Completion time = "``+ ` `                    ``completionTime(arr, n, p)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output :

```Completion time = 9
```

Time Complexity : O(n)

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