# C Program for Activity Selection Problem | Greedy Algo-1

• Last Updated : 22 Aug, 2019

You are given n activities with their start and finish times. Select the maximum number of activities that can be performed by a single person, assuming that a person can only work on a single activity at a time.
Example:

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```Example 1 : Consider the following 3 activities sorted by finish time.
start[]  =  {10, 12, 20};
finish[] =  {20, 25, 30};
A person can perform at most two activities. The
maximum set of activities that can be executed
is {0, 2} [ These are indexes in start[] and
finish[] ]

Example 2 : Consider the following 6 activities
sorted by by finish time.
start[]  =  {1, 3, 0, 5, 8, 5};
finish[] =  {2, 4, 6, 7, 9, 9};
A person can perform at most four activities. The
maximum set of activities that can be executed
is {0, 1, 3, 4} [ These are indexes in start[] and
finish[] ]
```

## C++

 `// C++ program for activity selection problem.``// The following implementation assumes that the activities``// are already sorted according to their finish time``#include `` ` `// Prints a maximum set of activities that can be done by a single``// person, one at a time.``// n   -->  Total number of activities``// s[] -->  An array that contains start time of all activities``// f[] -->  An array that contains finish time of all activities``void` `printMaxActivities(``int` `s[], ``int` `f[], ``int` `n)``{``    ``int` `i, j;`` ` `    ``printf``(``"Following activities are selected n"``);`` ` `    ``// The first activity always gets selected``    ``i = 0;``    ``printf``(``"%d "``, i);`` ` `    ``// Consider rest of the activities``    ``for` `(j = 1; j < n; j++) {``        ``// If this activity has start time greater than or``        ``// equal to the finish time of previously selected``        ``// activity, then select it``        ``if` `(s[j] >= f[i]) {``            ``printf``(``"%d "``, j);``            ``i = j;``        ``}``    ``}``}`` ` `// driver program to test above function``int` `main()``{``    ``int` `s[] = { 1, 3, 0, 5, 8, 5 };``    ``int` `f[] = { 2, 4, 6, 7, 9, 9 };``    ``int` `n = ``sizeof``(s) / ``sizeof``(s[0]);``    ``printMaxActivities(s, f, n);``    ``return` `0;``}`
Output:
```Following activities are selected n0 1 3 4
```

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