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# Standing Ovation: Google codejam 2015 Qualification round problem

Every year Google conducts a competition called Codejam for programming enthusiasts. The qualification round for the year 2015 got finished yesterday. In this post, I am going to explain the solution approach for the first problem.

You can read the problem from this link. I will try to give the abridged problem statement below.

There are a group of N people in a stadium with their shyness levels represented by a String S of digits(0-9). The digit at each index represents the number of persons with shyness level equal to the index.

For example S[0] represents the number persons with shyness ‘0’.
S[1] represent the number of persons with shyness ‘1’ and so on.

Persons with shyness level ‘L’ will not standup and clap until ‘L’ other people standup and clap before them. For example, in order for a person with shyness level 3 to standup and clap, 3 other people must do that before him.

Now the problem is to find whether the all the people in the given group gives a standing ovation on its own? or What is the minimum number of people to add to the group to ensure that they give a standing ovation.

Let us consider a couple of simple examples.

1. [1, 1, 2]: There is one person with shyness level 0 who will clap first. After seeing him, the second person with shyness 1 will also clap. Then the last two people with shyness 2 will also clap. So there is no need to add any people to give a standing ovation.
2. [1, 0, 1]: In this case, we need one person with shyness 0 or 1 to be added to the group because the last person will not standup until 2 other people clap.

A simple greedy approach will solve the problem. Here is the algorithm.

Start at the first person and count the number of people stood up before him. If this count is not enough for the current person to standup, count the difference needed.  Accordingly increment the people currently standing.

Here is the C++ implementation of the above approach. This runs in O(n) time.