9th JLTi Code Jam – Nov 2017
Whole JLT is buzzing with Recreational Committee (RC) aka Fun Ministry election 2018. It is more palpable in JLTi where a fierce competition is taking place between two candidates representing Millennial Party and Traditional Party. In this two-party system, Millennial Party is claiming that they know the magic as to how people can be entertained while the Traditional Party cannot stop laughing at them saying they are just inexperienced kids incapable of running the massive Fun Ministry.
This time, voting mechanism has changed. Instead of one person one vote that was how it worked till last year, one person’s vote weight would now equal to the number of years he/she is working at JLT. For example, I am working here for 4 years and hence my vote would count as 4. If somebody is working for just 1 year, his/her vote weight would be 1. For obvious reason, Millennial Party is unhappy about this new legislation that was recently passed by the incumbent Traditional Party. They call it unfair. But law is law.
Voting stopped on 10th Nov 2017 and counting votes would commence on 13th Nov 2017 followed by the announcement of result on the same day.
Being a member of the existing RC, my concern is little different. I am worried about a tie, and if that happens, what would be the next course of action.
Hence, I am checking the possibility of a tie. Manually doing so is quite problematic, if not impossible, for several hundred employees that we have in Singapore. Being the only software engineer in the existing RC, I am tasked to write a program that would take vote weight of each of the voters as input and output whether a tie is possible.
Input:
20 10 4 6
Output: Possible
Explanation: Vote weight 20 and 10 – sounds familiar? Anyway, we see that if the first voter votes for one candidate and the rests for another – a tie is inevitable.
Input:
8 7 2 5 16
Output: Not Possible
Explanation: As you see the total vote weight for the above 5 voters is 38. For a tie to happen each candidate should get a vote count of 19. However, we can see, no way a vote count of 19 is possible here.
Input:
5, 6, 7, 3
Output: Not Possible
Explanation: We see, the total vote weight for the above 4 voters is 21. An odd number cannot be divided by two.
Task: Given a list of vote weight, one for each voter, we need to find whether a tie is possible. We are assuming that all voters in the input would vote for sure.
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