Nope, he's spot on. The decision player one gets to make is at the beginning.

imagine a 6 element list... 

The beginning, bottom row is the number, top row is odds/evens flag.

 1 0 1 0 1 0
 5 1 3 4 5 3

player 1 totals the odds and the evens, decides to go for odds,
and takes from the left

 1 0 1 0 1 0
   1 3 4 5 3

player 2 has to take an even element, lets say from the right

 1 0 1 0 1 0
   1 3 4 5

player 1 is after odds, so takes from the right

 1 0 1 0 1 0
   1 3 4

hmmm, player two is left with the evens again.

 1 0 1 0 1 0
   1 3

player 1 sticks with odds

 1 0 1 0 1 0
   1

player two takes the last even element, and loses

[download]

If you try this strategy on the example numbers the OP gave, you win by ten, which is as good as strategy in the OP's example code.

---
my name's not Keith, and I'm not reasonable.

In reply to Re^3: Puzzle: Given an array of integers, find the best sequence of pop / shift... by reasonablekeith
in thread Puzzle: Given an array of integers, find the best sequence of pop / shift... by TedPride

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