Yes exactly! I just stumbled upon your solution while looking through the comments for what values I should be outputting, and comparing answers to my code.
Your updated solution works great! I'm not sure I get the "math" side of it, as my solution is definitely brute force (recursively remove tokens until we can't anymore).
I get that your code sorts, then returns either the total number of chips / 2 OR the total number of chips in the smaller stacks, whichever is smaller.
Math-wise, I'll take a stab at understanding it.
If we have 1 larger / infinite stack, then we are limited by the other two. We pair a chip from the smaller stacks with one from the large stack everyday, and that's our total. That's the second half of the return statement.
However, if we have stacks that are somewhat even, we can rotate through and deplete them evenly. We code this by taking the total number of chips, and dividing them by the daily chip requirement (we can use the bitwise operator >> here since the daily chip requirement is 2).
I approached it like this: There's a case where it's better to split the highest stack in half to divvy between the lower 2 stacks, this only happens when the lower 2 stacks add up to be greater than the highest stack, then there are these "leftovers" in the stacks that you can combine.
I ended up with something looking like high/2 + (mid + low)/2
which can be reduced mathematically to (high + mid + low)/2
There was a point where it stopped making sense and I was just reducing the math and logic. This was a solution that helped me dev.to/juanrodriguezarc/comment/12o94
For further actions, you may consider blocking this person and/or reporting abuse
We're a place where coders share, stay up-to-date and grow their careers.
Yes exactly! I just stumbled upon your solution while looking through the comments for what values I should be outputting, and comparing answers to my code.
Your updated solution works great! I'm not sure I get the "math" side of it, as my solution is definitely brute force (recursively remove tokens until we can't anymore).
I get that your code sorts, then returns either the total number of chips / 2 OR the total number of chips in the smaller stacks, whichever is smaller.
Math-wise, I'll take a stab at understanding it.
If we have 1 larger / infinite stack, then we are limited by the other two. We pair a chip from the smaller stacks with one from the large stack everyday, and that's our total. That's the second half of the return statement.
However, if we have stacks that are somewhat even, we can rotate through and deplete them evenly. We code this by taking the total number of chips, and dividing them by the daily chip requirement (we can use the bitwise operator >> here since the daily chip requirement is 2).
I approached it like this: There's a case where it's better to split the highest stack in half to divvy between the lower 2 stacks, this only happens when the lower 2 stacks add up to be greater than the highest stack, then there are these "leftovers" in the stacks that you can combine.
I ended up with something looking like
high/2 + (mid + low)/2which can be reduced mathematically to
(high + mid + low)/2There was a point where it stopped making sense and I was just reducing the math and logic. This was a solution that helped me dev.to/juanrodriguezarc/comment/12o94