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oh, TryAdd exists.
lol, replacing one of the conditions with TryAdd bumps me performance-wise from 83rd to 93rd percentile, but kicks me down memory-wise from 49th to 8th percentile.
and the difference is about a megabyte =D
and all the other solutions are basically this same one, just written slightly differently.
okay.
well. i still like the solution =D -
I'd go for a sorting plus binary search approach that would automatically tackle the bonus question, too.
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@Fast-Nop
bonus question?
(tries to scroll the description)
oh.
...yeah, i haven't slept all night... again.
if i had, maybe i would have noticed that. -
Grumm18242yWhat about just using the array ?
public int[] TwoSum(int[] nums, int target) {
Dictionary<int, int> remainsAtIndex = new Dictionary<int, int>();
for (int c = 0; c < nums.Length; c++)
{
int index = Array.IndexOf(nums, target - nums[c], c+1);
if (index > -1)
return new int[] { c, index };
}
return new int[] { 0, 0 };
}
(It checks all the test data in 192 ms)
Oh, yeah it will not get you the bonus question either. IndexOf is also O(n) -
@Grumm array.IndexOf does linear search though.
dictionary does it in 153 ms, and I was going for speed -
Looks pretty smart although I'm generally not a fan of allocating when not necessary
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@12bitfloat
...and where do I do that? i mean, i could do bruteforce two nested loops just adding up together all the combinations, which would still allocate 3 vars, but ints, so maybe that would be faster, allocation-wise, except the whole solution would then be shitslow, so... i'm sorry, but i absolutely don't understand what you mean.
Related Rants
What do you think about my solution to two sum?
https://leetcode.com/problems/...
It took me about 10 seconds to realize it can be solved this way, and then FUCKING HALF AN HOUR until i finally wrote the actual code in a way that worked as it should...
...i really should sleep more. and get examined for brain decomposition or something.
question
leetcode