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Arranging Coins

@Evilgit overflow

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I got the same idea at the first look. Glad to see someone has posted that.

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@sujiths52 Use -b+sqrt(b^2 - 4ac)/2*a is correct too.

I just found my solution intuition.

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@SJD_Dream Sigh....that's not fun....

No one has replied

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python 1 line accepted solution:

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Excellent solution..Thank you!

I think we can skip the floor function because the int cast will do

Why do we need the + 1 here, without +1, we will get Time Exceed Limit. Can anyone help me understand the corner case about that? I assume there is something related to the stopping condition while (low < high)

@jasonshieh said in C++_9ms_Accepted:

class Solution { public: int arrangeCoins(int n) { if(n <= 1) return n; long left = 0, right = n; while(left <= right){ long mid = left + (right - left)/2; long sum = ((1 + mid) * mid)/2; if(sum <= n){left = mid + 1;} else {right = mid - 1;} } return left - 1; } };

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Best code world！

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@zebra1831 to prevent integer overflow. left + right could overflow, whereas right - left couldn't.

nice solution

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