When we are finding distance b/w P and Q (x2-x1)^2 + (y2-y1)^2 whole underoot.(for 2d case). Now according to constraint of co-ordinates (this value (x2-x1)^2 + (y2-y1)^2) can exceed max value possible to store)?
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Its guaranteed that co-ordinates of Q never exceed 2*{10}^{9} during entire time.
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As specified in the constraints that the absolute value of the coordinates can be at max 2 *10^9 .
Now if we take x2=2 *10^9, x1=-2 *10^9, y2=2 *10^9, y1=-2 *10^9, so it’s pretty clear the value
(x2-x1)^2 + (y2-y1)^2 turns out to be 32 *10^18 which does not fit into long type integers . So yes, it would overflow.
Use long double.
As that is 64 bit then i think it can also store same as c++ upto 2^64-1,thats nearly 18*10^18.
Will double not suffice? Did you try that?
Thats true, but long double afaik is more precise than double. I never trust floating points so thought better safe than sorry xD
How can long double store more than 64 bits i.e >10^18
long double or even double can be used. They have a huge range (around 1.7e^{308}) (check here) even though they are 64 bits since they use floating point representation, i.e. have some bits reserved for exponent and some bits for mantissa.