Sticks length not required . Think in coordinate geometry in 3D .
Also for Checkin and Checkout question - Make a flag array associated with
each element of the list and initialize the every element to *{1}* and
traverse the checkout list (checkout[i]) . Now do binary search for an entry
in the checkin list for the time just greater than a given checkout time i.e
checkin[j] > checkout[i] and make the flag[j] = flag[j] -1 . Now keep a
counter *count *to count the number of elements in the party at a given
point . When you traverse the list add flag[i] to *count *i.e* *count + =
flag[i] . Find the maximum value of count and at the that moment there would
be max no of people in the party .
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