Suppose that the output (how many seconds passed between the arrival of the first type and second types of monkeys) is ~S~. Then we know that monkeys of the first type spent ~S~ seconds in the garden and monkeys of the second type spent ~T-S~ seconds.

We can calculate how many coconuts the first type can gather in ~S~ seconds. Let this amount be ~X(S)~.

Also, we can calculate how many coconuts the second type can open in ~T-S~ seconds (assuming that there is an infinite number on the ground). Let this amount be ~Y(S)~.

If ~X(S) > Y(S)~, this means that the actual output is less than ~S~ because monkeys of the second type do not have enough time to open all the coconuts.

If ~X(S) \le Y(S)~, this means that the output can be ~S~ or more because there is time to open all coconuts.

The correct output is the largest ~S~ for which ~X(S) \le Y(S)~. We can use binary search to find it.