The complexity of the algorithm described by the recurrence T(n) = 4T(n/3) + n^2 can be determined by using the Master Theorem.

According to the Master Theorem, if T(n) = aT(n/b) + f(n) where a >= 1 and b > 1, then the following three cases can occur:

If f(n) = O(n^logb(a)) then T(n) = O(n^logb(a))

If f(n) = O(n^logb(a) * logn) then T(n) = O(n^logb(a) * logn)

If f(n) = O(n^logb(a)) and f(n) is not O(n^(logb(a) * logn)) then T(n) = O(n^logb(a) * log n)

In this case, a = 4, b = 3, and f(n) = n^2. Since log3(4) = 1.63, it can be seen that n^2 = O(n^1.63), and since n^2 = O(n^1.63 * logn), the recurrence falls into the second case of the Master Theorem.

Therefore, the time complexity of the algorithm is O(n^1.63 * logn).
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