Let ~S_x~ denote the set of all the possible joke intervals that can be found at the party if we observe person ~x~ and all their direct or indirect subordinates (the subtree of person ~x~).

Let us assume that for a person ~x~ we have calculated the values of sets ~S~ for each direct subordinate node to that person. Then we can calculate ~S_x~ using dynamic programming.

Let ~Q_l~ denote the set of all right sides of the interval of jokes that can be heard in the subtree of person ~x~ and whose left side of the interval is equal to ~l~. We iterate in the descending order over all the possible left sides ~l~ and for each ~l~ we iterate over all the possible right sides ~r~ so that there is a child of node ~x~ that has the interval ~[l, r]~ in its set and we calculate the following relation:

$$\displaystyle Q_l = Q_l \cup Q_{r+1}$$

A special case when ~l = v_x~, then simply means:

$$\displaystyle Q_l = \{v_x\} \cup Q_{l+1}$$

The time complexity of this algorithm is ~\mathcal O(NK^3)~ where ~K~ is the number of different jokes, and ~N~ is the number of nodes. A bitset data structure can be used in order to accelerate calculating the union operation by ~32~ times.