Suppose you have 1 room that you want to rent out. (AirBnb style) You want to maximize profits that you will get by renting it out.

For example: Given intervals: [[1, 10], [2, 5], [7, 20], [23, 30]] - you could rent it out [2, 5], [7, 20] and [23, 30]

Another example: ([[1, 2], [1, 11], [5, 8], [4, 33], [18, 72]]) = 66

Note: Start and end times of the interval are inclusive

I implemented a brute force solution to this problem. First I sort by start time, then I create all possible subsets and take the longest possible value. This works. But I want to do this using Dynamic programming.

This problem screams DP, but I am not able to figure out if this has an optimal substructure.

My recurrence relation: f(i, j) = f(i + 1, j) or f(i + nums[i], j) if nums[i].start > f[i].end

Can someone help me figure out the thought behind the dp solution for the sub-problem?

Note: this problem is slightly different from job scheduling.


The most common subproblems where a particular subsequence should be selected are parametrized by the index of the last element selected. The classical, simple and brilliant example is Kadane's algorithm.

Let the given intervals are $I_1, I_2,\cdots, I_n$, where $I_j=[l_j, r_j]$.

The subproblem $DP[i]$, where $1\le i\le n$ is the maximum rent if the last interval rented is the $i$-th interval. The answer is $\max_iDP[i]$.

We could add an artificial base case, $DP[0]=0$, which says that the maximum rent is 0 if there is no interval rented.

The recurrence relation is $$DP[i] = \text{Length}(I_i) + \max_{r_j\lt l_i} DP[j].$$

That is it, basically. There is a minor problem with the above recurrence relation, however. When we compute $DP[i]$, not all of $DP[j]$ with $r_j\lt l_i$ has been computed. I will leave it for you to resolve this minor obstacle.

  • $\begingroup$ In fact, that obstacle can be ignored, if we use recursion and memorization, although it might slow down the computation. If we use iterative approach, that obstacle becomes critical. $\endgroup$
    – John L.
    Apr 5 '20 at 23:55
  • $\begingroup$ The running time of the iterative approach is $O(n^2)$. There is an obvious approach to reduce the computation time in general. To obtain $DP[i]$, we only need $DP[j]$'s such that $I_j$ is immediate before $I_i$, that is, there is no other interval between $I_j$ and $I_i$ that is disjoint with both of them. $\endgroup$
    – John L.
    Apr 7 '20 at 0:18

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