This problem is based off of the following problem on stack overflow: https://stackoverflow.com/questions/64797299/greedy-algorithm-to-maximize-score. The second answer is incorrect because the answer $[3,4,1,6]$ does not satisfy the constraint that games must be played at or before their deadline, and game $1$ is played on day $3$, after its deadline of day $2$. The first answer doesn't provide pseudocode and I'm not sure about the running time for the improved approach.

Here is the problem: There are $n$ fruits, and each has an associated energy level, stored in the array e[1...n]. Also, each fruit has an expiry date, stored in the array d[1...n]. Fruit i must be picked up before d[i] days from today. Today is 0 days from today. Both e[1...n] and d[1...n] only store positive values. A person wishes to maximize the total energy of the fruits they can take subject to the constraint that they take at most one fruit per day.

How would one design and prove a greedy algorithm (or a dynamic programming algorithm) for this problem that runs in time $O(n\log n)$?

If possible, could the algorithm avoid using disjoint set forests? It can be similar to the fast implementation of the algorithm described in CLRS (problem 16-4) that is used to solve this problem: A task-scheduling problem as a matroid (CLRS book), but I'd like to have a more detailed description of how the faster method involving disjoint set forests works.

I've found an algorithm that works (at least for the many tests I've done) in $O(n^2)$ time. It essentially involves computing the maximum energy that can be obtained assuming $i$ days have passed for $0\leq i\leq n-1$. Since it may update previous selections, it is arguably not a greedy algorithm.

  • $\begingroup$ Some general tips on greedy algorithms here: cs.stackexchange.com/q/59964/755 $\endgroup$
    – D.W.
    Commented Jan 27, 2022 at 6:03
  • $\begingroup$ @D.W. I've modified the question again. Even though I said I came up with a dynamic programming algorithm, I think the algorithm is still a greedy algorithm. $\endgroup$ Commented Jan 29, 2022 at 15:02
  • $\begingroup$ @D.W. thanks for the comments. Next time I'll make sure to ask clearer questions. $\endgroup$ Commented Jan 29, 2022 at 23:01
  • 2
    $\begingroup$ There is nothing in the Stack Overflow question claims that a greedy algorithm is possible; the only indication of anything related to a greedy algorithm is the tag "greedy". Meanwhile, the accepted answer is baloney (it describes itself as "greedy" but actually appears to be exhaustive search). I recommend that you don't trust anything you read on Stack Overflow about algorithms -- the quality there is variable. $\endgroup$
    – D.W.
    Commented Jan 30, 2022 at 3:17
  • 1
    $\begingroup$ This is the unit-time scheduling problem. It is already featured in this site <cs.stackexchange.com/questions/139254/…>. $\endgroup$
    – pcpthm
    Commented Jan 30, 2022 at 4:20

2 Answers 2


This problem is basically the problem of unit-time job sequencing with deadlines and profits as appeared on Geeksforgeeks, if we consider each fruit with its expiry date and energy as a one-day job with corresponding deadline and profit.

Basic idea

The basic ideas to solve this problem are

  • A fruit with higher energy level is prioritized over another fruit with lower energy level, and
  • a fruit should be picked up on the latest day among the remaining available days so as to leave more available days for fruits with more stringent expiry dates.

A greedy algorithm

  1. Let the total energy be 0.
  2. Obtain a sorted list of all fruits. A fruit with higher energy level precedes another fruit with lower energy level.
  3. For each fruit $f$ in the sorted list, try finding $d$, the latest available date before the expiry date of $f$.
    • If there is no such date, do nothing (this fruit is ignored).
    • Otherwise, pick up this fruit on date $d$. Add the energy level of $f$ to the total energy. Mark date $d$ as unavailable.
  4. Return the total energy.

Proof of correctness

Agreement(k), $0\le k\le n.$ There is an optimal scheduling that agrees with the algorithm on the scheduling of the first $m$ fruits (in the sorted list of all fruits) at steps 3.

The correctness of the algorithm is the proposition Agreement(n). Let us prove Agreement(k) for all $k$.

Proof. The base case, when $k=0$, is of course correct.

As the induction hypothesis, assume the proposition is correct for $k$, i.e., there is an optimal scheduling $\mathcal O$ that agrees with the algorithm on the scheduling of the first $k$ fruits.

Suppose the algorithm has just finished processing the first $k$ fruits at step 3. Consider the next iteration of step 3, when the next fruit, $g$ is processed.

  • If there is no available date before the expiry date of $g$, the algorithm will not pick up $g$. This is what $\mathcal O$ has to do as well.
  • Otherwise, the algorithm will pick up fruit $g$ on date $d$, the latest available date before the expiry date of fruit $g$.
    • If $\mathcal O$ does not pick up any fruit on date $d$,
      • If $\mathcal O$ does pick up $g$, modify $\mathcal O$ so that $\mathcal O$ pick up $g$ on date $d$.
      • Otherwise, $\mathcal O$ does not pick up $g$. Modify $\mathcal O$ so that $\mathcal O$ picks up $g$ on date $d$, which will increase the total energy, which is impossible. So this case cannot happen.
    • Otherwise, $\mathcal O$ picks up some fruit $g'$ on date $d$. Note that $g'$ cannot be one of the first $k$ fruits.
      • If $g'$ is $g$, we are done.
      • Otherwise, $g'$ is not $g$.
        • If $\mathcal O$ pick up fruit $g$ on some date $d'$, modify $\mathcal O$ so that $\mathcal O$ picks up $g$ on date $d$ and picks up $g'$ on date $d'$. This modification is valid since both $g$ and $g'$ expires later than date $d$ and date $d'$ must be earlier than date $d$.
        • Otherwise $\mathcal O$ does not pick up fruit $g$. Modify $\mathcal O$ so that $\mathcal O$ picks up $g$ instead of $g'$ on date $d$. Since the list of fruits are sorted primarily by energy level, the energy level of $g$ is no less than that of $g'$. So after modification, $\mathcal O$ should be optimal still.

In all cases, $\mathcal O$ with possible modification is an optimal scheduling that agrees with the algorithm on the scheduling of the first $k+1$ fruits. $\quad\checkmark$

Implementation with $O(n\log n)$ time-complexity

  • Use a sorting algorithm with $O(n\log n)$ time-complexity such as merge sort to sort all fruits.
  • How can we track available days and find some particular available day?
    • We can use a balanced binary search tree where each node represents some consecutive available days.
    • We can also use disjoint-set data structure (DSDS). Make a DSDS out of the first $n$ days. Maintain a partition of the set where each part consists of some consecutive days such that the only available day among them is the earliest one. Initially each day constitutes its own part of the partition. Whenever a day is used to pick up a fruit, the two parts near that day are combined into one part. Thanks to the power of path compression, the scheduling takes $O(n\log n)$ time.

An implementation in Python

Here is an implementation in Python.

There is a minor tweak to boost performance. We will apply the greedy algorithm only for all fruits whose expiry dates are no less than $n$. All other fruits will be counted towards total energy. It is not hard to verify the correctness of this modification. (We do not have to use this tweak. We can just replace every expiry date that is no less than $n$ with $n$.)

def maximum_energy(n, energy, expiry):
    """Return the maximum total energy

    There are fruit 0, 1, ..., n-1.
    `energy[i]` is the energy level of fruit `i`
    `expiry[i]` is the expiry day of fruit `i`
    fruits = [(energy[i], expiry[i]) for i in range(n) if expiry[i] < n]
    fruits.sort()  # ordered primarily by energy level

    # disjoint-set data structure (DSDS)
    # day `parent[d]` is some day no later than day `d`.
    # for any `d < n`, day `d` is available iff `parent[d] == d`.
    parent = {i: i for i in range(n)}  # all days are available
    parent[-1] = -1  # day -1 is yesterday

    def find(date):
        """ return the latest available day no later than `date`"""
        if parent[date] != date:
            parent[date] = find(parent[date])
            return parent[date]
            return date

    # customized union method of DSDS. `date` was available.
    def use(date):
        # day `date` is not available any more.
        parent[date] = find(date - 1)

    total_energy = sum(energy[i] for i in range(n) if expiry[i] >= n)
    while fruits:
        energy, expiry = fruits.pop()  # from the end
        pick_up_day = find(expiry - 1)
        if pick_up_day >= 0:
            total_energy += energy

    return total_energy

def test():
    energy = [76, 66, 52, 51, 47, 23]
    expiry = [3, 2, 2, 4, 2, 1]
    max_energy = maximum_energy(6, energy, expiry)
    print("maximum energy:", max_energy)  # 245


We use a greedy algorithm. Sort all fruits in decreasing order of energy. Let $d_i$ be the deadline of the $i$-th fruit in the sorted array. Iterate through the fruits and schedule each fruit on the latest day less than or equal to $d_i$ that has not been taken yet (if it exists). This can be implemented with, for example, a balanced binary search tree (BBST) in $O(n \log n)$. It can also be implemented with disjoint sets.

We provide a (very rough) sketch of a proof that the algorithm is optimal. Let the set of fruits be $S$ and the set of fruits the greedy algorithm takes be $G$. As elements have already been placed at the latest possible time, we cannot take an additional element $i \in S \setminus G$ without removing at least one element $j \in G$ (or else we would have already taken it). Moreover, removing an element $j \in G$ would only free up one space for a different element to take. Since this element would have had to be after $j$ in the order, it would have had less value then $j$ and so the answer would be no better.


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