1052MediumArrayDynamic Programming

Choosing Which Tasks to Take

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Trains the technique from

LeetCode 2140Solving Questions With Brainpower

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A shift offers tasks in a fixed order, given as tasks, where tasks[i] = [points, rest].

Work through the tasks from the first to the last. At each task you may either take it, earning its points and then skipping the next rest tasks entirely, or leave it and move straight on to the next.

Return the most points the shift can earn.

Examples

Example 1

Input: tasks = [[1, 5], [9, 1], [1, 1]]
Output: 9

Leaving the first task and taking the second earns nine, which beats taking the first for one and then being skipped past everything else.

Example 2

Input: tasks = [[5, 1], [5, 1], [5, 1]]
Output: 10

Each task skips the one after it, so the first and the third can both be taken, earning ten.

Example 3

Input: tasks = [[10, 1]]
Output: 10

One task, so take it: the rest period runs off the end of the shift and costs nothing.

Constraints

1 <= tasks.length <= 10^5
tasks[i].length == 2
1 <= tasks[i][0] <= 10^5
1 <= tasks[i][1] <= 10^5

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def most_points(tasks: list[list[int]]) -> int:

Java

public long mostPoints(int[][] tasks)

The intended approach

Time O(n) over the tasks.Space O(n) for the table of bests.

Nothing about the past matters once a task is reached: whatever was taken earlier, the best still to come depends only on which task you stand at. So work out, for each task, the most that can be earned from there onwards. Settle those from the last task backwards, so both answers a task needs are already known. Taking the task earns its points plus the best from the task after its rest period, or nothing more when that lands past the end. Leaving it gives the best from the very next task. The larger of the two is the answer for that task. The best from the first task is the answer for the whole shift. Working backwards is what turns a search over every choice into one pass with one number remembered per task.

Pythonentry point: most_points

def most_points(tasks):
    n = len(tasks)
    best = [0] * (n + 1)
    for at in range(n - 1, -1, -1):
        points = tasks[at][0]
        rest = tasks[at][1]
        after = at + rest + 1
        take = points + (best[after] if after <= n else 0)
        leave = best[at + 1]
        best[at] = take if take > leave else leave
    return best[0]

The obvious version

Search the choices outright. From a task, work out both branches by calling the same routine again, and keep the better. Nothing is remembered between calls, so the same task is worked out again and again, which costs a walk over every combination of choices rather than one pass.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_most_points(tasks):
    n = len(tasks)
    def walk(at):
        if at >= n:
            return 0
        take = tasks[at][0] + walk(at + tasks[at][1] + 1)
        leave = walk(at + 1)
        return take if take > leave else leave
    return walk(0)

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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