0323HardArrayDynamic Programming

Cheapest Shift Plan For A Restoration Queue

Tracked in this browser only

Trains the technique from

LeetCode 1335Minimum Difficulty of a Job Schedule

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 film archive restores reels in a fixed queue order. jobDifficulty[i] is the wear rating of the reel at position i, and the queue must be handled in exactly d shifts.

A shift takes a contiguous run of reels from the front of what is left, and every shift has to take at least one reel. The technician on a shift is billed for the single worst reel they touched, so a shift costs the largest wear rating inside its run. The bill for the whole plan is the sum over the d shifts.

Return the smallest possible bill. If the queue cannot be cut into d non-empty runs at all, return -1.

Examples

Example 1

Input: jobDifficulty = [4, 1, 9, 2], d = 2
Output: 11

The plan [4, 1, 9] then [2] uses two contiguous non-empty runs and is billed 9 + 2 = 11.

Example 2

Input: jobDifficulty = [7, 3, 5, 8, 2], d = 3
Output: 17

The plan [7], [3, 5, 8], [2] is billed 7 + 8 + 2 = 17, and each of its three runs is non-empty and contiguous.

Example 3

Input: jobDifficulty = [6, 2], d = 3
Output: -1

Only two reels are queued, so three non-empty runs cannot be formed and the sentinel is returned.

Example 4

Input: jobDifficulty = [5, 5, 5], d = 3
Output: 15

Each shift must take exactly one reel and every reel is rated 5, so the bill is 5 + 5 + 5.

Constraints

1 <= jobDifficulty.length <= 300
0 <= jobDifficulty[i] <= 1000
1 <= d <= 10

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 cheapest_shift_plan(reels: list[int], shifts: int) -> int:

Java

public int cheapestShiftPlan(int[] reels, int shifts)

The intended approach

Time O(d * n^2)Space O(n)

Build the answer one shift-count at a time. Row k holds, for every prefix length, the least bill for covering that prefix with k shifts; extending to row k+1 tries every start for the new final run, walking the start leftwards so the run's maximum accumulates in a single variable. Only the previous row is needed, so two rows suffice.

Pythonentry point: cheapest_shift_plan

def cheapest_shift_plan(reels, shifts):
    # Layered DP: table[k][i] = least billing for the first i reels split into k
    # shifts. Extending a layer scans candidate block starts right to left so the
    # block maximum is carried along instead of recomputed.
    # Time O(shifts * n^2). Space O(n) -- two rows of length n + 1.
    n = len(reels)
    if n < shifts:
        return -1
    INF = float("inf")
    prev = [INF] * (n + 1)
    prev[0] = 0
    for k in range(1, shifts + 1):
        cur = [INF] * (n + 1)
        top = n - (shifts - k)
        for i in range(k, top + 1):
            best = INF
            block_max = 0
            j = i - 1
            while j >= k - 1:
                if reels[j] > block_max:
                    block_max = reels[j]
                if prev[j] < INF:
                    cand = prev[j] + block_max
                    if cand < best:
                        best = cand
                j -= 1
            cur[i] = best
        prev = cur
    return prev[n]

The obvious version

Enumerate every choice of d-1 cut positions, add up the run maxima, keep the smallest.

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_cheapest_shift_plan(reels, shifts):
    from itertools import combinations
    n = len(reels)
    if n < shifts:
        return -1
    best = None
    for cuts in combinations(range(1, n), shifts - 1):
        edges = (0,) + cuts + (n,)
        total = 0
        for a, b in zip(edges, edges[1:]):
            total += max(reels[a:b])
        if best is None or total < best:
            best = total
    return best

How this problem was checked

Reference solution executed against 8 test cases — 8 passed
4 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in