0326MediumArrayMathDynamic Programming

Best Dial Offset Score

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Trains the technique from

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 calibration dial has n slots arranged in a circle, and slot i holds the signed reading nums[i]. A read head lines up with one slot and then walks all the way round, giving heavier credit to later stops.

Starting the head at slot k, the sequence it visits is B with B[i] = nums[(i + k) % n] for i from 0 to n - 1, and the score of that start is

score(k) = 0 * B[0] + 1 * B[1] + 2 * B[2] + ... + (n - 1) * B[n - 1]

Return the largest score(k) over the n possible starting slots 0 <= k < n. Readings may be negative, so the best start is not always slot 0.

Examples

Example 1

Input: nums = [4, -1, 6, 2]
Output: 22

Starting at slot 1 the head reads [-1, 6, 2, 4] and scores 0*(-1) + 1*6 + 2*2 + 3*4 = 22, which is the reported value.

Example 2

Input: nums = [7]
Output: 0

A one-slot dial has a single start and the only weight is 0, so the score is 0.

Example 3

Input: nums = [-3, -8, -2]
Output: -8

Every start scores below zero. Starting at slot 1 the head reads [-8, -2, -3] for 0*(-8) + 1*(-2) + 2*(-3) = -8, the value returned.

Example 4

Input: nums = [0, 5, -5, 9, 1]
Output: 31

Starting at slot 2 the head reads [-5, 9, 1, 0, 5] and scores 0*(-5) + 1*9 + 2*1 + 3*0 + 4*5 = 31, the value returned.

Constraints

n == nums.length
1 <= n <= 10^5
-100 <= nums[i] <= 100

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 best_dial_score(readings: list[int]) -> int:

Java

public int bestDialScore(int[] readings)

The intended approach

Time O(n)Space O(1)

Compute the running total and the score of the slot-0 start in one pass. Shifting the start forward by one lowers the weight of every reading by one and moves the reading that had weight 0 to weight n-1, so the next score is the previous score minus the running total plus n times the reading that just wrapped. Track the maximum as the offsets are stepped through.

Pythonentry point: best_dial_score

def best_dial_score(readings):
    # Rolling the window by one slot pulls every weight down by one, except the
    # reading that wraps from weight 0 round to weight n-1.
    # Time O(n). Space O(1).
    n = len(readings)
    total = 0
    score = 0
    for i, v in enumerate(readings):
        total += v
        score += i * v
    best = score
    for k in range(n - 1):
        score = score - total + n * readings[k]
        if score > best:
            best = score
    return best

The obvious version

For each of the n starts, walk the whole circle and add up the weighted readings; keep the largest.

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_best_dial_score(readings):
    n = len(readings)
    best = None
    for k in range(n):
        s = 0
        for i in range(n):
            s += i * readings[(i + k) % n]
        if best is None or s > best:
            best = s
    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

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