1103MediumArrayGreedySortingCounting

Nudging Dials Apart

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Trains the technique from

LeetCode 945Minimum Increment to Make Array Unique

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 row of dials shows whole-number readings, and several dials may show the same reading. One nudge picks a single dial and raises its reading by 1.

Return the fewest nudges that leave no two dials showing the same reading.

Examples

Example 1

Input: dials = [0, 0]
Output: 1

One of the two dials climbs to 1, which takes a single nudge.

Example 2

Input: dials = [2, 2, 2, 2, 2]
Output: 10

Leave one dial where it is and lift the other four to 3, 4, 5 and 6, costing 1 plus 2 plus 3 plus 4.

Example 3

Input: dials = [1, 2, 3, 4, 5]
Output: 0

No two dials already share a reading, so nothing has to move.

Constraints

1 <= dials.length <= 10^5
0 <= dials[i] <= 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 min_increment_for_unique(dials: list[int]) -> int:

Java

public int minIncrementForUnique(int[] dials)

The intended approach

Time O(n log n) for the sort plus one pass.Space O(n) for the sorted order.

Nudges only push readings upwards, so the dial showing the smallest reading never needs to move at all, and once it is settled the next-smallest only has to clear it. Sorting the readings turns the whole thing into one pass with a single number carried along: the lowest reading still free. Take the readings in order. A reading already above that mark is left alone and becomes the new mark. A reading at or below it has to climb to one above the mark, which costs the difference, and that lands it on the new mark. Every climb here is forced, because the readings below have already taken every value underneath, so no cheaper arrangement exists. Starting the mark below every possible reading keeps the very first dial free of charge.

Pythonentry point: min_increment_for_unique

def min_increment_for_unique(dials):
    order = sorted(dials)
    nudges = 0
    mark = -1
    for value in order:
        if value > mark:
            mark = value
        else:
            mark += 1
            nudges += mark - value
    return nudges

The obvious version

Tally how many dials sit on each reading and then sweep the readings upwards carrying the overflow. A reading holding several dials keeps one and hands the rest to the next reading up, and each handed-on dial costs one nudge per step it is carried. Costs the whole span of readings rather than a sort.

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_min_increment_for_unique(dials):
    tally = {}
    top = 0
    for value in dials:
        tally[value] = tally.get(value, 0) + 1
        if value > top:
            top = value
    nudges = 0
    carry = 0
    slot = 0
    limit = top + len(dials) + 1
    while slot <= limit:
        here = tally.get(slot, 0) + carry
        if here > 0:
            carry = here - 1
            nudges += carry
        else:
            carry = 0
        slot += 1
    return nudges

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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