0363MediumMathGreedy

One Wheel Exchange

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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 mechanical counter shows the value num on a row of digit wheels.

Maintenance is allowed one adjustment: pick two wheels and exchange the digits sitting in them. Making no exchange at all is also allowed.

Return the largest value the row can show once the adjustment is done. The row never gains or loses wheels, and the row is read as an ordinary number, so any leading zeros it ends up showing contribute nothing to the value.

Examples

Example 1

Input: num = 1993
Output: 9913

Exchanging the 1 in the first wheel with the 9 in the third wheel leaves the row reading 9913.

Example 2

Input: num = 87654
Output: 87654

Every exchange available here gives a value below 87654, so the row is left as it is.

Example 3

Input: num = 4020
Output: 4200

Exchanging the 0 in the second wheel with the 2 in the third wheel leaves the row reading 4200.

Example 4

Input: num = 0
Output: 0

A single wheel has nothing to exchange with, so the reading stays 0.

Constraints

0 <= num <= 10^8

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 maximum_swap(num: int) -> int:

Java

public int maximumSwap(int num)

The intended approach

Time O(d) where d is the number of digits, at most 9Space O(d) for the digit list and the ten-entry position table

Record the last slot in which each digit value appears. Then walk the wheels from the front and, at the first slot that some larger digit value still occupies further right, bring in the largest such value from the last slot that holds it and stop. Exchanging into the earliest possible slot outweighs anything gained further right, and taking the last copy of the chosen value moves the smaller digit as far right as it goes.

Pythonentry point: maximum_swap

def maximum_swap(num):
    # d <= 9 digits. One right-to-left pass records, for each digit value,
    # the last slot it appears in; one left-to-right pass finds the exchange.
    # O(d) time, O(d) space for the digit list.
    digits = list(str(num))
    last = {}
    for i, ch in enumerate(digits):
        last[ch] = i
    for i, ch in enumerate(digits):
        for value in "987654321":
            if value <= ch:
                break
            if value in last and last[value] > i:
                j = last[value]
                digits[i], digits[j] = digits[j], digits[i]
                return int("".join(digits))
    return num

The obvious version

Try every pair of wheels, read off the value each exchange produces, and keep the largest along with the untouched reading.

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_maximum_swap(num):
    digits = list(str(num))
    best = num
    for i in range(len(digits)):
        for j in range(i + 1, len(digits)):
            trial = digits[:]
            trial[i], trial[j] = trial[j], trial[i]
            value = int("".join(trial))
            if value > best:
                best = value
    return best

How this problem was checked

Reference solution executed against 13 test cases — 13 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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