1109EasyMathGreedy

The Best Single Swap on a Six-Nine Counter

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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 counter shows the number num, and every one of its digits is a 6 or a 9. At most one digit may be swapped, turning a 6 into a 9 or a 9 into a 6. Leaving the counter untouched is allowed.

Return the largest number the counter can be made to show.

Examples

Example 1

Input: num = 66
Output: 96

Lifting the leading digit gives 96, which beats lifting the second one.

Example 2

Input: num = 9696
Output: 9996

The leading digit is already a nine, so the earliest six is the second digit, and lifting it gives 9996.

Example 3

Input: num = 999
Output: 999

Every digit is already a nine, so the counter is best left untouched.

Constraints

1 <= num <= 10^4
every digit of num is a 6 or a 9

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 maximum69_number(num: int) -> int:

Java

public int maximum69Number(int num)

The intended approach

Time O(d) over the digits of num.Space O(d) for the digits.

Turning a `9` into a `6` only ever makes the number smaller, so the swap worth making is a `6` becoming a `9`. Every digit place is worth more than all the places to its right put together, so lifting the leftmost `6` beats lifting any `6` further along, whatever the rest of the digits look like. So read the digits from the front, change the first `6` found to a `9`, and stop. If no `6` is there at all, every digit is already a `9` and the counter is left as it stands, which the allowance for doing nothing permits.

Pythonentry point: maximum69_number

def maximum69_number(num):
    digits = list(str(num))
    for i in range(len(digits)):
        if digits[i] == "6":
            digits[i] = "9"
            break
    return int("".join(digits))

The obvious version

Try everything: for each digit place, write both a `6` and a `9` there in turn, read the result as a number, and keep the largest found alongside the untouched counter. Costs twice the digit count of whole numbers built and compared.

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_maximum69_number(num):
    text = str(num)
    best = num
    for i in range(len(text)):
        for ch in ("6", "9"):
            value = int(text[:i] + ch + text[i + 1:])
            if value > best:
                best = value
    return best

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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