1177EasySortingHeap (Priority Queue)

Swapping Digits That Share Their Oddness

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Trains the technique from

LeetCode 2231Largest Number After Digit Swaps by Parity

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. Any two of its digits may be swapped, as often as you like, provided the two are both odd or both even.

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

Examples

Example 1

Input: num = 1357
Output: 7531

Every digit is odd, so they can all be swapped with each other and falling order is largest.

Example 2

Input: num = 2468
Output: 8642

Every digit is even, so again they arrange into falling order.

Example 3

Input: num = 1122
Output: 1122

The two odd digits are both 1 and the two even digits are both 2, so no swap changes anything.

Constraints

1 <= num <= 10^9

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 largest_integer(num: int) -> int:

Java

public int largestInteger(int num)

The intended approach

Time O(d log d) over the digits.Space O(d) for the two groups.

Swapping pairs freely within a group lets the digits of that group be arranged in any order at all, since any rearrangement is a run of pairwise swaps. So the odd digits can be shuffled among the places that currently hold odd digits, and the even digits among the places holding even digits, but no digit ever crosses between the two. That means each place's oddness is fixed for good, and the two groups are settled quite separately. To make the number largest, the biggest digits should land in the earliest places available to them, so hand out the odd digits in falling order to the odd places as they come, and likewise for the even digits. Walking the original digits once and drawing from whichever group matches keeps the places in order and needs only the two groups sorted.

Pythonentry point: largest_integer

def largest_integer(num):
    digits = list(str(num))
    odds = []
    evens = []
    for ch in digits:
        if (ord(ch) - 48) % 2:
            odds.append(ch)
        else:
            evens.append(ch)
    odds.sort()
    evens.sort()
    out = []
    for ch in digits:
        if (ord(ch) - 48) % 2:
            out.append(odds.pop())
        else:
            out.append(evens.pop())
    return int("".join(out))

The obvious version

List every arrangement of the odd digits among the odd places and every arrangement of the even digits among the even places, build the number each pairing gives and keep the largest. Costs the factorials of the two group sizes multiplied together.

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_largest_integer(num):
    from itertools import permutations
    digits = list(str(num))
    odd_spots = []
    even_spots = []
    for i in range(len(digits)):
        if (ord(digits[i]) - 48) % 2:
            odd_spots.append(i)
        else:
            even_spots.append(i)
    best = num
    odd_pool = [digits[i] for i in odd_spots]
    even_pool = [digits[i] for i in even_spots]
    for odd_order in permutations(odd_pool):
        for even_order in permutations(even_pool):
            body = [""] * len(digits)
            for k in range(len(odd_spots)):
                body[odd_spots[k]] = odd_order[k]
            for k in range(len(even_spots)):
                body[even_spots[k]] = even_order[k]
            value = int("".join(body))
            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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