Here is a step-by-step solution to the problem: 1. Initialize a variable `i` to 0. This variable will be used to iterate over the string `s`. 2. Start a while loop that continues until `i` is less than the length of `s` minus 1. This is because we are comparing each digit with its adjacent digit, so we don't need to check the last digit. 3. Inside the loop, check if the current digit and the next digit have the same parity. You can do this by converting each digit to an integer, taking the modulus 2 of each, and checking if they are equal. If they are, this means they are both even or both odd. 4. If the current digit is greater than the next digit, swap them. You can do this by slicing the string to get the part before the current digit, the next digit, the current digit, and the part after the next digit, and concatenating these parts together. 5. If a swap was made, break the loop. This is because we can only make at most one swap. 6. If no swap was made, increment `i` by 1 to move on to the next digit. 7. After the loop, return the string `s`. This will be the lexicographically smallest string that can be obtained after swapping adjacent digits with the same parity at most once. Here is a Python code snippet that implements this solution: ```python def smallestString(s): i = 0 while i s[i+1]: s = s[:i] + s[i+1] + s[i] + s[i+2:] break i += 1 return s ``` For example, if you call `smallestString("45320")`, it will return `"43520"`.

Question

Here is a step-by-step solution to the problem:

1. Initialize a variable `i` to 0. This variable will be used to iterate over the string `s`.

2. Start a while loop that continues until `i` is less than the length of `s` minus 1. This is because we are comparing each digit with its adjacent digit, so we don't need to check the last digit.

3. Inside the loop, check if the current digit and the next digit have the same parity. You can do this by converting each digit to an integer, taking the modulus 2 of each, and checking if they are equal. If they are, this means they are both even or both odd.

4. If the current digit is greater than the next digit, swap them. You can do this by slicing the string to get the part before the current digit, the next digit, the current digit, and the part after the next digit, and concatenating these parts together.

5. If a swap was made, break the loop. This is because we can only make at most one swap.

6. If no swap was made, increment `i` by 1 to move on to the next digit.

7. After the loop, return the string `s`. This will be the lexicographically smallest string that can be obtained after swapping adjacent digits with the same parity at most once.

Here is a Python code snippet that implements this solution:

```python
def smallestString(s):
    i = 0
    while i < len(s) - 1:
        if int(s[i]) % 2 == int(s[i+1]) % 2 and s[i] > s[i+1]:
            s = s[:i] + s[i+1] + s[i] + s[i+2:]
            break
        i += 1
    return s
```

For example, if you call `smallestString("45320")`, it will return `"43520"`.

Knowee AI · Accepted Answer