Problem Description

You are given an integer num. The task is to rearrange the digits of num in such a way that the resulting number is the smallest possible value and doesn't have any leading zeros. Importantly, you must maintain the sign of the original number; if the input number is negative, the result should also be negative, and the same for a positive number. The requirement is to return this minimum possible rearreganged number.

Intuition

To solve this problem, we can utilize a counting sort-like approach since we're dealing with individual digits which are only from 0 to 9. Here’s the intuition broken down into steps:

1. Handling Zero: If the number is 0, return it directly since 0 cannot be rearranged to get a smaller number.

2. Digit Counting: Count how many times each digit from 0 to 9 appears in the number using a list where the index corresponds to the digit and the value at that index represents the count.

3. Handle Negative Numbers: If the number is negative, to get the smallest possible number after rearrangement, we should start the number with the largest digit available and proceed to the smallest. Since leading zeros are not allowed, we only arrange the nonzero digits.

4. Handle Positive Numbers: a. If there are any zeros (as recorded in the digit counting step), make sure to place a nonzero digit at the beginning to avoid leading zeros. This is achieved by looking for the first nonzero count from 1 to 9, adding it to the answer string, and decreasing its count. b. After placing the first nonzero digit, append the remaining digits starting from 0 to 9 to ensure a minimal value.

5. Returning the Result: Combine the arranged digits into a single integer, preserving the sign, and return this integer as the answer.

This approach effectively sorts the digits to form the smallest (or largest if negative) possible number without using any extra sorting function, and the counting steps ensure that no leading zeros will appear in the final arrangement.

Learn more about Math and Sorting patterns.

Solution Approach

The solution follows a direct approach, implementing the intuition we described earlier. Here's how the code breaks down the problem and applies an algorithm to solve it:

1. Initialization and Handling of Zero:

   • If num is 0, we return 0 immediately as we cannot rearrange it for a smaller value.
   • Initialize a counter array cnt with 10 zeroes, which will serve to count the occurrences of each digit (0-9).
   • Record the sign of num in a boolean neg and use abs(num) to work with the absolute value, simplifying our logic for rearrangement.

2. Counting Digits:

   • A while loop is used to extract each digit from num using divmod(num, 10), which provides quotient and remainder - the latter being the digit we're counting.
   • Increment the count of the extracted digit in the cnt array.

3. Constructing the Smallest/Largest Number Based on Sign:

   • Initialize an empty string ans to build the final digit string.
   • If the number is negative (neg is True), iterate from 9 down to 0:
     • For each digit i, if the count cnt[i] is non-zero, append the digit i converted to a string, multiplied by its count, to ans. This assembles the largest possible number from the digits.
   • If the number is positive (neg is False):
     • First, find and place a non-zero digit if there are any zeros to avoid leading zeros. Iterate from 1 to 9 and on finding a non-zero count, append that digit to ans and decrement its count.
     • Next, append the rest of the digits in ascending order to build the smallest possible number. Iterate from 0 to 9 and for each digit i, if the count cnt[i] is non-zero, append the digit multiplied by its count to ans.

4. Finalizing the Answer:

   • Since ans is a string, it is converted back to an integer. The sign is reintroduced by negating the result if the original number was negative.

In summary, the code uses the counting sort technique to tally the occurrences of each digit and then rearranges the digits according to whether the number is positive or negative, to return the smallest possible value without leading zeros. It leverages simple array manipulation and string concatenation to avoid more complex sorting algorithms or additional data structures.

Example Walkthrough

Consider the integer num = 310. We need to find the smallest possible value by rearranging num's digits without leading zeros.

1. Handling Zero:

   • num is not 0, so we proceed with the solution.

2. Digit Counting:

   • We count the occurrences of each digit:
     • 3 appears once.
     • 1 appears once.
     • 0 appears once.
   • The counter array cnt is thus [1, 1, 0, 1, 0, 0, 0, 0, 0, 0].

3. Handle Positive Numbers:

   • Since num is positive, we follow the steps for positive numbers. a) We pick the smallest nonzero digit to avoid a leading zero. This is the digit 1.
   • We place 1 at the beginning, and the cnt array updates to [1, 0, 0, 1, 0, 0, 0, 0, 0, 0]. b) Now we append the rest of the digits sorted in ascending order. The next smallest digit is 0, followed by 3.
   • We append these to get the number 103.

4. Returning the Result:

   • Now, we combine the digits to form the integer 103, which is the smallest variation of 310 with no leading zero.

On the other hand, consider the integer num = -310. For negative numbers, the smallest value is achieved by arranging digits in reverse order, from largest to smallest.

1. Handling Zero:

   • num is not 0, so we proceed with the solution.

2. Digit Counting:

   • The occurrences of each digit are the same as above, and the counter array cnt doesn't change.

3. Handle Negative Numbers:

   • Since num is negative, the largest digit, which is 3, goes first, followed by 1, then 0.
   • We get the negative number -310 since rearranging the digits in descending order gives the same number. We don't need to change anything for this specific example, as it is already in the required form.

Thus, by following the counting sort-like approach, we can efficiently determine the smallest possible rearrangement of num while preserving the original sign and avoiding any leading zeros.

Solution Implementation

Time and Space Complexity

The time complexity of the provided solution is O(n + 10) which simplifies to O(n) where n represents the number of digits in the integer num. This is because we first iterate through all digits of the input num to count them (which contributes to O(n)), and then we iterate through the count array of 10 digits (which contributes to O(10) but constants are omitted in Big O notation).

The space complexity of the solution is O(10), which simplifies to O(1) - constant space complexity. This is because the count array cnt has a fixed size of 10, regardless of the size of num. The string ans also doesn't count towards space complexity in this analysis as it simply corresponds to the size of the output, which is not considered additional space used by the algorithm.

Learn more about how to find time and space complexity quickly using problem constraints.