Problem Description

You are given an array of positive integers nums. Your task is to create a new array answer that contains all the individual digits from each number in nums, preserving their original order.

When you separate a number into its digits, you extract each digit while maintaining their left-to-right sequence. For instance, if you have the number 10921, separating it gives you the array [1, 0, 9, 2, 1].

The process works as follows:

• Take each number from the input array nums
• Break it down into its individual digits
• Add these digits to the result array in the same order they appear
• Continue this for all numbers in nums

For example, if nums = [13, 25, 83], the output would be [1, 3, 2, 5, 8, 3] because:

• 13 becomes [1, 3]
• 25 becomes [2, 5]
• 83 becomes [8, 3]
• Combining them in order gives [1, 3, 2, 5, 8, 3]

The solution extracts digits from each number by repeatedly taking the remainder when dividing by 10 (which gives the rightmost digit) and then dividing by 10 to remove that digit. Since this process extracts digits from right to left, the digits are collected in a temporary list and then reversed before adding to the final answer.

Intuition

The key insight is that we need to extract individual digits from each number while preserving their order. The most straightforward way to extract digits from a number is through mathematical operations.

When we think about how to get individual digits from a number like 123, we can use the modulo operation. The expression 123 % 10 gives us 3 (the last digit), and 123 // 10 gives us 12 (removing the last digit). We can repeat this process: 12 % 10 gives 2, and 12 // 10 gives 1. Finally, 1 % 10 gives 1, and 1 // 10 gives 0, which tells us we're done.

However, this approach extracts digits from right to left (we get 3, then 2, then 1), but we need them in left-to-right order (1, 2, 3). This is why we collect the digits in a temporary list and reverse it before adding to our final answer.

The pattern becomes clear:

1. For each number, repeatedly use % 10 to get the rightmost digit
2. Use // 10 to remove that digit from the number
3. Continue until the number becomes 0
4. Since we collected digits in reverse order, flip them back to get the correct sequence
5. Add these digits to our result array

This mathematical approach is efficient and doesn't require converting numbers to strings, making it a clean solution for digit extraction.

Solution Approach

The solution implements a simulation approach where we process each number in the array and extract its digits one by one.

Here's how the algorithm works step by step:

1. Initialize the result array: We create an empty list ans that will store all the separated digits.

2. Iterate through each number: We loop through each integer x in the input array nums.

3. Extract digits from each number: For each number x, we:

   • Create a temporary list t to store the digits of the current number
   • Use a while loop that continues as long as x is not zero
   • In each iteration:
     • Extract the rightmost digit using x % 10 and append it to t
     • Remove the rightmost digit by integer division: x //= 10

4. Handle the digit order: Since the modulo operation extracts digits from right to left, but we need them from left to right, we reverse the temporary list using t[::-1].

5. Add to result: We extend the answer array with the reversed digits using ans.extend(t[::-1]).

Let's trace through an example with nums = [123, 45]:

• For 123:
  • 123 % 10 = 3, t = [3], x = 12
  • 12 % 10 = 2, t = [3, 2], x = 1
  • 1 % 10 = 1, t = [3, 2, 1], x = 0
  • Reverse t to get [1, 2, 3] and add to ans
• For 45:
  • 45 % 10 = 5, t = [5], x = 4
  • 4 % 10 = 4, t = [5, 4], x = 0
  • Reverse t to get [4, 5] and add to ans
• Final result: [1, 2, 3, 4, 5]

The time complexity is O(n * d) where n is the number of elements in nums and d is the average number of digits per number. The space complexity is O(d) for the temporary list used to store digits of each number.

Example Walkthrough

Let's walk through a small example with nums = [52, 7, 406] to illustrate the solution approach.

Step 1: Process the first number (52)

• Initialize temporary list t = []
• Extract digits from right to left:
  • 52 % 10 = 2, add to t: t = [2], update x = 52 // 10 = 5
  • 5 % 10 = 5, add to t: t = [2, 5], update x = 5 // 10 = 0
  • Loop ends since x = 0
• Reverse t to get correct order: [2, 5] → [5, 2]
• Add to result: ans = [5, 2]

Step 2: Process the second number (7)

• Initialize temporary list t = []
• Extract digits:
  • 7 % 10 = 7, add to t: t = [7], update x = 7 // 10 = 0
  • Loop ends since x = 0
• Reverse t: [7] → [7] (single digit stays the same)
• Add to result: ans = [5, 2, 7]

Step 3: Process the third number (406)

• Initialize temporary list t = []
• Extract digits from right to left:
  • 406 % 10 = 6, add to t: t = [6], update x = 406 // 10 = 40
  • 40 % 10 = 0, add to t: t = [6, 0], update x = 40 // 10 = 4
  • 4 % 10 = 4, add to t: t = [6, 0, 4], update x = 4 // 10 = 0
  • Loop ends since x = 0
• Reverse t to get correct order: [6, 0, 4] → [4, 0, 6]
• Add to result: ans = [5, 2, 7, 4, 0, 6]

Final Result: [5, 2, 7, 4, 0, 6]

The key insight is that we extract digits mathematically using modulo and integer division operations, which naturally gives us digits from right to left. By reversing each number's digits before adding them to our answer, we maintain the correct left-to-right order as required.

Solution Implementation

Time and Space Complexity

The time complexity is O(n × log₁₀ M), where n is the length of the array nums and M is the maximum value in the array nums. This is because:

• The outer loop iterates through all n elements in the array
• For each number x, the inner while loop executes log₁₀(x) times (the number of digits in x)
• In the worst case, when x = M, this becomes log₁₀ M iterations
• Therefore, the total time complexity is O(n × log₁₀ M)

The space complexity is O(n × log₁₀ M), where n is the length of the array nums and M is the maximum value in the array nums. This is because:

• The temporary list t stores at most log₁₀ M digits for each number
• The final answer list ans stores all separated digits from all numbers, which totals to O(n × log₁₀ M) in the worst case
• The space used by t is reused for each number, but the overall space requirement is dominated by the ans list

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

Common Pitfalls

1. Handling Zero Values

The most critical pitfall in this solution is that it fails to handle the number 0 correctly. When the input contains 0, the while loop condition while number > 0 immediately evaluates to false, causing the algorithm to skip the zero entirely instead of adding it as a digit.

Example of the bug:

• Input: nums = [30, 0, 105]
• Expected output: [3, 0, 0, 1, 0, 5]
• Actual buggy output: [3, 0, 1, 0, 5] (the standalone 0 is missing)

Solution: Check if the number is zero before entering the loop and handle it as a special case:

for number in nums:
    digits_stack = []
  
    # Special case: handle zero
    if number == 0:
        result.append(0)
        continue
  
    # Extract digits for non-zero numbers
    while number > 0:
        digit = number % 10
        digits_stack.append(digit)
        number //= 10
  
    result.extend(digits_stack[::-1])

2. Alternative Approach Pitfall: String Conversion

Many developers might think to use string conversion as a simpler approach:

for number in nums:
    result.extend([int(d) for d in str(number)])

While this works and is more concise, it has potential downsides:

• Performance overhead: String conversion and parsing can be slower for large numbers
• Memory overhead: Creates intermediate string objects
• Type conversion costs: Converting from int to string and back to int for each digit

However, for most practical cases, the string approach is actually cleaner and less error-prone than the mathematical approach, especially considering the zero-handling issue.

3. Integer Division Operator Confusion

Using regular division / instead of integer division // would cause type errors since regular division returns a float in Python 3:

# Wrong: number = number / 10  # Results in float
# Correct: number = number // 10  # Results in integer

This would lead to infinite loops or type errors when trying to use modulo on floats.