LeetCode 2553 - Separate the Digits in an Array
This problem asks us to take an array of positive integers, nums, and transform it into another array, answer, where each element of nums is broken down into its constituent digits in order.
Difficulty: 🟢 Easy
Topics: Array, Simulation
Solution
Problem Understanding
This problem asks us to take an array of positive integers, nums, and transform it into another array, answer, where each element of nums is broken down into its constituent digits in order. The key requirement is that the order of the digits must reflect the original numbers’ order in the array. For example, if a number is 10921, the digits should appear as [1,0,9,2,1]. The input array can contain up to 1000 integers, each with a value up to 105, meaning no number exceeds 6 digits. The output should be a single flattened list containing all digits.
The problem guarantees that nums is non-empty and all values are positive integers. Edge cases include numbers with a single digit, very large numbers with the maximum number of digits, and arrays with only one number. These are important because naive approaches may fail to handle single-digit numbers or might incorrectly reverse the digits of larger numbers.
Approaches
A straightforward approach is to iterate over each integer in nums, convert it to a string, then iterate over the string to convert each character back to an integer. This works correctly because string conversion preserves digit order, and iterating over each character ensures we get all digits. The approach is efficient enough for the given constraints because even the largest numbers contain at most 6 digits.
A more optimal approach, which avoids converting to strings, uses mathematical operations. For each number, we can extract digits by repeatedly dividing by powers of 10. However, because we want to preserve the original digit order, this method requires temporarily storing digits in reverse order and then reversing them before appending to the result array. While this avoids string conversion, it adds complexity without significant performance gain for the given constraints.
| Approach | Time Complexity | Space Complexity | Notes |
|---|---|---|---|
| Brute Force (String Conversion) | O(n * k) | O(n * k) | n = number of integers, k = max digits per integer; simple, easy to implement |
| Optimal (Mathematical Extraction) | O(n * k) | O(n * k) | Slightly more complex; avoids string conversion but requires reversing digits |
Algorithm Walkthrough
- Initialize an empty list called
answerto store the resulting digits. - Iterate over each number in
nums. - Convert the current number to a string to get its digits in order.
- Iterate over each character in the string, convert it to an integer, and append it to
answer. - After processing all numbers, return
answer.
Why it works: The string conversion guarantees that digits are extracted in the same order they appear in the number. Appending digits sequentially for each number preserves the overall order in the array. The algorithm’s invariant is that each number’s digits are fully processed before moving to the next number, ensuring the final array accurately represents the flattened digit sequence.
Python Solution
from typing import List
class Solution:
def separateDigits(self, nums: List[int]) -> List[int]:
answer = []
for num in nums:
for digit in str(num):
answer.append(int(digit))
return answer
In this implementation, we first create an empty list answer. For each number in nums, we convert it to a string so that each digit is a separate character. We then iterate over these characters, convert each back to an integer, and append it to answer. Finally, we return the fully constructed list. This directly mirrors the algorithm steps and ensures correct order preservation.
Go Solution
func separateDigits(nums []int) []int {
var answer []int
for _, num := range nums {
digits := []int{}
n := num
for n > 0 {
digits = append([]int{n % 10}, digits...)
n /= 10
}
if num == 0 { // edge case for zero
digits = append(digits, 0)
}
answer = append(answer, digits...)
}
return answer
}
In Go, we handle digit extraction mathematically. For each number, we repeatedly take the remainder modulo 10 to extract the least significant digit and prepend it to a temporary slice digits to preserve order. After all digits of the number are extracted, we append digits to answer. This approach avoids string conversion but requires careful handling of slices to maintain order.
Worked Examples
Example 1: nums = [13,25,83,77]
| Step | num | digits extracted | answer |
|---|---|---|---|
| 1 | 13 | [1,3] | [1,3] |
| 2 | 25 | [2,5] | [1,3,2,5] |
| 3 | 83 | [8,3] | [1,3,2,5,8,3] |
| 4 | 77 | [7,7] | [1,3,2,5,8,3,7,7] |
Example 2: nums = [7,1,3,9]
| Step | num | digits extracted | answer |
|---|---|---|---|
| 1 | 7 | [7] | [7] |
| 2 | 1 | [1] | [7,1] |
| 3 | 3 | [3] | [7,1,3] |
| 4 | 9 | [9] | [7,1,3,9] |
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(n * k) | For each of the n numbers, we process up to k digits, where k ≤ 6 |
| Space | O(n * k) | The result array stores all digits of all numbers, up to n * k elements |
The reasoning is that converting numbers to strings or extracting digits mathematically both require processing each digit individually, making the algorithm linear in the total number of digits.
Test Cases
# Provided examples
assert Solution().separateDigits([13,25,83,77]) == [1,3,2,5,8,3,7,7] # multiple-digit numbers
assert Solution().separateDigits([7,1,3,9]) == [7,1,3,9] # single-digit numbers
# Boundary cases
assert Solution().separateDigits([1]) == [1] # single element
assert Solution().separateDigits([105]) == [1,0,5] # largest 3-digit number
assert Solution().separateDigits([100000]) == [1,0,0,0,0,0] # largest 6-digit number
# Mixed sizes
assert Solution().separateDigits([9,100,25,7]) == [9,1,0,0,2,5,7]
| Test | Why |
|---|---|
| [13,25,83,77] | Validates multiple-digit numbers and order preservation |
| [7,1,3,9] | Single-digit numbers should remain unchanged |
| [1] | Minimal input size |
| [105] | Three-digit number extraction |
| [100000] | Maximum allowed number size |
| [9,100,25,7] | Mixed sizes validate correct flattening |
Edge Cases
One important edge case is when the number is a single digit, which should be returned as is. Another edge case is the maximum allowed number, 100000, which has six digits and tests that the algorithm handles large numbers correctly. A third edge case is when the array contains both single-digit and multi-digit numbers, which ensures the algorithm correctly concatenates digits while maintaining order. In all these scenarios, the string conversion or careful mathematical extraction handles the digits correctly without loss or reversal, guaranteeing accurate output.