How do you solve String to Integer (atoi) on LeetCode?

String to Integer (atoi) (LeetCode #8) can be solved using Brute Force or Optimal Solution (Single Pass). The optimal approach is Optimal Solution (Single Pass) with O(n) time complexity and O(1) space complexity. Key insight: The key observation is that we can process the string in a single pass, which reduces unnecessary checks and improves efficiency.

What is the brute force solution for String to Integer (atoi)?

The brute force approach for String to Integer (atoi) is Brute Force, with O(n) time complexity and O(1) space complexity. The brute-force approach involves parsing the string character by character, checking for whitespace, sign, and digits. We will build the integer value step by step, but this method can be inefficient due to repeated checks. The optimal Optimal Solution (Single Pass) approach improves this to O(n).

What algorithm or data structure is used to solve String to Integer (atoi)?

String to Integer (atoi) uses the following concepts: String. The recommended patterns to study are: Hash Map, Array, Two Pointers.

What are the interview tips for String to Integer (atoi)?

When you first see this problem, clarify the input constraints and expected output format. Communicate your approach clearly, breaking down the steps you plan to take to convert the string to an integer. Mention edge cases such as empty strings, strings with only whitespace, or strings that contain non-digit characters.

What are common mistakes when solving String to Integer (atoi)?

Students often forget to handle leading whitespace correctly, leading to incorrect results. Another common mistake is not checking for overflow when calculating the result, which can lead to incorrect outputs.

String to Integer (atoi)

Medium

StringHash MapArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution (Single Pass)★
Time	O(n)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution still follows the same logic but is implemented in a single pass through the string. We carefully manage the integer conversion and handle overflow without needing to check conditions multiple times.

⚙️

Algorithm

5 steps

1Step 1: Initialize an index, result, and sign.
2Step 2: Skip leading whitespace.
3Step 3: Determine the sign of the number.
4Step 4: Convert characters to digits while checking for overflow in a single loop.
5Step 5: Return the final result after applying the sign.

solution.py29 lines

1# Full working Python code with comments
2
3def myAtoi(s: str) -> int:
4    index = 0
5    result = 0
6    sign = 1
7    INT_MAX = 2**31 - 1
8    INT_MIN = -2**31
9
10    # Step 2: Skip leading whitespace
11    while index < len(s) and s[index] == ' ':
12        index += 1
13
14    # Step 3: Check for sign
15    if index < len(s) and (s[index] == '-' or s[index] == '+'):
16        sign = -1 if s[index] == '-' else 1
17        index += 1
18
19    # Step 4: Read digits and check for overflow
20    while index < len(s) and s[index].isdigit():
21        digit = int(s[index])
22        # Check for overflow
23        if result > (INT_MAX - digit) // 10:
24            return INT_MAX if sign == 1 else INT_MIN
25        result = result * 10 + digit
26        index += 1
27
28    return result * sign
29

ℹ

Complexity note: The time complexity remains O(n) as we still traverse the string once. The space complexity is O(1) since we only use a few variables for computation.

1The key observation is that we can process the string in a single pass, which reduces unnecessary checks and improves efficiency.
2Understanding how to handle overflow is crucial, as it helps avoid errors when converting large numbers.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.