**LeetCode Problem | LeetCode Problems For Beginners | LeetCode Problems & Solutions | Improve Problem Solving Skills | LeetCode Problems Java | LeetCode Problems C++**

Hello **Programmers/Coders,** Today we are going to share ** solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python**. At Each Problem with Successful submission with

**all Test Cases Passed,**you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the **String to Integer (atoi)** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

Use “Ctrl+F” To Find Any Questions Answer. & For Mobile User, You Just Need To Click On Three dots In Your Browser & You Will Get A “Find” Option There. Use These Option to Get Any Random Questions Answer.

**About LeetCode**

*LeetCode* is one of the most well-known online judge platforms to help you enhance your skills, expand your knowledge and prepare for technical interviews.

LeetCode is for **software engineers who are looking to practice technical questions and advance their skills**. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either *Easy*, *Medium*, or *Hard*.

**LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:**

- Mathematics/Basic Logical Based Questions
- Arrays
- Strings
- Hash Table
- Dynamic Programming
- Stack & Queue
- Trees & Graphs
- Greedy Algorithms
- Breadth-First Search
- Depth-First Search
- Sorting & Searching
- BST (Binary Search Tree)
- Database
- Linked List
- Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

** Link for the Problem** – String to Integer (atoi)– LeetCode Problem

String to Integer (atoi)– LeetCode Problem

**Problem:**

Implement the `myAtoi(string s)`

function, which converts a string to a 32-bit signed integer (similar to C/C++’s `atoi`

function).

The algorithm for `myAtoi(string s)`

is as follows:

- Read in and ignore any leading whitespace.
- Check if the next character (if not already at the end of the string) is
`'-'`

or`'+'`

. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present. - Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored.
- Convert these digits into an integer (i.e.
`"123" -> 123`

,`"0032" -> 32`

). If no digits were read, then the integer is`0`

. Change the sign as necessary (from step 2). - If the integer is out of the 32-bit signed integer range
`[-2`

, then clamp the integer so that it remains in the range. Specifically, integers less than^{31}, 2^{31}- 1]`-2`

should be clamped to^{31}`-2`

, and integers greater than^{31}`2`

should be clamped to^{31}- 1`2`

.^{31}- 1 - Return the integer as the final result.

**Note:**

- Only the space character
`' '`

is considered a whitespace character. **Do not ignore**any characters other than the leading whitespace or the rest of the string after the digits.

**Example 1:**

Input:s = "42"Output:42Explanation:The underlined characters are what is read in, the caret is the current reader position. Step 1: "42" (no characters read because there is no leading whitespace) ^ Step 2: "42" (no characters read because there is neither a '-' nor '+') ^ Step 3: "42" ("42" is read in) ^ The parsed integer is 42. Since 42 is in the range [-2^{31}, 2^{31}- 1], the final result is 42.

**Example 2:**

Input:s = " -42"Output:-42Explanation:Step 1: " -42" (leading whitespace is read and ignored) ^ Step 2: " -42" ('-' is read, so the result should be negative) ^ Step 3: " -42" ("42" is read in) ^ The parsed integer is -42. Since -42 is in the range [-2^{31}, 2^{31}- 1], the final result is -42.

**Example 3:**

Input:s = "4193 with words"Output:4193Explanation:Step 1: "4193 with words" (no characters read because there is no leading whitespace) ^ Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+') ^ Step 3: "4193 with words" ("4193" is read in; reading stops because the next character is a non-digit) ^ The parsed integer is 4193. Since 4193 is in the range [-2^{31}, 2^{31}- 1], the final result is 4193.

**Constraints:**

`0 <= s.length <= 200`

`s`

consists of English letters (lower-case and upper-case), digits (`0-9`

),`' '`

,`'+'`

,`'-'`

, and`'.'`

.

String to Integer (atoi)– LeetCode Solutions

class Solution { public: int myAtoi(string s) { trim(s); if (s.empty()) return 0; const int sign = s[0] == '-' ? -1 : 1; if (s[0] == '+' || s[0] == '-') s = s.substr(1); long num = 0; for (const char c : s) { if (!isdigit(c)) break; num = num * 10 + (c - '0'); if (sign * num < INT_MIN) return INT_MIN; if (sign * num > INT_MAX) return INT_MAX; } return sign * num; } private: void trim(string& s) { s.erase(0, s.find_first_not_of(' ')); s.erase(s.find_last_not_of(' ') + 1); } };

class Solution { public int myAtoi(String s) { s = s.strip(); if (s.isEmpty()) return 0; final int sign = s.charAt(0) == '-' ? -1 : 1; if (s.charAt(0) == '+' || s.charAt(0) == '-') s = s.substring(1); long num = 0; for (final char c : s.toCharArray()) { if (!Character.isDigit(c)) break; num = num * 10 + (c - '0'); if (sign * num <= Integer.MIN_VALUE) return Integer.MIN_VALUE; if (sign * num >= Integer.MAX_VALUE) return Integer.MAX_VALUE; } return sign * (int) num; } }

class Solution: def myAtoi(self, s: str) -> int: s = s.strip() if not s: return 0 sign = -1 if s[0] == '-' else 1 if s[0] in {'-', '+'}: s = s[1:] num = 0 for c in s: if not c.isdigit(): break num = num * 10 + ord(c) - ord('0') if sign * num <= -2**31: return -2**31 if sign * num >= 2**31 - 1: return 2**31 - 1 return sign * num