String to Integer atoi in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this atoi implementation with input " -42abc"?

Consider the following Java method that converts a string to an integer similar to atoi. What will be printed when calling myAtoi(" -42abc")?

DSA Java

public class Solution {
    public static int myAtoi(String s) {
        int i = 0, sign = 1, result = 0;
        int n = s.length();
        while (i < n && s.charAt(i) == ' ') i++;
        if (i < n && (s.charAt(i) == '+' || s.charAt(i) == '-')) {
            sign = (s.charAt(i) == '-') ? -1 : 1;
            i++;
        }
        while (i < n && Character.isDigit(s.charAt(i))) {
            int digit = s.charAt(i) - '0';
            if (result > (Integer.MAX_VALUE - digit) / 10) {
                return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
            }
            result = result * 10 + digit;
            i++;
        }
        return result * sign;
    }

    public static void main(String[] args) {
        System.out.println(myAtoi("  -42abc"));
    }
}

A -4200

B 0

C -42

D 42

💡 Hint

Focus on how the method skips spaces, reads the sign, and stops reading digits at the first non-digit character.

The method skips leading spaces, reads the '-' sign, then reads digits '4' and '2'. It stops at 'a' which is not a digit. So the result is -42.

❓ Predict Output

intermediate

What does this atoi variant return for input "4193 with words"?

Given this Java method similar to atoi, what will be the output when calling myAtoi("4193 with words")?

DSA Java

public class Solution {
    public static int myAtoi(String s) {
        int i = 0, sign = 1, result = 0;
        int n = s.length();
        while (i < n && s.charAt(i) == ' ') i++;
        if (i < n && (s.charAt(i) == '+' || s.charAt(i) == '-')) {
            sign = (s.charAt(i) == '-') ? -1 : 1;
            i++;
        }
        while (i < n && Character.isDigit(s.charAt(i))) {
            int digit = s.charAt(i) - '0';
            if (result > (Integer.MAX_VALUE - digit) / 10) {
                return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
            }
            result = result * 10 + digit;
            i++;
        }
        return result * sign;
    }

    public static void main(String[] args) {
        System.out.println(myAtoi("4193 with words"));
    }
}

A 4193

B 0

C 4193 with words

D -4193

💡 Hint

The method stops reading digits once it encounters a non-digit character.

The method reads digits '4', '1', '9', '3' and stops at the space after '3'. So it returns 4193.

❓ Predict Output

advanced

What is the output when input is "-91283472332" causing overflow?

What will the following Java atoi-like method return when called with myAtoi("-91283472332")?

DSA Java

public class Solution {
    public static int myAtoi(String s) {
        int i = 0, sign = 1, result = 0;
        int n = s.length();
        while (i < n && s.charAt(i) == ' ') i++;
        if (i < n && (s.charAt(i) == '+' || s.charAt(i) == '-')) {
            sign = (s.charAt(i) == '-') ? -1 : 1;
            i++;
        }
        while (i < n && Character.isDigit(s.charAt(i))) {
            int digit = s.charAt(i) - '0';
            if (result > (Integer.MAX_VALUE - digit) / 10) {
                return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
            }
            result = result * 10 + digit;
            i++;
        }
        return result * sign;
    }

    public static void main(String[] args) {
        System.out.println(myAtoi("-91283472332"));
    }
}

A 91283472332

B 0

C 2147483647

D -2147483648

💡 Hint

Check how the method handles overflow by returning min or max integer values.

The number is less than Integer.MIN_VALUE, so the method returns Integer.MIN_VALUE (-2147483648).

❓ Predict Output

advanced

What is the output for input "+-12" in this atoi implementation?

What will the following Java method print when called with myAtoi("+-12")?

DSA Java

public class Solution {
    public static int myAtoi(String s) {
        int i = 0, sign = 1, result = 0;
        int n = s.length();
        while (i < n && s.charAt(i) == ' ') i++;
        if (i < n && (s.charAt(i) == '+' || s.charAt(i) == '-')) {
            sign = (s.charAt(i) == '-') ? -1 : 1;
            i++;
        }
        while (i < n && Character.isDigit(s.charAt(i))) {
            int digit = s.charAt(i) - '0';
            if (result > (Integer.MAX_VALUE - digit) / 10) {
                return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
            }
            result = result * 10 + digit;
            i++;
        }
        return result * sign;
    }

    public static void main(String[] args) {
        System.out.println(myAtoi("+-12"));
    }
}

A 0

B -12

C 12

D Exception

💡 Hint

The method only accepts one optional sign character before digits.

The method reads '+' as sign, then sees '-' which is not a digit, so it stops parsing and returns 0.

🧠 Conceptual

expert

Which step is crucial to prevent integer overflow in atoi implementation?

In an atoi implementation converting string to 32-bit signed integer, which step is essential to avoid integer overflow during digit accumulation?

A Multiply result by 10 and add next_digit without any checks

B Check if current result > (Integer.MAX_VALUE - next_digit) / 10 before adding next_digit

C Convert the entire string to long first, then cast to int

D Ignore overflow and rely on Java's integer wrap-around behavior

💡 Hint

Think about how to detect if adding a digit will exceed the integer limit before it happens.

Checking if result > (Integer.MAX_VALUE - digit) / 10 ensures no overflow occurs when multiplying by 10 and adding digit.

Practice

1. What does the atoi function do when given a string input?

easy

A. Counts the number of characters in the string

B. Converts the string to a floating-point number

C. Reverses the string characters

D. Converts the string to an integer number

Solution

1. Step 1: Understand the purpose of atoi

   The atoi function stands for ASCII to integer and converts a string representing a number into an integer value.

2. Step 2: Compare options with function behavior

   Only Converts the string to an integer number matches the behavior of converting string to integer. Other options describe different operations.

3. Final Answer:

   Converts the string to an integer number -> Option D

4. Quick Check:

   atoi converts string to integer [OK]

Hint: Remember atoi means ASCII to integer conversion [OK]

Common Mistakes:

• Thinking atoi converts to float
• Confusing atoi with string length functions
• Assuming it reverses the string

2. Which of the following Java code snippets correctly converts a string s to an integer using Integer.parseInt?

easy

A. int num = parseInt.Integer(s);

B. int num = Integer.parseInt(s);

C. int num = Integer.valueOf(s).toInt();

D. int num = Integer.toString(s);

Solution

1. Step 1: Identify correct method syntax

   Java's correct method to convert string to int is Integer.parseInt(string).

2. Step 2: Check each option

   int num = Integer.parseInt(s); uses correct syntax. int num = Integer.toString(s); tries to convert int to string incorrectly. int num = Integer.valueOf(s).toInt(); uses invalid method chaining. int num = parseInt.Integer(s); has wrong method call order.

3. Final Answer:

   int num = Integer.parseInt(s); -> Option B

4. Quick Check:

   Correct Java syntax is Integer.parseInt(s) [OK]

Hint: Use Integer.parseInt(string) to convert string to int [OK]

Common Mistakes:

• Using Integer.toString instead of parseInt
• Incorrect method chaining like valueOf().toInt()
• Wrong method call order

3. What will be the output of the following Java code?

String s = "  -42abc";
int result = 0;
int i = 0;
boolean negative = false;

// Skip leading spaces
while (i < s.length() && s.charAt(i) == ' ') i++;

// Check sign
if (i < s.length() && (s.charAt(i) == '-' || s.charAt(i) == '+')) {
    negative = s.charAt(i) == '-';
    i++;
}

// Convert digits
while (i < s.length() && Character.isDigit(s.charAt(i))) {
    result = result * 10 + (s.charAt(i) - '0');
    i++;
}

if (negative) result = -result;
System.out.println(result);

medium

A. -42

B. 42

C. 0

D. NumberFormatException

Solution

1. Step 1: Skip spaces and detect sign

   The code skips leading spaces, finds '-', sets negative true, and moves index forward.

2. Step 2: Parse digits until non-digit

   It reads '4' and '2', converts to integer 42, stops at 'a'. Then applies negative sign.

3. Final Answer:

   -42 -> Option A

4. Quick Check:

   Parsed number with sign is -42 [OK]

Hint: Skip spaces, check sign, parse digits until non-digit [OK]

Common Mistakes:

• Not skipping spaces before parsing
• Ignoring sign character
• Parsing non-digit characters causing errors

4. The following code attempts to convert a string to integer but throws an exception. What is the main error?

String s = "123abc";
int num = Integer.parseInt(s);
System.out.println(num);

medium

A. parseInt method does not exist in Integer class

B. Missing import for Integer class

C. String contains non-digit characters causing NumberFormatException

D. Variable num is not declared

Solution

1. Step 1: Understand Integer.parseInt behavior

   Integer.parseInt expects a string with only digits (and optional sign). Non-digit characters cause NumberFormatException.

2. Step 2: Analyze input string

   String "123abc" contains letters after digits, causing the exception.

3. Final Answer:

   String contains non-digit characters causing NumberFormatException -> Option C

4. Quick Check:

   Non-digit chars cause NumberFormatException [OK]

Hint: parseInt fails if string has letters or symbols [OK]

Common Mistakes:

• Assuming parseInt ignores trailing letters
• Forgetting to handle exceptions
• Thinking parseInt needs imports

5. You want to implement a custom atoi function in Java that handles leading spaces, optional '+' or '-' sign, digits, and stops parsing at the first non-digit character. Which approach below correctly handles integer overflow during conversion?

hard

A. Use a long variable to accumulate and check if it exceeds Integer.MAX_VALUE or Integer.MIN_VALUE before casting

B. Ignore overflow and cast directly to int after parsing all digits

C. Use Integer.parseInt inside a try-catch block to catch overflow exceptions

D. Stop parsing digits once the number of digits exceeds 10

Solution

1. Step 1: Understand overflow risk in atoi

   Parsing digits can exceed int limits, causing incorrect results or errors.

2. Step 2: Use long to detect overflow before casting

   Accumulating in a long lets you check if value goes beyond int range, then clamp or handle accordingly.

3. Step 3: Evaluate other options

   Ignore overflow and cast directly to int after parsing all digits ignores overflow, causing wrong results. Use Integer.parseInt inside a try-catch block to catch overflow exceptions is invalid because parseInt throws NumberFormatException, not overflow exception. Stop parsing digits once the number of digits exceeds 10 is unreliable as digit count alone doesn't guarantee overflow.

4. Final Answer:

   Use a long variable to accumulate and check if it exceeds Integer.MAX_VALUE or Integer.MIN_VALUE before casting -> Option A

5. Quick Check:

   Use long to detect overflow before int cast [OK]

Hint: Use long to detect overflow before converting to int [OK]

Common Mistakes:

• Ignoring overflow and casting directly
• Relying on digit count only
• Misusing try-catch for overflow