String to Integer atoi in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to convert the first character of the string to an integer digit.

DSA Java

int digit = str.charAt(0) - '[1]';

Drag options to blanks, or click blank then click option'

A'9'

B'1'

C'a'

D'0'

Attempts:

3 left

💡 Hint

Common Mistakes

Using '1' instead of '0' causes wrong digit calculation.

Using character 'a' is incorrect for digits.

Subtracting the character '0' from a digit character converts it to its integer value.

2fill in blank

medium

Complete the code to check if the current character is a digit.

DSA Java

if (str.charAt(i) [1] '0' && str.charAt(i) <= '9') {

Drag options to blanks, or click blank then click option'

A==

B<=

C>=

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' instead of '>=' causes logic error.

Using '==' checks equality, not range.

The character must be greater than or equal to '0' and less than or equal to '9' to be a digit.

3fill in blank

hard

Fix the error in the code that updates the result integer with the new digit.

DSA Java

result = result [1] 10 + (str.charAt(i) - '0');

Drag options to blanks, or click blank then click option'

A+

B*

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '*' causes incorrect number formation.

Using '-' or '/' breaks the logic.

Multiply the current result by 10 before adding the new digit to shift digits left.

4fill in blank

hard

Fill both blanks to correctly handle the sign and update the index after reading it.

DSA Java

int sign = [1];
if (str.charAt(i) == '+' || str.charAt(i) == '-') {
    sign = (str.charAt(i) == '-') ? -1 : 1;
    i [2] 1;
}

Drag options to blanks, or click blank then click option'

A1

B-1

C+=

D-=

Attempts:

3 left

💡 Hint

Common Mistakes

Setting sign to -1 by default is wrong.

Using '-=' decreases index incorrectly.

Initialize sign as 1 (positive). Increment index i by 1 after reading sign character.

5fill in blank

hard

Fill all three blanks to complete the loop that parses digits and stops on non-digit.

DSA Java

while (i < str.length() && str.charAt(i) [1] '0' && str.charAt(i) [2] '9') {
    result = result [3] 10 + (str.charAt(i) - '0');
    i++;
}

Drag options to blanks, or click blank then click option'

A>=

B<=

C*

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '*' for result update.

Using wrong comparison operators.

The loop continues while character is between '0' and '9'. Multiply result by 10 before 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