Check if Number is Even or Odd Using Bits in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a number is even using bitwise AND.

DSA C++

bool isEven(int num) {
    return (num & [1]) == 0;
}

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

A1

B0

C2

Dnum

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 as mask always returns true.

Using 2 checks the second bit, not the last bit.

The least significant bit (LSB) of an even number is 0. Using bitwise AND with 1 checks this bit.

2fill in blank

medium

Complete the code to print "Even" if number is even, else "Odd".

DSA C++

if ((num & [1]) == 0) {
    std::cout << "Even";
} else {
    std::cout << "Odd";
}

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

Anum

B0

C1

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 as mask causes condition to always be true.

Using 2 checks wrong bit.

Bitwise AND with 1 checks the last bit to determine even or odd.

3fill in blank

hard

Fix the error in the function that returns true if number is odd using bits.

DSA C++

bool isOdd(int num) {
    return (num & [1]) == 1;
}

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

Anum

B1

C2

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 as mask always returns false.

Using 2 checks wrong bit.

To check if a number is odd, bitwise AND with 1 should equal 1.

4fill in blank

hard

Fill both blanks to create a function that returns "Even" or "Odd" using bits.

DSA C++

std::string checkEvenOdd(int num) {
    if ((num [1] [2]) == 0) {
        return "Even";
    } else {
        return "Odd";
    }
}

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

A&

B|

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using bitwise OR (|) instead of AND (&).

Using 0 as mask causes condition to always be true.

Use bitwise AND (&) with 1 to check the least significant bit.

5fill in blank

hard

Fill all three blanks to create a concise function returning true if number is even using bits.

DSA C++

bool isEven(int num) {
    return (num [1] [2]) [3] 0;
}

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

A&

B1

C==

D|

Attempts:

3 left

💡 Hint

Common Mistakes

Using bitwise OR (|) instead of AND (&).

Using != instead of ==.

Use bitwise AND (&) with 1 and compare (==) to 0 to check if number is even.

Practice

1. Which bit operation can be used to check if a number is even or odd?

easy

A. Check the least significant bit using AND with 1

B. Shift the number right by 1

C. XOR the number with 1

D. Invert all bits of the number

Solution

1. Step 1: Understand the least significant bit (LSB)

   The LSB of a binary number determines if it is even or odd. If LSB is 0, number is even; if 1, number is odd.

2. Step 2: Use AND operation with 1 to check LSB

   Performing bitwise AND with 1 isolates the LSB. Result 0 means even, 1 means odd.

3. Final Answer:

   Check the least significant bit using AND with 1 -> Option A

4. Quick Check:

   LSB & 1 = 0 (even) or 1 (odd) [OK]

Hint: Use n & 1 to check even or odd quickly [OK]

Common Mistakes:

• Using XOR instead of AND to check LSB
• Checking the most significant bit instead
• Using bit shift without checking bits

2. Which of the following C++ code snippets correctly checks if num is odd using bits?

easy

A. if ((num & 1) == 0) { /* odd */ }

B. if ((num >> 1) == 1) { /* odd */ }

C. if ((num ^ 1) == 1) { /* odd */ }

D. if ((num & 1) == 1) { /* odd */ }

Solution

1. Step 1: Understand bitwise AND with 1

   Using (num & 1) extracts the least significant bit. If it equals 1, number is odd.

2. Step 2: Check condition for odd number

   The condition (num & 1) == 1 correctly identifies odd numbers.

3. Final Answer:

   if ((num & 1) == 1) { /* odd */ } -> Option D

4. Quick Check:

   (num & 1) == 1 means odd [OK]

Hint: Use (num & 1) == 1 to check odd numbers [OK]

Common Mistakes:

• Using == 0 to check odd (actually checks even)
• Using bit shift without comparison
• Using XOR incorrectly for odd check

3. What is the output of the following C++ code?

int num = 14;
if (num & 1) {
    std::cout << "Odd";
} else {
    std::cout << "Even";
}

medium

A. Odd

B. Even

C. 14

D. Error

Solution

1. Step 1: Evaluate bitwise AND of num and 1

   14 in binary is 1110. The least significant bit is 0. So, 14 & 1 = 0.

2. Step 2: Check the if condition

   Since (num & 1) is 0, the else block executes, printing "Even".

3. Final Answer:

   Even -> Option B

4. Quick Check:

   14 & 1 = 0 means Even [OK]

Hint: Check LSB: 0 means even, 1 means odd [OK]

Common Mistakes:

• Assuming 14 is odd because it ends with 4
• Confusing bitwise AND with arithmetic AND
• Printing num instead of "Even" or "Odd"

4. Identify the error in the following code snippet that checks if a number is odd:

int num = 7;
if (num & 1 == 1) {
    std::cout << "Odd";
} else {
    std::cout << "Even";
}

medium

A. Incorrect variable type for num

B. Using == instead of =

C. Missing parentheses around (num & 1)

D. No error, code is correct

Solution

1. Step 1: Analyze operator precedence in condition

   In C++, == has higher precedence than &. So, num & 1 == 1 is parsed as num & (1 == 1), which is num & 1. This is not the intended comparison.

2. Step 2: Identify the error

   Without parentheses, the expression does not explicitly compare (num & 1) to 1, which can cause confusion or unintended behavior. Parentheses are needed: (num & 1) == 1.

3. Final Answer:

   Missing parentheses around (num & 1) -> Option C

4. Quick Check:

   Operator precedence requires parentheses for clarity [OK]

Hint: Use parentheses: (num & 1) == 1 to avoid precedence issues [OK]

Common Mistakes:

• Misinterpreting operator precedence (== > &)
• Thinking parentheses are not needed
• Believing the code works correctly without parentheses

5. You want to write a function in C++ that returns true if a number is even and false if odd, using bitwise operations. Which of the following implementations is correct?

hard

A. bool isEven(int n) { return (n & 1) == 0; }

B. bool isEven(int n) { return (n & 1) == 1; }

C. bool isEven(int n) { return (n >> 1) == 0; }

D. bool isEven(int n) { return !(n ^ 1); }

Solution

1. Step 1: Understand even number bit pattern

   Even numbers have least significant bit 0. So (n & 1) == 0 means even.

2. Step 2: Check each function implementation

   bool isEven(int n) { return (n & 1) == 1; } returns true for odd numbers, so incorrect. bool isEven(int n) { return (n & 1) == 0; } correctly returns true for even numbers. Options C and D do not correctly check evenness.

3. Final Answer:

   bool isEven(int n) { return (n & 1) == 0; } -> Option A

4. Quick Check:

   Even if LSB is 0 [OK]

Hint: Return (n & 1) == 0 for even check [OK]

Common Mistakes:

• Returning true for odd numbers by mistake
• Using bit shift incorrectly for even check
• Using XOR without understanding its effect