Check if Number is Even or Odd Using Bits in DSA C

Mental Model

Every number in computers is stored in bits, and the last bit tells if a number is even or odd. If the last bit is 0, the number is even; if 1, it is odd.

Analogy: Think of a row of light switches representing bits. The last switch controls whether the number is even or odd. If the last switch is off (0), the number is even; if on (1), the number is odd.

Number bits:  ... b3 b2 b1 b0
                  ↑
               last bit

Dry Run Walkthrough

Input: number = 5

Goal: Determine if 5 is even or odd using bit operations

Step 1: Perform bitwise AND of number 5 with 1

5 in bits: 0101
1 in bits:  0001
AND result:  0001

Why: AND with 1 isolates the last bit to check if it is 0 or 1

Step 2: Check if AND result is 0 or 1

AND result = 1

Why: If result is 1, number is odd; if 0, number is even

Result:

5 is odd because last bit is 1

Annotated Code

DSA C

#include <stdio.h>

// Function to check even or odd using bitwise AND
const char* checkEvenOdd(int num) {
    if ((num & 1) == 0) {
        return "Even";
    } else {
        return "Odd";
    }
}

int main() {
    int number = 5;
    printf("%d is %s\n", number, checkEvenOdd(number));
    return 0;
}

if ((num & 1) == 0) {

Check last bit by ANDing with 1 to determine even or odd

return "Even";

Return Even if last bit is 0

return "Odd";

Return Odd if last bit is 1

Output

5 is Odd

Complexity Analysis

Time: O(1) because bitwise AND is a single operation regardless of number size

Space: O(1) because no extra memory is used besides input and output

vs Alternative: Compared to using modulo (%) operator which also runs in O(1), bitwise AND is often faster and more direct at hardware level

Edge Cases

number = 0

Returns Even because 0's last bit is 0

DSA C

if ((num & 1) == 0) {

number = -2

Returns Even because bitwise AND checks last bit regardless of sign

DSA C

if ((num & 1) == 0) {

number = 1

Returns Odd because last bit is 1

DSA C

if ((num & 1) == 0) {

When to Use This Pattern

When you need to quickly check if a number is even or odd without division, use bitwise AND with 1 because it directly inspects the last bit.

Common Mistakes

Mistake: Using modulo operator (%) instead of bitwise AND for checking even/odd
Fix: Use (num & 1) == 0 for faster and simpler check

Mistake: Checking if (num & 1) == 1 to decide even, which is reversed
Fix: Remember (num & 1) == 0 means even, else odd

Summary

Checks if a number is even or odd by looking at its last bit using bitwise AND.

Use when you want a fast, simple way to test evenness without division or modulo.

The last bit of a number tells if it is even (0) or odd (1).

Practice

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

easy

A. Shift the number right by 1

B. Check the least significant bit using AND with 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: 0 means even, 1 means odd.

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

   Performing bitwise AND with 1 keeps only the LSB. If result is 0, number is even; if 1, odd.

3. Final Answer:

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

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 an integer n is odd using bitwise operators?

easy

A. if ((n & 1) == 1) printf("Odd");

B. if ((n >> 1) == 1) printf("Odd");

C. if ((n & 1) == 0) printf("Odd");

D. if ((n ^ 1) == 1) printf("Odd");

Solution

1. Step 1: Understand bitwise AND with 1

   Using n & 1 isolates the least significant bit. If it equals 1, the number is odd.

2. Step 2: Check condition for odd number

   The correct condition is (n & 1) == 1 to print "Odd".

3. Final Answer:

   if ((n & 1) == 1) printf("Odd"); -> Option A

4. Quick Check:

   n & 1 == 1 means odd [OK]

Hint: Use (n & 1) == 1 to detect odd numbers [OK]

Common Mistakes:

• Checking for == 0 to print odd (actually even)
• Using right shift instead of AND
• Using XOR incorrectly for odd check

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

int n = 14;
if (n & 1)
    printf("Odd");
else
    printf("Even");

medium

A. Odd

B. Compilation error

C. Even

D. 14

Solution

1. Step 1: Evaluate n & 1 for n = 14

   Binary of 14 is 1110. The least significant bit is 0. So, 14 & 1 = 0.

2. Step 2: Check the if condition

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

3. Final Answer:

   Even -> Option C

4. Quick Check:

   14 is even, so output is 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 logical AND
• Expecting output "14" instead of even/odd

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

int n = 7;
if (n & 1 == 1)
    printf("Odd");
else
    printf("Even");

medium

A. Missing parentheses around (n & 1)

B. Using == instead of = in if condition

C. Incorrect use of bitwise OR instead of AND

D. No error, code works fine

Solution

1. Step 1: Analyze operator precedence in condition

   In C, == has higher precedence than &. So expression is evaluated as n & (1 == 1).

2. Step 2: Evaluate the expression and fix

   Since (1 == 1) is true (1), expression becomes n & 1, which is correct but confusing. To avoid this, parentheses should be used: (n & 1) == 1.

3. Final Answer:

   Missing parentheses around (n & 1) -> Option A

4. Quick Check:

   Use (n & 1) == 1 to ensure correct evaluation [OK]

Hint: Use parentheses: (n & 1) == 1 to avoid precedence errors [OK]

Common Mistakes:

• Ignoring operator precedence causing wrong evaluation
• Using assignment = instead of comparison ==
• Confusing bitwise AND with logical AND

5. You want to write a function in C that returns 1 if a number is odd and 0 if even, using bitwise operators. Which of the following implementations is correct and efficient?

int isOdd(int n) {
    // Choose one
}

hard

A. return (n >> 1) & 1;

B. return !(n & 1);

C. return n ^ 1;

D. return (n & 1) == 1 ? 1 : 0;

Solution

1. Step 1: Understand the goal of the function

   The function should return 1 if n is odd, 0 if even, using bitwise operations.

2. Step 2: Evaluate each option

   return (n & 1) == 1 ? 1 : 0; correctly returns 1 if LSB is 1 (odd), else 0. return (n >> 1) & 1; returns the second least significant bit, incorrect. return n ^ 1; returns n XOR 1, which changes the number, not a boolean. return !(n & 1); returns 1 if even, opposite of requirement.

3. Final Answer:

   return (n & 1) == 1 ? 1 : 0; -> Option D

4. Quick Check:

   Use (n & 1) == 1 to detect odd and return 1 [OK]

Hint: Return (n & 1) == 1 for odd check function [OK]

Common Mistakes:

• Using bit shift instead of checking LSB
• Returning modified number instead of boolean
• Negating the condition incorrectly