Reverse Bits of a Number in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does it mean to reverse the bits of a number?

Reversing the bits means flipping the order of all bits in the binary form of the number, so the first bit becomes the last, the second becomes the second last, and so on.

beginner

How many bits are typically considered when reversing bits of an integer in C?

Usually 32 bits are considered for a standard unsigned int in C, but it depends on the system and the integer size.

beginner

What is the purpose of using bitwise AND (&) with 1 in bit reversal?

Bitwise AND with 1 extracts the least significant bit (rightmost bit) of the number, which is used to get each bit one by one during reversal.

intermediate

Explain the role of left shift (<<) in the bit reversal process.

Left shift moves bits to the left, making space for the next bit to be added at the right end of the reversed number.

intermediate

Why do we right shift (>>) the original number during bit reversal?

Right shifting the original number moves the next bit to the least significant position so it can be extracted and added to the reversed number.

What is the first step in reversing bits of a number?

A Right shift the reversed number

B Extract the least significant bit using & 1

C Left shift the number by 1

D Add 1 to the number

We start by extracting the least significant bit to add it to the reversed number.

How many times do you repeat the bit extraction and shifting to reverse a 32-bit number?

A 8 times

B 16 times

C 32 times

D 64 times

You repeat the process 32 times to cover all bits in a 32-bit number.

What does the left shift operator (<<) do in the reversed number during bit reversal?

A Moves bits to the right

B Resets bits

C Extracts bits

D Moves bits to the left

Left shift moves bits to the left to make space for the next bit.

Why do we right shift the original number during bit reversal?

A To discard the least significant bit after processing

B To add bits to the reversed number

C To reset the reversed number

D To multiply the number

Right shifting discards the bit already processed and brings the next bit to the least significant position.

If the input number is 13 (binary 00000000000000000000000000001101), what is the reversed 8-bit output?

A 10110000 (decimal 176)

B 10110000 (decimal 13)

C 00001101 (decimal 13)

D 11010000 (decimal 208)

Reversing bits of 00001101 gives 10110000 which is decimal 176.

Describe the step-by-step process to reverse the bits of a 32-bit unsigned integer.

Think about how you move bits from the original number to the reversed number one by one.

Explain why bitwise operations are efficient for reversing bits compared to string or array methods.

Consider how computers handle bits natively.

Practice

1. What does reversing the bits of a number mean in simple terms?

easy

A. Shifting bits to the left by one position

B. Changing all 0s to 1s and 1s to 0s

C. Flipping the order of bits from left to right

D. Adding 1 to the number

Solution

1. Step 1: Understand bit order

   Bits are arranged from left (most significant) to right (least significant).

2. Step 2: Reverse means flip order

   Reversing bits means the leftmost bit becomes rightmost, second leftmost becomes second rightmost, and so on.

3. Final Answer:

   Flipping the order of bits from left to right -> Option C

4. Quick Check:

   Reverse bits = flip order [OK]

Hint: Reverse bits means flip their order, not change values [OK]

Common Mistakes:

• Confusing reversing bits with bitwise NOT
• Thinking reversing bits adds or subtracts from number
• Mixing bit shifting with reversing bits

2. Which of the following C code snippets correctly reverses the bits of an 8-bit unsigned integer n?

easy

A. unsigned char reverseBits(unsigned char n) { return n << 1; }

B. unsigned char reverseBits(unsigned char n) { unsigned char rev = 0; for(int i=0; i<8; i++) { rev |= (n & 1) << i; n >>= 1; } return rev; }

C. unsigned char reverseBits(unsigned char n) { return ~n; }

D. unsigned char reverseBits(unsigned char n) { unsigned char rev = 0; for(int i=0; i<8; i++) { rev <<= 1; rev |= (n & 1); n >>= 1; } return rev; }

Solution

1. Step 1: Analyze the correct loop logic

   The loop shifts rev left by 1, ORs with (n & 1), then shifts n right by 1. This reverses bits correctly.

2. Step 2: Check distractor options

   rev |= (n & 1) << i; n >>= 1; copies bits in same order. return ~n; is bitwise NOT, not reverse. return n << 1; shifts bits left, not reverse.

3. Final Answer:

   rev <<= 1; rev |= (n & 1); n >>= 1; -> Option D

4. Quick Check:

   Shift rev left + add bit from n = reverse bits [OK]

Hint: Shift rev left, add n's LSB, shift n right each step [OK]

Common Mistakes:

• Using bitwise NOT (~) instead of reversing bits
• Not shifting rev before adding new bit
• Shifting bits incorrectly causing wrong order

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

unsigned char reverseBits(unsigned char n) {
    unsigned char rev = 0;
    for(int i = 0; i < 8; i++) {
        rev <<= 1;
        rev |= (n & 1);
        n >>= 1;
    }
    return rev;
}

int main() {
    unsigned char x = 13; // binary 00001101
    unsigned char y = reverseBits(x);
    printf("%u", y);
    return 0;
}

medium

A. 11

B. 176

C. 101

D. 13

Solution

1. Step 1: Convert 13 to binary and reverse bits

   13 in 8-bit binary is 00001101. Reversing bits gives 10110000.

2. Step 2: Convert reversed binary to decimal

   10110000 binary = 128 + 32 + 16 = 176 decimal.

3. Final Answer:

   176 -> Option B

4. Quick Check:

   Reverse 00001101 = 10110000 = 176 [OK]

Hint: Reverse bits visually, then convert binary to decimal [OK]

Common Mistakes:

• Confusing reversed bits with original number
• Calculating decimal incorrectly from binary
• Ignoring leading zeros in binary

4. Identify the error in the following C function that attempts to reverse bits of a 4-bit number:

unsigned char reverseBits4(unsigned char n) {
    unsigned char rev = 0;
    for(int i = 0; i < 4; i++) {
        rev |= (n & 1) << i;
        n >>= 1;
    }
    return rev;
}

medium

A. It does not shift rev left before adding bits, so bits are not reversed

B. It shifts n left instead of right

C. It uses 8-bit loop instead of 4-bit

D. It returns rev without masking bits

Solution

1. Step 1: Analyze bit addition logic

   The code adds bits to rev using rev |= (n & 1) << i, which places bits in original order, not reversed.

2. Step 2: Understand correct reversing method

   To reverse bits, rev should be shifted left before adding new bit, not ORed at position i.

3. Final Answer:

   It does not shift rev left before adding bits, so bits are not reversed -> Option A

4. Quick Check:

   Shift rev left before adding bit to reverse [OK]

Hint: Shift rev left before OR to reverse bits correctly [OK]

Common Mistakes:

• Adding bits at same positions without shifting rev
• Confusing left and right shifts on n
• Looping wrong number of times

5. You want to reverse the bits of a 16-bit unsigned integer and then check if the reversed number is a palindrome in binary (reads the same forwards and backwards). Which approach is correct?

hard

A. Reverse bits using a loop, then compare reversed number with original

B. Check if original number equals reversed number without reversing bits

C. Reverse bits using a loop, then reverse again and compare with original

D. Shift bits left by 8 and compare with original

Solution

1. Step 1: Compute reversed bits

   Reverse bits using a loop to obtain the reversed number rev.

2. Step 2: Check if reversed is palindrome

   The reversed number is a palindrome if it reads the same forwards and backwards, i.e., rev == original (since reverse(rev) == original always holds for a correct reverse).

3. Final Answer:

   Reverse bits using a loop, then compare reversed number with original -> Option A

4. Quick Check:

   rev == original means rev is palindrome [OK]

Hint: rev == original checks if reversed bits form a palindrome [OK]

Common Mistakes:

• Thinking double reverse checks palindrome (it only verifies the reverse function)
• Trying to check palindrome without computing reversed bits
• Confusing bit reversal with simple shifts