Reverse Bits of a Number in DSA Go - Time & Space Complexity

Time Complexity: Reverse Bits of a Number

O(1)

Understanding Time Complexity

We want to understand how the time needed to reverse bits in a number changes as the number size changes.

Specifically, how does the number of steps grow when we reverse bits one by one?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func reverseBits(num uint32) uint32 {
    var result uint32 = 0
    for i := 0; i < 32; i++ {
        result <<= 1
        result |= num & 1
        num >>= 1
    }
    return result
}
    

This code reverses the bits of a 32-bit unsigned integer by shifting and masking each bit.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop runs 32 times, once for each bit.
• How many times: Exactly 32 times, fixed for 32-bit numbers.

How Execution Grows With Input

Since the loop runs a fixed 32 times regardless of the number's value, the operations stay the same.

Input Size (bits)	Approx. Operations
10	32 (fixed)
100	32 (fixed)
1000	32 (fixed)

Pattern observation: The number of steps does not grow with input size because the bit length is fixed.

Final Time Complexity

Time Complexity: O(1)

This means the time to reverse bits stays constant no matter what number you use, because the bit count is fixed.

Common Mistake

[X] Wrong: "The time grows with the value of the number because bigger numbers have more bits to reverse."

[OK] Correct: The number of bits is fixed (32 bits here), so the loop always runs the same number of times regardless of the number's size.

Interview Connect

Understanding fixed-size operations like bit reversal helps you explain why some algorithms run in constant time, which is a valuable skill in interviews.

Self-Check

"What if we changed the input to a 64-bit number? How would the time complexity change?"

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. Flipping the order of bits from left to right

C. Adding 1 to the number

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

Solution

1. Step 1: Understand bit representation

   Numbers are stored as bits (0s and 1s) in a sequence from left (most significant) to right (least significant).

2. Step 2: Define reversing bits

   Reversing bits means taking the bit sequence and flipping it so the first bit becomes last, second becomes second last, and so on.

3. Final Answer:

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

4. Quick Check:

   Reverse bits = flip order [OK]

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

Common Mistakes:

• Confusing reversing bits with bitwise NOT (flip 0s and 1s)
• Thinking reversing bits means adding or subtracting
• Mixing reversing bits with shifting bits

2. Which Go code snippet correctly reverses bits of a 8-bit unsigned integer n?

easy

A. var rev uint8 = 0
for i := 0; i < 8; i++ {
rev = rev >> 1
rev |= n & 1
n <<= 1
}

B. var rev uint8 = 0
for i := 0; i < 8; i++ {
rev |= (n & 1) << i
n >>= 1
}

C. var rev uint8 = 0
for i := 0; i < 8; i++ {
rev += n & 1
n <<= 1
}

D. var rev uint8 = 0
for i := 0; i < 8; i++ {
rev <<= 1
rev |= n & 1
n >>= 1
}

Solution

1. Step 1: Analyze bit reversal logic

   We shift rev left to make space, add the least significant bit of n, then shift n right to process next bit.

2. Step 2: Check code correctness

   var rev uint8 = 0 for i := 0; i < 8; i++ { rev <<= 1 rev |= n & 1 n >>= 1 } correctly shifts rev left, ORs with n's last bit, then shifts n right. Others either shift wrong direction or add bits incorrectly.

3. Final Answer:

   The code in A correctly reverses bits -> Option D

4. Quick Check:

   Shift rev left + add n's LSB + shift n right = correct [OK]

Hint: Shift rev left, add n's last bit, shift n right each loop [OK]

Common Mistakes:

• Shifting rev right instead of left
• Shifting n left instead of right
• Adding bits without shifting rev

3. What is the output of this Go code snippet?

func reverseBits(n uint8) uint8 {
  var rev uint8 = 0
  for i := 0; i < 8; i++ {
    rev <<= 1
    rev |= n & 1
    n >>= 1
  }
  return rev
}

fmt.Println(reverseBits(13))

Note: 13 in binary is 00001101.

medium

A. 176

B. 11

C. 101

D. 208

Solution

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

   13 in 8 bits: 00001101. Reversing bits: 10110000.

2. Step 2: Convert reversed bits back to decimal

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

3. Final Answer:

   176 -> Option A

4. Quick Check:

   Reverse(00001101) = 10110000 = 176 [OK]

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

Common Mistakes:

• Confusing bit order and reversing incorrectly
• Calculating decimal from wrong binary
• Ignoring leading zeros in 8-bit representation

4. Find the error in this Go function that tries to reverse bits of a 4-bit number:

func reverse4Bits(n uint8) uint8 {
  var rev uint8 = 0
  for i := 0; i < 4; i++ {
    rev |= (n & 1) << i
    n >>= 1
  }
  return rev
}

medium

A. The shift operator should be >> instead of <<

B. The loop should run 8 times for 4-bit numbers

C. The bits are not reversed; they are copied in the same order

D. Variable rev should be int, not uint8

Solution

1. Step 1: Understand the code logic

   The code shifts bits of n right, then sets rev bits by shifting left by i, which copies bits in same order.

2. Step 2: Identify why bits are not reversed

   Reversing bits requires shifting rev left each time and adding n's last bit, not shifting by i.

3. Final Answer:

   The bits are not reversed; they are copied in the same order -> Option C

4. Quick Check:

   Shifting rev by i copies bits, not reverses [OK]

Hint: Shift rev left each loop to reverse bits, not shift by i [OK]

Common Mistakes:

• Confusing bit copying with reversing
• Wrong loop count for bit size
• Incorrect variable types for bits

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

hard

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

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

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

D. Shift bits left and right alternately without reversing

Solution

1. Step 1: Reverse bits of the number

   Use a loop to reverse all 16 bits to get reversed number.

2. Step 2: Check if reversed number is a palindrome

   To verify if reversed reads the same forwards and backwards, compute reverse(reversed) and check if it equals reversed. Since reverse(reversed) = original, this is equivalent to comparing reversed with original.

3. Final Answer:

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

4. Quick Check:

   reversed == original iff reversed is palindrome [OK]

Hint: Reverse once, then compare reversed with original [OK]

Common Mistakes:

• Double reversing and comparing to original (always true, doesn't check palindrome)
• Trying to check equality without computing the reverse
• Using shifting without proper bit reversal