Reverse Bits of a Number in DSA Javascript - 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 changes as the number size grows.

How does the number of steps grow when we reverse bits of bigger numbers?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function reverseBits(n) {
  let result = 0;
  for (let i = 0; i < 32; i++) {
    result <<= 1;
    result |= (n & 1);
    n >>= 1;
  }
  return result;
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A for-loop running 32 times, processing one bit each time.
• How many times: Exactly 32 times, once for each bit in the number.

How Execution Grows With Input

Since the loop runs a fixed 32 times, the steps do not grow with the number's value.

Input Size (n)	Approx. Operations
10 (small number)	32 steps
100 (larger number)	32 steps
1000 (even larger)	32 steps

Pattern observation: The number of steps stays the same regardless of input size because the bit count is fixed.

Final Time Complexity

Time Complexity: O(1)

This means the time to reverse bits is constant and does not increase with the input number.

Common Mistake

[X] Wrong: "Reversing bits takes longer for bigger numbers because they have more digits."

[OK] Correct: The number of bits is fixed (32 bits here), so the steps are always the same no matter the number's size.

Interview Connect

Understanding fixed-time operations like bit manipulation shows you can reason about low-level efficiency, a useful skill in many coding challenges.

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. Flipping the order of all bits from left to right

B. Changing all 0 bits to 1 bits

C. Adding 1 to the number

D. Multiplying the number by 2

Solution

1. Step 1: Understand bits as binary digits

   Bits are the 0s and 1s that make up a number in binary form.

2. Step 2: Define reversing bits

   Reversing bits means reading the bits from right to left instead of left to right.

3. Final Answer:

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

4. Quick Check:

   Reverse bits = flip order [OK]

Hint: Think of bits like beads on a string flipped end to end [OK]

Common Mistakes:

• Confusing reversing bits with flipping bits (0 to 1)
• Thinking reversing bits changes the number by adding or multiplying
• Assuming reversing bits means reversing digits in decimal

2. Which of the following JavaScript code snippets correctly reverses the bits of an 8-bit number n?

easy

A. let result = 0; for(let i=0; i<8; i++) { result += (n & 1) << i; n <<= 1; }

B. let result = 0; for(let i=0; i<8; i++) { result |= (n & 1) << i; n >>= 1; }

C. let result = 0; for(let i=0; i<8; i++) { result <<= 1; result |= (n & 1); n >>= 1; }

D. let result = 0; for(let i=7; i>0; i--) { result |= (n & 1) << i; n >>= 1; }

Solution

1. Step 1: Analyze the loop and bit operations

   let result = 0; for(let i=0; i<8; i++) { result <<= 1; result |= (n & 1); n >>= 1; } shifts result left, adds least significant bit of n, then shifts n right. This reverses bits correctly.

2. Step 2: Check other options for errors

   let result = 0; for(let i=0; i<8; i++) { result |= (n & 1) << i; n >>= 1; } does not reverse bits, it copies bits in same order. let result = 0; for(let i=7; i>0; i--) { result |= (n & 1) << i; n >>= 1; } shifts bits incorrectly. let result = 0; for(let i=0; i<8; i++) { result += (n & 1) << i; n <<= 1; } shifts n left, which is wrong.

3. Final Answer:

   let result = 0; for(let i=0; i<8; i++) { result <<= 1; result |= (n & 1); n >>= 1; } -> Option C

4. Quick Check:

   Shift result left + add bit from n right [OK]

Hint: Shift result left, add n's right bit, shift n right [OK]

Common Mistakes:

• Shifting n left instead of right
• Not shifting result before adding bit
• Using wrong loop direction or bit positions

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

function reverseBits(n) {
  let result = 0;
  for(let i = 0; i < 4; i++) {
    result <<= 1;
    result |= (n & 1);
    n >>= 1;
  }
  return result;
}

console.log(reverseBits(13));

medium

A. 14

B. 11

C. 7

D. 13

Solution

1. Step 1: Convert 13 to 4-bit binary

   13 in binary (4 bits) is 1101.

2. Step 2: Reverse bits step-by-step

   Loop 4 times: result builds as follows:
   i=0: result=0<<1=0 + (13&1=1) =1
   n=13>>1=6 (0110)
   i=1: result=1<<1=2 + (6&1=0) =2
   n=6>>1=3 (0011)
   i=2: result=2<<1=4 + (3&1=1) =5
   n=3>>1=1 (0001)
   i=3: result=5<<1=10 + (1&1=1) =11

3. Final Answer:

   11 -> Option B

4. Quick Check:

   Reverse 1101 = 1011 (binary 11 decimal) [OK]

Hint: Trace bit shifts and OR operations carefully [OK]

Common Mistakes:

• Confusing decimal and binary values
• Not shifting result before OR
• Looping wrong number of times

4. Find the error in this bit reversal function and choose the corrected version:

function reverseBits(n) {
  let result = 0;
  for(let i = 0; i < 8; i++) {
    result |= (n & 1);
    result <<= 1;
    n >>= 1;
  }
  return result;
}

medium

A. Change loop to run 16 times

B. Shift n left instead of right

C. Initialize result with 1 instead of 0

D. Swap the order: shift result left before OR operation

Solution

1. Step 1: Identify order of operations

   Current code ORs bit then shifts result left, causing last bit to be lost.

2. Step 2: Correct by shifting result left before OR

   Shift result left first, then OR with current bit to build reversed bits correctly.

3. Final Answer:

   Swap the order: shift result left before OR operation -> Option D

4. Quick Check:

   Shift before OR to keep all bits [OK]

Hint: Always shift result left before adding new bit [OK]

Common Mistakes:

• Shifting result after OR loses last bit
• Shifting n left instead of right
• Wrong loop count for bit size

5. You want to reverse the bits of a 16-bit number 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 check if reversed number equals its own reverse

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

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

D. Reverse bits using a loop, then convert to decimal and check if even

Solution

1. Step 1: Reverse bits of the number

   Use a loop to reverse all 16 bits correctly.

2. Step 2: Check palindrome by comparing reversed number with its own reverse

   If reversed number equals its reverse, it is a binary palindrome.

3. Final Answer:

   Reverse bits using a loop, then check if reversed number equals its own reverse -> Option A

4. Quick Check:

   Palindrome check = reversed == reverse(reversed) [OK]

Hint: Check palindrome by comparing reversed number to its reverse [OK]

Common Mistakes:

• Comparing reversed number directly to original (wrong for palindrome)
• Checking decimal evenness instead of binary palindrome
• Reversing bits twice unnecessarily