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DSA Javaprogramming~10 mins

Reverse Bits of a Number in DSA Java - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to shift bits to the right by one position.

DSA Java
int shifted = n [1] 1;
Drag options to blanks, or click blank then click option'
A>>
B<<
C&
D|
Attempts:
3 left
💡 Hint
Common Mistakes
Using left shift '<<' instead of right shift '>>'.
Using bitwise AND '&' instead of shift.
2fill in blank
medium

Complete the code to extract the least significant bit of n.

DSA Java
int bit = n [1] 1;
Drag options to blanks, or click blank then click option'
A|
B^
C&
D>>
Attempts:
3 left
💡 Hint
Common Mistakes
Using OR '|' instead of AND '&'.
Using XOR '^' which toggles bits.
3fill in blank
hard

Fix the error in the code to add the extracted bit to the reversed number.

DSA Java
reversed = (reversed [1] 1) [2] bit;
Drag options to blanks, or click blank then click option'
A<<
B>>
C|
D&
Attempts:
3 left
💡 Hint
Common Mistakes
Using right shift '>>' instead of left shift '<<'.
Using AND '&' instead of OR '|'.
4fill in blank
hard

Fill both blanks to complete the loop that reverses bits of n.

DSA Java
for (int i = 0; i < [1]; i++) {
    int bit = n [2] 1;
    n = n >>> 1;
    reversed = (reversed << 1) | bit;
}
Drag options to blanks, or click blank then click option'
A32
B16
C&
D|
Attempts:
3 left
💡 Hint
Common Mistakes
Looping 16 times instead of 32.
Using '|' instead of '&' to extract bit.
5fill in blank
hard

Fill all three blanks to complete the reverseBits method.

DSA Java
public int reverseBits(int n) {
    int reversed = 0;
    for (int i = 0; i < [1]; i++) {
        int bit = n [2] 1;
        n = n >>> 1;
        reversed = (reversed [3] 1) | bit;
    }
    return reversed;
}
Drag options to blanks, or click blank then click option'
A32
B&
C<<
D>>
Attempts:
3 left
💡 Hint
Common Mistakes
Using '>>' instead of '<<' to shift reversed.
Looping fewer than 32 times.
Using '|' instead of '&' to extract bit.

Practice

(1/5)
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 bits from 0 to 1

Solution

  1. Step 1: Understand bit order

    Bits are the 0s and 1s that make up a number. Reversing means flipping their order.
  2. Step 2: Explain reversing bits

    If the bits are 1010, reversing them gives 0101, flipping left to right.
  3. Final Answer:

    Flipping the order of bits from left to right -> Option B
  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 means adding or subtracting
  • Mixing bit shifting with bit reversing
2. Which Java syntax correctly reverses bits of an integer n using a loop?
easy
A. for (int i = 0; i < 32; i++) { result <<= 1; result |= (n & 1); n >>= 1; }
B. for (int i = 0; i < 32; i++) { result |= (n & 1) << i; n >>= 1; }
C. for (int i = 0; i < 32; i++) { result += (n & 1) << i; n <<= 1; }
D. for (int i = 0; i < 32; i++) { result <<= 1; result |= (n >> i) & 1; }

Solution

  1. Step 1: Analyze bit extraction and shifting

    for (int i = 0; i < 32; i++) { result <<= 1; result |= (n & 1); n >>= 1; } extracts the least significant bit (n & 1), shifts result left, then adds the bit, and shifts n right.
  2. Step 2: Check correctness of loop operations

    for (int i = 0; i < 32; i++) { result <<= 1; result |= (n & 1); n >>= 1; } correctly reverses bits by building result from left to right while moving through n's bits right to left.
  3. Final Answer:

    for (int i = 0; i < 32; i++) { result <<= 1; result |= (n & 1); n >>= 1; } -> Option A
  4. Quick Check:

    Shift result left, add bit from n, shift n right [OK]
Hint: Shift result left before adding extracted bit [OK]
Common Mistakes:
  • Shifting n left instead of right
  • Adding bits without shifting result
  • Using incorrect bit extraction
3. What is the output of this Java code snippet?
int n = 13; // binary 00000000000000000000000000001101
int result = 0;
for (int i = 0; i < 4; i++) {
  result <<= 1;
  result |= (n & 1);
  n >>= 1;
}
System.out.println(result);
medium
A. 11
B. 13
C. 7
D. 14

Solution

  1. Step 1: Trace each loop iteration

    Initial n=13 (binary 1101). Loop runs 4 times:
    i=0: result=0<<1=0; result|=1 (n&1=1); result=1; n=6
    i=1: result=1<<1=2; result|=0 (6&1=0); result=2; n=3
    i=2: result=2<<1=4; result|=1 (3&1=1); result=5; n=1
    i=3: result=5<<1=10; result|=1 (1&1=1); result=11; n=0
  2. Step 2: Convert final result to decimal

    Result is 11 decimal after reversing 4 bits of 13.
  3. Final Answer:

    11 -> Option A
  4. Quick Check:

    Reversed 1101 bits = 1011 = 11 decimal [OK]
Hint: Reverse bits step-by-step, track result and n carefully [OK]
Common Mistakes:
  • Not limiting loop to 4 bits
  • Mixing bit order in result
  • Confusing decimal and binary values
4. Find the error in this Java code that tries to reverse bits of a 8-bit number n:
int result = 0;
for (int i = 0; i <= 8; i++) {
  result <<= 1;
  result |= (n & 1);
  n >>= 1;
}
medium
A. Result should be shifted right, not left
B. Bitwise OR operator is used incorrectly
C. Loop runs one extra time causing wrong result
D. Variable n should not be shifted inside the loop

Solution

  1. Step 1: Check loop range for 8 bits

    The loop runs from i=0 to i=8 inclusive, which is 9 iterations, but 8 bits need only 8 iterations.
  2. Step 2: Understand effect of extra iteration

    Extra iteration shifts result one more time and adds an extra bit, causing incorrect reversed value.
  3. Final Answer:

    Loop runs one extra time causing wrong result -> Option C
  4. Quick Check:

    Loop should run 8 times, not 9 [OK]
Hint: Use < 8, not <= 8 for 8-bit loops [OK]
Common Mistakes:
  • Using <= instead of < in loop condition
  • Shifting result right instead of left
  • Not shifting n inside loop
5. Given a 16-bit unsigned integer, how would you reverse its bits and return the result in Java? Which approach is best?
hard
A. Shift n left 16 times and OR with result
B. Convert number to binary string, reverse string, parse back to int
C. Use recursion to reverse bits one by one without loops
D. Use a loop shifting result left and extracting bits from n right, running 16 times

Solution

  1. Step 1: Consider efficiency and correctness

    Looping 16 times, shifting result left and extracting bits from n right is efficient and direct for 16-bit reversal.
  2. Step 2: Evaluate other options

    Converting to string is slower and error-prone; recursion adds overhead; shifting n left 16 times loses bits.
  3. Final Answer:

    Use a loop shifting result left and extracting bits from n right, running 16 times -> Option D
  4. Quick Check:

    Loop + bit shifts = best for fixed bit reversal [OK]
Hint: Loop 16 times, shift result left, add n's right bit [OK]
Common Mistakes:
  • Using string conversion for bit operations
  • Recursion without base case or efficiency
  • Shifting n left losing original bits
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