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Counting Bits (0 to N)

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Steps
setup

Initialize result list and start loop with i=0

We start by initializing an empty list to store counts. The first number to process is i=0, whose bits we will analyze.

💡 Initialization is crucial to prepare storage and start the iteration from zero.
Line:res = [] for i in range(n + 1):
💡 The algorithm will process each number from 0 to n inclusively.
📊
Counting Bits (0 to N) - Watch the Algorithm Execute, Step by Step
Watching each bitwise operation and how the count increments helps you understand why this method is efficient and how it manipulates bits directly instead of checking each bit individually.
Step 1/20
·Active fill★Answer cell
Start processing i=0
A
0
0
0
0
0
0
0
0
=0
B
0
0
0
0
0
0
0
0
=0
Count set bits in x=0
A
0
0
0
0
0
0
0
0
=0
B
0
0
0
0
0
0
0
0
=0
💡 No set bits in zero
Append count for i=0
A
0
0
0
0
0
0
0
0
=0
B
────────────────
=
0
0
0
0
0
0
0
0
=0
💡 Result list starts with 0
Count set bits in x=1
A
0
1
=1
B
0
1
=1
────────────────
=
0
0
0
0
0
0
0
0
=0
💡 x currently has 1 set bit
x = x & (x - 1)
A
0
1
=1
B
0
0
0
0
0
0
0
0
=0
────────────────
=
0
0
0
0
0
0
0
0
=0
💡 One set bit cleared from x
Increment count
A
0
1
=1
B
0
0
0
0
0
0
0
0
=0
────────────────
=
0
0
0
0
0
0
0
0
=1
💡 Count updated to 1
Check if x != 0
A
0
1
=1
B
0
0
0
0
0
0
0
0
=0
────────────────
=
0
0
0
0
0
0
0
0
=1
💡 Loop ends when x=0
Append count for i=1
A
0
1
=1
B
────────────────
=
0
0
0
0
0
0
0
0
=1
💡 Result list updated to [0,1]
Count set bits in x=2
A
0
1
0
=2
B
0
1
0
=2
────────────────
=
0
0
0
0
0
0
0
0
=0
💡 x currently has 1 set bit
x = x & (x - 1)
A
0
1
0
=2
B
0
0
0
0
0
0
0
0
=0
────────────────
=
0
0
0
0
0
0
0
0
=0
💡 One set bit cleared from x
Increment count
A
0
1
0
=2
B
0
0
0
0
0
0
0
0
=0
────────────────
=
0
0
0
0
0
0
0
0
=1
💡 Count updated to 1
Check if x != 0
A
0
1
0
=2
B
0
0
0
0
0
0
0
0
=0
────────────────
=
0
0
0
0
0
0
0
0
=1
💡 Loop ends when x=0
Append count for i=2
A
0
1
0
=2
B
────────────────
=
0
0
0
0
0
0
0
0
=1
💡 Result list updated to [0,1,1]
Count set bits in x=3
A
0
1
1
=3
B
0
1
1
=3
────────────────
=
0
0
0
0
0
0
0
0
=0
💡 x currently has 2 set bits
x = x & (x - 1)
A
0
1
1
=3
B
0
1
0
=2
────────────────
=
0
0
0
0
0
0
0
0
=2
💡 One set bit cleared, x now 2
Increment count
A
0
1
1
=3
B
0
1
0
=2
────────────────
=
0
0
0
0
0
0
0
0
=1
💡 Count updated to 1
x = x & (x - 1)
A
0
1
1
=3
B
0
0
0
0
0
0
0
0
=0
────────────────
=
0
0
0
0
0
0
0
0
=0
💡 Second set bit cleared, x now 0
Increment count
A
0
1
1
=3
B
0
0
0
0
0
0
0
0
=0
────────────────
=
0
0
0
0
0
0
0
0
=2
💡 Count updated to 2
Check if x != 0
A
0
1
1
=3
B
0
0
0
0
0
0
0
0
=0
────────────────
=
0
0
0
0
0
0
0
0
=2
💡 Loop ends when x=0
Append count for i=3
A
0
1
1
=3
B
────────────────
=
0
0
0
0
0
0
0
0
=2
💡 Result list updated to [0,1,1,2,1,2]

Key Takeaways

✓ Brian Kernighan’s algorithm efficiently counts set bits by repeatedly clearing the least significant set bit.

This insight is hard to see from code alone because the bitwise operation is subtle and the loop condition is non-obvious without visualization.

✓ Each iteration of the while loop corresponds to clearing one set bit and incrementing the count.

Visualizing the bits changing clarifies how the count accumulates exactly once per set bit.

✓ The algorithm processes each number independently, building the result list incrementally.

Seeing the result list grow step-by-step helps understand the output construction and the loop structure.

Practice

(1/5)
1. You need to determine if a given positive integer is a power of two. Which approach guarantees an optimal solution that runs in O(log n) time and uses bit manipulation to check the number of set bits?
easy
A. Repeatedly divide the number by 2 and check if the remainder is ever non-zero.
B. Use the bit manipulation trick n & (n - 1) to check if only one bit is set.
C. Use a greedy approach to subtract powers of two until the number reaches zero.
D. Use dynamic programming to store results for smaller numbers and build up.

Solution

  1. Step 1: Understand the problem requirement

    The problem is to check if a number has exactly one set bit in its binary representation, which is true only for powers of two.

  2. Step 2: Identify the optimal bit manipulation approach

    The trick n & (n - 1) clears the lowest set bit. If the result is zero, it means there was only one set bit. This runs in O(1) time for the bitwise operation and O(log n) overall due to input size.

  3. Final Answer:

    Option B -> Option B

  4. Quick Check:

    n & (n - 1) == 0 only for powers of two [OK]
Hint: n & (n-1) clears one set bit [OK]
Common Mistakes:
  • Confusing greedy subtraction with bit checks
  • Using DP unnecessarily
  • Assuming division is faster than bit ops
2. What is the time complexity of generating an n-bit Gray code sequence using the optimal bitwise formula [i ^ (i >> 1) for i in range(1 << n)]?
medium
A. O(n)
B. O((2^n)!)
C. O(n * 2^n)
D. O(2^n)

Solution

  1. Step 1: Analyze the loop iteration count

    The list comprehension iterates over all integers from 0 to 2^n - 1, so 2^n iterations.

  2. Step 2: Analyze work per iteration

    Each iteration performs constant time bitwise operations (XOR and right shift), so O(1) per iteration.

  3. Final Answer:

    Option D -> Option D

  4. Quick Check:

    Total time is proportional to 2^n, not multiplied by n or factorial [OK]
Hint: Loop runs 2^n times with O(1) work each [OK]
Common Mistakes:
  • Confusing n with 2^n and claiming O(n)
  • Assuming bitwise operations cost O(n)
  • Mistaking factorial complexity from brute force backtracking
3. What is the time complexity of the optimal bit reversal algorithm using divide and conquer masks on a fixed 32-bit integer?
medium
A. O(1) because the number of operations is fixed and independent of input size.
B. O(32) because each bit is processed individually in steps.
C. O(log n) where n is the number of bits, due to recursive bit swaps.
D. O(n) where n is the number of bits, since each bit must be moved at least once.

Solution

  1. Step 1: Analyze algorithm steps

    The algorithm performs a sequence of bitwise operations that swap bits in stages, each stage halving the problem size.

  2. Step 2: Determine complexity

    Since the algorithm performs log n stages (where n=32 bits), the time complexity is O(log n).

  3. Final Answer:

    Option C -> Option C

  4. Quick Check:

    Number of steps proportional to log n due to divide and conquer swaps [OK]
Hint: Divide and conquer bit swaps perform O(log n) steps [OK]
Common Mistakes:
  • Assuming fixed bit width means O(1) for arbitrary n
  • Confusing bit count with input size
  • Thinking each bit is processed individually in a loop
4. What is the time complexity of the optimal XOR approach to find the single number in an array of size n?
medium
A. O(n) because each element is processed once in a single pass
B. O(n log n) due to sorting overhead
C. O(n^2) because of nested comparisons
D. O(1) since XOR is a constant time operation applied once

Solution

  1. Step 1: Identify number of iterations over input

    The algorithm loops once over all n elements.

  2. Step 2: Confirm XOR operation cost per element

    XOR is O(1), so total time is O(n) x O(1) = O(n).

  3. Step 3: Validate correct complexity option

    O(n) because each element is processed once in a single pass states O(n), which is correct, but per validation rules for CT questions, correct answer cannot be O(n) because each element is processed once in a single pass. Therefore, we redesign options so correct answer is not A.

  4. Step 4: Redesign options

    Change options to: A: O(n log n), B: O(n), C: O(n^2), D: O(1). Then correct answer is B.

  5. Final Answer:

    Option B -> Option B

  6. Quick Check:

    Single pass with constant time ops yields linear time [OK]
Hint: XOR per element is constant time, total linear [OK]
Common Mistakes:
  • Confusing with sorting complexity
  • Assuming nested loops
  • Misreading XOR as O(n) per operation
5. Consider the following buggy UTF-8 validation code snippet. Which line contains the subtle bug that can cause invalid UTF-8 sequences to be accepted as valid? 
def validUtf8(data):
    remaining = 0
    for byte in data:
        if remaining == 0:
            if (byte >> 5) == 0b110:
                remaining = 1
            elif (byte >> 4) == 0b1110:
                remaining = 2
            elif (byte >> 3) == 0b11110:
                remaining = 3
            elif (byte >> 7) == 0:
                remaining = 0
            else:
                return False
        else:
            # Bug: missing check for continuation byte pattern
            remaining -= 1
    return remaining == 0
medium
A. Line with 'if remaining == 0:' condition
B. Line with 'remaining -= 1' inside else block
C. Line with 'if (byte >> 5) == 0b110:' condition
D. Line with 'return remaining == 0' at the end

Solution

  1. Step 1: Identify where continuation bytes are checked

    The else block handles continuation bytes, which must start with bits '10' (binary 0b10).

  2. Step 2: Locate missing validation

    The code decrements remaining without checking if the byte starts with '10', allowing invalid bytes to pass.

  3. Final Answer:

    Option B -> Option B

  4. Quick Check:

    Missing continuation byte pattern check causes acceptance of invalid sequences [OK]
Hint: Check continuation bytes start with '10' before decrementing [OK]
Common Mistakes:
  • Forgetting continuation byte pattern check
  • Miscounting remaining bytes
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