Bit Manipulation (Easy)Master bitwise operations including AND, OR, XOR, and bit shifting.

Check if K-th Bit is Set

Problem Statement

Given a number n, check if the k-th bit (0-indexed) is set (1) or not (0).

Solution

def is_kth_bit_set(n, k):
    return (n & (1 << k)) != 0

Set K-th Bit

Problem Statement

Given a number n, set the k-th bit to 1.

Solution

def set_kth_bit(n, k):
    return n | (1 << k)

Clear K-th Bit

Problem Statement

Given a number n, clear the k-th bit (set it to 0).

Solution

def clear_kth_bit(n, k):
    return n & ~(1 << k)

Toggle K-th Bit

Problem Statement

Given a number n, toggle the k-th bit (0 to 1, or 1 to 0).

Solution

def toggle_kth_bit(n, k):
    return n ^ (1 << k)

Count Set Bits

Problem Statement

Count the number of 1s in the binary representation of a number.

Solution

def count_set_bits(n):
    count = 0
    while n:
        n &= (n - 1)
        count += 1
    return count

Check Power of Two

Problem Statement

Determine if a number is a power of two.

Solution

def is_power_of_two(n):
    return n > 0 and (n & (n - 1)) == 0

Find Single Number

Problem Statement

Given an array where every element appears twice except for one, find that single one.

Solution

def find_single_number(arr):
    res = 0
    for num in arr:
        res ^= num
    return res

Swap Two Numbers

Problem Statement

Swap two numbers without using a temporary variable.

Solution

def swap_xor(a, b):
    a = a ^ b
    b = a ^ b
    a = a ^ b
    return a, b

Isolate Rightmost Set Bit

Problem Statement

Find the number with only the rightmost set bit of n set.

Solution

def isolate_rightmost_bit(n):
    return n & -n

Binary to Decimal

Problem Statement

Convert a binary string to a decimal integer.

Solution

def binary_to_decimal(s):
    return int(s, 2)

Check if Number is Power of Four

Problem Statement

Determine if a given integer is a power of four.

Solution

def is_power_of_four(n):
    return n > 0 and (n & (n - 1)) == 0 and (n & 0x55555555) != 0

Count Bits to Flip to Convert A to B

Problem Statement

Find the number of bits required to flip to convert integer A to integer B.

Solution

def bits_to_flip(a, b):
    diff = a ^ b
    count = 0
    while diff:
        diff &= (diff - 1)
        count += 1
    return count

Swap Odd and Even Bits

Problem Statement

Swap all even bits with odd bits in an integer (e.g., bit 0 with bit 1, bit 2 with bit 3).

Solution

def swap_bits(n):
    even_bits = n & 0xAAAAAAAA
    odd_bits = n & 0x55555555
    return (even_bits >> 1) | (odd_bits << 1)

Find Absolute Value without Branching

Problem Statement

Calculate the absolute value of an integer without using any if-statements or ternary operators.

Solution

def bitwise_abs(n):
    mask = n >> 31
    return (n + mask) ^ mask

Check if Number is Multiple of 3

Problem Statement

Check if a number is a multiple of 3 using bitwise operations.

Solution

def is_multiple_of_3(n):
    if n < 0: n = abs(n)
    if n == 0: return True
    if n == 3: return True
    
    odd_count = 0
    even_count = 0
    while n:
        if n & 1: odd_count += 1
        n >>= 1
        if n & 1: even_count += 1
        n >>= 1
    return is_multiple_of_3(abs(odd_count - even_count))

Reverse Bits of a 32-bit Integer

Problem Statement

Reverse the bits of a given 32-bit unsigned integer.

Solution

def reverse_bits(n):
    res = 0
    for _ in range(32):
        res = (res << 1) | (n & 1)
        n >>= 1
    return res

Find Next Higher Power of 2

Problem Statement

Find the smallest power of 2 that is greater than or equal to n.

Solution

def next_power_of_2(n):
    if n == 0: return 1
    n -= 1
    n |= n >> 1
    n |= n >> 2
    n |= n >> 4
    n |= n >> 8
    n |= n >> 16
    return n + 1

Compute Modulo N without %

Problem Statement

Compute n % d where d is a power of 2 (d = 2^k).

Solution

def bitwise_mod(n, d):
    # Only works if d is a power of 2
    return n & (d - 1)

Check if Two Integers have Opposite Signs

Problem Statement

Determine if two integers have opposite signs without using branching.

Solution

def opposite_signs(x, y):
    return (x ^ y) < 0

Find Position of Rightmost Set Bit

Problem Statement

Find the position of the only set bit in a number that is a power of 2.

Solution

import math
def pos_rightmost_bit(n):
    if not (n > 0 and (n & (n - 1)) == 0):
        return -1
    return int(math.log2(n)) + 1

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Bit Manipulation (Easy) / Check if K-th Bit is Set

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Check if K-th Bit is Set

Given a number n, check if the k-th bit (0-indexed) is set (1) or not (0).

1
Create a mask by shifting 1 to the left by k positions (1 << k).
2
Perform a bitwise AND between n and the mask.
3
If the result is non-zero, the bit is set.

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