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▎Bitwise Tricks in Python

▎1. Check Even/Odd

def is_odd(n: int) -> bool:
    return bool(n & 1)

# Usage:
n = 7
print("Odd" if is_odd(n) else "Even")  # Output: Odd

Try: “Check if a Number is Odd or Even” on GeeksforGeeks (GfG)

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▎2. Swap Two Numbers Without a Temp Variable

def xor_swap(a: int, b: int) -> tuple[int, int]:
    a ^= b
    b ^= a
    a ^= b
    return a, b

# Usage:
x, y = 5, 9
x, y = xor_swap(x, y)
print(x, y)  # Output: 9 5

Note: In real Python, you can simply use x, y = y, x.

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▎3. Turn Off the Rightmost Set Bit

def clear_lowest_bit(n: int) -> int:
    return n & (n - 1)

# Usage:
n = 0b10110  # 22
print(bin(clear_lowest_bit(n)))  # Output: 0b10100 (20)

Try: “Count Set Bits in an Integer” on GfG (use this in a loop)

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▎4. Isolate the Rightmost Set Bit

def lowest_bit(n: int) -> int:
    return n & -n

# Usage:
n = 0b10110  # 22
print(bin(lowest_bit(n)))  # Output: 0b10 (2)

Try: “Find Position of Rightmost Different Bit” on GfG

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▎5. Check Power of Two

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

# Usage:
for x in [1, 2, 3, 16, 18]:
    print(x, is_power_of_two(x))

Try: “Check if a Number is a Power of Two” on GfG

---

▎6. Count Set Bits (Brian Kernighan’s Algorithm)

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

# Usage:
print(count_set_bits(29))  # Output: 4, since 29 is 11101₂

Try: “Count Set Bits” on GfG (Python)

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▎7. Add Two Numbers Without ‘+’

def add(a: int, b: int) -> int:
    MASK = 0xFFFFFFFF
    while b != 0:
        carry = (a & b) & MASK
        a = (a ^ b) & MASK
        b = (carry << 1) & MASK
    
    return a if a <= 0x7FFFFFFF else ~(a ^ MASK)

# Usage:
print(add(15, 27))  # Output: 42
print(add(-5, 3))   # Output: -2

Try: “Sum of Two Integers” on LeetCode (Python)

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▎8. Test Opposite Signs

def opposite_signs(a: int, b: int) -> bool:
    return (a ^ b) < 0

# Usage:
print(opposite_signs(5, -3))   # Output: True
print(opposite_signs(-4, -2))  # Output: False

Try: “Check if Two Numbers Have Opposite Signs” on GfG

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▎9. Reverse Bits of a 32-bit Integer

def reverse_bits(n: int) -> int:
    result = 0
    for _ in range(32):
        result = (result << 1) | (n & 1)
        n >>= 1
    return result & 0xFFFFFFFF

# Usage:
x = 0b00000010100101000001111010011100
print(bin(reverse_bits(x)))  

Try: “Reverse Bits” on LeetCode (Python)

---

▎10. Gray Code Conversion

def to_gray(n: int) -> int:
    return n ^ (n >> 1)

def gray_to_binary(g: int) -> int:
    mask = g
    while mask:
        mask >>= 1
        g ^= mask
    return g

# Usage:
for i in range(8):
    g = to_gray(i)
    print(f"{i:3} → Gray {g:3} → Back {gray_to_binary(g):3}")

Try: “Gray Code” on LeetCode (Python)

---

▎🎯 Next Steps:

1. Pick one or two of these tricks.

2. Find the corresponding GfG problem in Python and implement it.

3. Time yourself and compare it to a straightforward solution.

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can u solve this question
▎Title: 201. Bitwise AND of Numbers Range

Difficulty: Medium

Denoscription:

Given two integers left and right that represent the range [left, right], return the bitwise AND of all numbers in this range, inclusive.

Example 1:

Input: left = 5, right = 7  
Output: 4

Explanation:

5 = 101
6 = 110
7 = 111

Bitwise AND of all numbers in the range [5, 7] is 4 (binary 100).

Example 2:

Input: left = 0, right = 0  
Output: 0

Example 3:

Input: left = 1, right = 2147483647  
Output: 0

Explanation:
There is no common bit position that stays 1 throughout the range, so the result is 0.

Constraints:

0 <= left <= right <= 2^31 - 1

can u solve this question? with bit manipulation in constant extra space and linear time complexity
▎Title: 137. Single Number II

Difficulty: Medium

Denoscription:

Given an integer array nums where every element appears three times except for one, which appears exactly once, find the single element and return it.

You must implement a solution with a linear runtime complexity and use only constant extra space.

Example 1:
Input: nums = [2,2,3,2]
Output: 3

Example 2:
Input: nums = [0,1,0,1,0,1,99]
Output: 99

Constraints:

• 1 <= nums.length <= 3 * 10^4

• -2^31 <= nums[i] <= 2^31 - 1

• Each element in nums appears exactly three times except for one element which appears once.