Leetcode with dani

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

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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) --- ▎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. --- ▎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) --- ▎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 --- ▎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) --- ▎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) --- ▎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 --- ▎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.

▎Bitwise Tricks and Techniques for Competitive Programming ▎1. Check Even/Odd • Trick: n & 1 • Why it works: The least significant bit (LSB) of an integer is 1 if it’s odd, 0 if it’s even. • Usage:    

    if (n & 1) // n is odd 
    else // n is even 
    

• Try it on: “Count Odd Numbers in an Interval Range” (easy) --- ▎2. Swap Two Numbers Without a Temp Variable • Trick: a ^= b; b ^= a; a ^= b; • Why it works: XORing twice cancels out, effectively swapping the values. • Usage:    

    a ^= b; 
    b ^= a; 
    a ^= b; 
    

• Try it on: “Swap Two Numbers” (basic exercise) --- ▎3. Turn Off the Rightmost Set Bit • Trick: n & (n - 1) • Why it works: Subtracting 1 flips all bits from the rightmost 1 to the end; ANDing clears that rightmost 1. • Usage:    

    n = n & (n - 1); // removes lowest set bit 
    

• Try it on: “Count Set Bits in an Integer” (you can loop with this trick) --- ▎4. Isolate the Rightmost Set Bit • Trick: n & -n • Why it works: The two’s complement of n (-n) has only the rightmost set bit in common with n. • Usage:    

    int lowbit = n & -n; 
    

• Try it on: “Power of Two” (check n & (n - 1) == 0, then use n & -n to identify which bit it is) --- ▎5. Check Power of Two • Trick: (n & (n - 1)) == 0 and n > 0 • Why it works: Powers of two have exactly one set bit; subtracting 1 clears that bit. • Usage:    

    bool isPow2 = (n > 0) && ((n & (n - 1)) == 0); 
    

• Try it on: “Check if a Number is a Power of Two” (classic) --- ▎6. Count Set Bits Faster (Brian Kernighan’s Algorithm) • Trick: Repeatedly do n &= (n - 1) until zero, counting iterations. • Why it works: Each operation removes one set bit, so the loop runs in O(# bits set). • Usage:    

    int count = 0; 
    while (n) { 
        n &= (n - 1); 
        ++count; 
    } 
    

• Try it on: “Count Set Bits in an Integer” (better than O(32) per bit) --- ▎7. Compute a + b Without Plus • Trick: Use XOR and AND with shifts:    

    while (b) { 
        unsigned carry = a & b; 
        a = a ^ b; 
        b = carry << 1; 
    } // a now holds the sum 
    

• Why it works: XOR gives sum without carry; AND + shift gives carry bits until none remain. • Try it on: LeetCode “Sum of Two Integers” --- ▎8. Test if Two Integers Have Opposite Signs • Trick: (a ^ b) < 0 • Why it works: The sign bit differs if XOR’s sign bit is 1. • Usage:    

    bool oppSigns = (a ^ b) < 0; 
    

• Try it on: “Check if Two Integers Have Opposite Signs” (easy) --- ▎9. Reverse Bits • Trick: Use lookup tables for small chunks (e.g., 8 bits) or iterative bitwise swaps:    

    n = (n >> 16) | (n << 16); 
    n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8); 
    // …and so on 
    

• Why it works: Repeatedly swap bit-groups to reverse order in log₂(width) steps. • Try it on: LeetCode “Reverse Bits” --- ▎10. Gray Code Conversion • Trick to get Gray from binary: g = n ^ (n >> 1) • Trick to get Binary from Gray: Loop with shifting and XOR. • Try it on: “Gray Code” generation problems ---

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