Applications

Most Used

Assume 0-based indexing, we can shift n too instead of 1 in the following tricks.

  • Set ith bit: n | (1 << i)
  • Unset ith bit: n & ~(1 << i)
  • Get ith bit, check if ith bit is set or unset: if n & (1 << i) > 0 then set, else unset
  • Toggle ith bit: n ^ (1 << i)
  • Unset the rightmost set bit: n & (n - 1)
  • Set the rightmost unset bit: n | (n + 1)
  • Keep only the rightmost set bit: n & ~(n - 1)
  • Check parity of a number: if(n & 1) then odd, else even
  • Check if a number is power of 2: n & (n - 1) will be 0 (falsely reports 0 as a power of 2). Use this instead: if (n && !(n & (n - 1)) == 0)
  • Swap two numbers with XOR: x = x ^ y; y = x ^ y; x = x ^ y;
  • Toggle case of a character in a string: str[i] ^= 32
  • Modulus division by a number (n) of the form 2k (d): n & (d-1), so in place of (n % 4) we can write (n & 3).

Some More Usages

  • Detect if two integers have opposite signs: (x ^ y) < 0
  • Check if two numbers are equal: a ^ b = 0 only if a, b are equal
  • Counting bits to be flipped to convert A to B (Bit Difference): This is number of set bits in A ^ B.
  • Multiplying with 3.5: 2x + x + x/2 = (x << 1) + x + (x >> 1)
  • Multiplying with 7: 8x - x = (x << 3) - x
  • Adding 1 to number without using + operator: xPlusOne = -(~x), since ~x = -(x+1) holds.
  • Checking is number is power of 4: single set bit having even number of 0s on its right side
  • Finding position of rightmost set bit: Do 2's complement, & with original number, take log2(ANDresult) = log2(n & -n) + 1
  • Find XOR of two number without using XOR operator: (x | y) & (~x | ~y)
  • XOR from [1 to n] efficiently: Link [XOR repeats on modulo 4] [based on n % 4 values can be n, 1, n+1, 0]
  • Rotating bits: Left = n << k | n >> (32-k) and Right = n >> k | n << (32-k)

Algorithms

  • Brian Kernighan’s Algorithm - to find number of set bits in binary representation of a number [link]
  • Russian Peasant Multiplication and Egyptian Multiplication (Multiply two numbers using bitwise operators) [link]
  • Binary Exponentiation [link]
  • Binary GCD [link]

Related Questions & Tricks

  • Power Set - all subsets [link]
  • Find Binary Representation of a decimal number [link]

Operations on Range:

  • Find XOR of numbers in range L to R [link] - use XOR’s modulo property for ranges ans = [L - R] = [1 - R] ^ [1 - (L-1)]
  • Find AND of numbers in range L to R [link]
    • Brute Force: TC = O(n), TLE on LC
    • Shifting Approach: needs only L and R as both of their common part is also common to all numbers in the range. Common part will yield the same as AND’s result (common part), rest will become 0 (shift out and shift back replacing it with 0s). TC = O(log n)
    • Keep Unsetting Rightmost Set Bit Approach: builds answer using the same concept as above but goto only set bits instead of shifting each bit. Do n & (n-1) and this unsets the rightmost set bit and then AND operation on this position with any other number will yield 0 at this position in the final answer, so we keep doing this to form ans till ans > left (in the last iteration ans & left will happen as left = ans-1). TC = O(popcount(n))

Forming two buckets trick:

  • Find duplicate and missing elements in an array [link]
  • Find two elements occuring odd numnber of times [link]

Counting Bits trick: usually we can count set bits in log n time using Brian-Kernighan’s Algorithm, but if we need to count set bits for each number in range [0 - n] we can do so in O(n) time using DP as dp[i] = dp[i/2] + i%2 where dp[0] = 0 is the base case problem video

  • this is based on the simple observation that for a number N shifting it rightwards by 1 (dividing by 2; halving) will either have equal set bits (if N is even) or 1 extra set bits (if N is odd) since shifting causes LSB to get lost

TC of Bitwise Algorithms

Everything we do on a binary representation of a number is log n time, since a number’s binary representation can contain atmost n bits.