Given an integer n and two bit positions p1 and p2 inside it, swap bits at the given positions. The given positions are from least significant bit (lsb). For example, the position for lsb is 0.
Input: n = 28, p1 = 0, p2 = 3 Output: 21 28 in binary is 11100. If we swap 0'th and 3rd digits, we get 10101 which is 21 in decimal. Input: n = 20, p1 = 2, p2 = 3 Output: 24
We strongly recommend you to minimize your browser and try this yourself first.
The idea is to first find the bits, then use XOR based swapping concept, i..e., to swap two numbers ‘x’ and ‘y’, we do x = x ^ y, y = y ^ x and x = x ^ y.
Below is the implementation of the above idea
Result = 21
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
- Swap all odd and even bits
- Swap every two bits in bytes
- Swap bits in a given number
- Count set bits in an integer
- Next greater integer having one more number of set bits
- Count set bits in an integer using Lookup Table
- Previous smaller integer having one less number of set bits
- Check if bits of a number has count of consecutive set bits in increasing order
- Toggle bits of a number except first and last bits
- Print numbers having first and last bits as the only set bits
- Swap two nibbles in a byte
- Bit manipulation | Swap Endianness of a number
- Swap three variables without using temporary variable
- Number of ways to swap two bit of s1 so that bitwise OR of s1 and s2 changes
- How to swap two numbers without using a temporary variable?