Toggle case of a string using Bitwise Operators
Given a string, write a function that returns toggle case of a string using the bitwise operators in place.
In ASCII codes, character ‘A’ is integer 65 = (0100 0001)2, while character ‘a’ is integer 97 = (0110 0001)2. Similarly, character ‘D’ is integer 68 = (0100 0100)2, while character ‘d’ is integer 100 = (0110 0100)2.
As we can see, only sixth least significant bit is different in ASCII code of ‘A’ and ‘a’. Similar behavior can be seen in ASCII code of ‘D’ and ‘d’. Therefore, we need to toggle this bit for toggling case.
Input : "GeekSfOrgEEKs" Output : "gEEKsFoRGeekS" Input : "StRinG" Output : "sTrINg"
The ASCII table is constructed in such way that the binary representation of lowercase letters is almost identical of binary representation of uppercase letters.
The integer with 6th LSB as 1 is 32 (0010 0000). Therefore, bitwise XORing of a character with 32 will toggle the 6th LSB of character and hence, will toggle its case. If character is upper case, it will be converted to lower case and vice versa.
Toggle case: cHErRy Original string: CheRrY
Time complexity: O(n) since one traversal of the string is required to complete all operations hence the overall time required by the algorithm is linear
Auxiliary Space: O(1) since no extra array is used so the space taken by the algorithm is constant
Thanks to Kumar Gaurav for improving the solution.
Similar Article :
Case conversion of a string using BitWise operators in C/C++
This article is contributed by Sanjay Kumar Ulsha from JNTUH College Of Engineering, Hyderabad. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks.