Bitwise comparisions

Question

#include <iostream>
using namespace std;

int main() {
    int n, a = 0xfffffff;
    cin >> n;
    while (n--) {
        string s;
        cin >> s;
        int c = 0;
        for (char ch : s)
            c |= 1 << (ch - 'a');
        a &= c;
    }
    cout << __builtin_popcount(a) << endl;
    return 0;
}

This code is to find if a character is present in all the inputted strings at least once. Can someone explain whats happening in this code. I am trying to learn bit wise operations in C++ and I am not able to understand whats going on here.

Answer 1

The code is not to determine if a particular character is present in all strings.

It is to find the number of characters that are present in all strings.

Here's the breakdown of the code.

    int n, a = 0xfffffff;
    cin >> n;

n is the input parameter from the user which determines the number of strings. Let's say n is 3

    while (n--) {
        string s;
        cin >> s;
        int c = 0;
        for (char ch : s)
            c |= 1 << (ch - 'a');
        a &= c;
    }

This is the main part of the code..

you get n strings and for each string you store what characters it is composed of, in a bit array

For eg, let's says first string is "stack". you loop through the characters here

    for (char ch : s)
        c |= 1 << (ch - 'a');

c is initialized to 0 for every string.

In this example "stack", lets see what happens to c

c = c | 1 << ('s'-'a') => c = c | 1 << 18 (18th bit of c is set to 1)

c = c | 1 << ('t'-'a') => c = c | 1 << 19 (19th bit of c is set to 1)

c = c | 1 << ('a'-'a') => c = c | 1 << 0 (0th bit of c is set to 1)

c = c | 1 << ('c'-'a') => c = c | 1 << 2 (2nd bit of c is set to 1)

c = c | 1 << ('k'-'a') => c = c | 1 << 10 (10th bit of c is set to 1)

note that 1 << n means that 1 is left shifted by n digits. so 1 << 3 is = 0001 << 3 = binary 1000 = 2^3 = 8 (In case you did not understand shifting)

now c is a integer whose 0,2,10,18,19 bits are set to 1. a is initialized to all 1's before the loop.

a &= c; this gets rid of all 1's except 0,2,10,18 and 19.

we continue this for all strings. and at the end, a will have 1 bit set in the position which is occupied in all strings..

For eg, if string two is "aaaaa"

computing c reveals that c has only its 0th bit set.

a &= c; this gets rid of all 1's except 0 (ie., it sets 2,10,18 and 19th bits to 0, since they don't occur in "aaaaa")

and string 3 is "zzzzza", at the end, only 0th bit of a will be set

Hence, the number of characters that occurs in all these strings is 1

Answer 2

There's not much going on in here, and breaking it down bit by bit reveals what it does:

#include <iostream>

using namespace std;

int main() {
  // Initialize a bitmask, here assumed to be 32-bits
  // which is probably "enough" for this case.
  int n, a = 0xfffffff;

  // Read in the number of strings to process
  cin >> n;

  // Assume n > 0
  while (n--) {
    string s;

    // Read in a string
    cin >> s;

    int c = 0;

    // For each character in this string
    for (char ch : s)
       // Turn on a bit on representing the character, where
       // 'a' is bit 0, 'b' is 1, etc.
       c |= 1 << (ch - 'a');

    // Apply this mask to a
    a &= c;
  }

  // Report on which bits are toggled
  cout << __builtin_popcount(a) << endl;

  return 0;
}

This is some seriously sloppy code on the whole. Any non lower case letters could cause undefined behaviour. Most compilers are 64-bit for modern machines so setting only 32 bits might be insufficient.

Note that when I use the word "assume" here I mean bad things will probably happen.