Bitmask Operations Leetcode S3120

impl Solution {
    pub fn number_of_special_chars(word: String) -> i32 {
        let (mut lower, mut upper) = (0u32, 0u32);

        for b in word.bytes() {
            if b.is_ascii_lowercase() {
                lower |= 1 << (b - b'a');
            } else if b.is_ascii_uppercase() {
                upper |= 1 << (b - b'A');
            }
        }

        (lower & upper).count_ones() as i32
    }
}

First, we will save the number of lower and upper in zeroes.

let (mut lower, mut upper) = (0u32, 0u32);

Then, we convert the string to bytes since we only need the ascii representation.

(b - b'a')

If the variable b value is c (ASCII 99), then it substracted by ‘a’ (ASCII 97) then the value is 2.

1 << (b - b'a')

We create the number 1 in binary, then move by (b - b'a') which is 2 (in ‘c’ example) . So from 001 it moves left the 1 twice, so the result is 100.

lower |= 1 << (b - b'a');

We use bitwise or that compares the bits, if either one is 1, the result is 1. For example, if the lower value is 101 (c and a exists) and the right side is 010 (b) it will be 111 (abc).

It goes the same to upper variable.

At the end, we check the lower and upper using Bitwise AND.

(lower & upper).count_ones() as i32

if the lower is 101 (a and c) and upper is 011 (b and a), the bitwise and result is 001 (only a). The .count_ones() function returns the number of equal value lower and upper.