Bit field

const A_LCASE = 97;
const A_UCASE = 65;
const ALL_26_BITS_SET = 67108863;

export function isPangram(input) {
  let phrasemask = 0;
  [...input].forEach((letter) => {
    if (letter >= 'a' && letter <= 'z')
      phrasemask |= 1 << (letter.charCodeAt(0) - A_LCASE);
    else if (letter >= 'A' && letter <= 'Z')
      phrasemask |= 1 << (letter.charCodeAt(0) - A_UCASE);
  });
  return phrasemask == ALL_26_BITS_SET;
}

This solution uses the ASCII value of the letter to set the corresponding bit position.

Some constants are defined for readability in the function. The ASCII value for a is 97. The ASCII value for A is 65. The value for all of the rightmost 26 bits being set is 67108863.

• Spread syntax is used to make an Array of the characters in the input.
• The Array method forEach loops through the characters and looks for a character being a through z or A through Z.
• If a letter is found, then its ASCII value is taken by the charCodeAt method.

• If the lowercase letter is subtracted by 97, then a will result in 0, because 97 minus 97 equals 0. z would result in 25, because 122 minus 97 equals 25. So a would have 1 shifted left 0 places (so not shifted at all) and z would have 1 shifted left 25 places.
• If the uppercase letter is subtracted by A, then A will result in 0, because 65 minus 65 equals 0. Z would result in 25, because 90 minus 65 equals 25. So A would have 1 shifted left 0 places (so not shifted at all) and Z would have 1 shifted left 25 places.

In that way, both a lowercase z and an uppercase Z can share the same position in the bit field.

So, for an integer, if the values for a and Z were both set, the bits would look like

      zyxwvutsrqponmlkjihgfedcba
00000010000000000000000000000001

We can use the bitwise OR operator to set the bit. After the loop completes, the function returns if the phrasemask value is the same value as when all 26 bits are set, which is 67108863.