Bit field

const A_LCASE = 97;
const A_UCASE = 65;

export function isIsogram(word) {
  let letter_flags = 0;

  for (const letter of [...word]) {
    if (letter >= 'a' && letter <= 'z') {
      if ((letter_flags & (1 << (letter.charCodeAt(0) - A_LCASE))) != 0)
        return false;
      else letter_flags |= 1 << (letter.charCodeAt(0) - A_LCASE);
    } else if (letter >= 'A' && letter <= 'Z') {
      if ((letter_flags & (1 << (letter.charCodeAt(0) - A_UCASE))) != 0)
        return false;
      else letter_flags |= 1 << (letter.charCodeAt(0) - A_UCASE);
    }
  }
  return true;
}

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.

• Spread syntax is used to make an Array of the characters in the word.
• The string loops through its 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 lower-cased z and an upper-cased Z can share the same position in the bit field.

So, for an unsigned thirty-two bit integer, if the values for a and Z were both set, the bits would look like

      zyxwvutsrqponmlkjihgfedcba
00000010000000000000000000000001

We can use the bitwise AND operator to check if a bit has already been set. If it has been set, we know the letter is duplicated and we can immediately return false. If it has not been set, we can use the bitwise OR operator to set the bit. If the loop completes without finding a duplicate letter (and returning false), the function returns true.