Check if the Sentence Is Pangram

easy bitmask string

Problem

Return whether sentence (lowercase) contains every letter of the English alphabet.

Inputsentence = "thequickbrownfoxjumpsoverthelazydog"

Outputtrue

All 26 letters present.

sentence

def check_if_pangram(sentence):
    mask = 0
    for c in sentence:
        mask |= 1 << (ord(c) - 97)
    return mask == (1 << 26) - 1

function checkIfPangram(sentence) {
  let m = 0;
  for (const c of sentence) m |= 1 << (c.charCodeAt(0) - 97);
  return m === (1 << 26) - 1;
}

class Solution {
    public boolean checkIfPangram(String sentence) {
        int m = 0;
        for (char c : sentence.toCharArray()) m |= 1 << (c - 'a');
        return m == (1 << 26) - 1;
    }
}

bool checkIfPangram(string sentence) {
    int m = 0;
    for (char c : sentence) m |= 1 << (c - 'a');
    return m == (1 << 26) - 1;
}

Explanation

A pangram uses all 26 letters at least once. The neat trick here is to track which letters we have seen using a single integer as a 26-bit bitmask instead of a set or array.

We start with mask = 0. For each character c we compute its position in the alphabet with ord(c) - 97 (so 'a' is 0, 'b' is 1, and so on) and turn on that bit using mask |= 1 << (ord(c) - 97). Seeing the same letter again just re-sets a bit that is already on, which is harmless.

At the end, every letter present corresponds to one set bit. The sentence is a pangram exactly when all 26 bits are on. The value with bits 0 through 25 set is (1 << 26) - 1, so we simply compare mask to that.

This is fast and uses constant extra space, since one int holds the whole letter set.

Example: for "thequickbrownfoxjumpsoverthelazydog" every letter appears, so the final mask equals (1 << 26) - 1 and the answer is true.

Time: O(n) Space: O(1)