Append Characters to String to Make Subsequence - Greedy - Two Pointers [JS]

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Description 

Solution: Greedy

Since we can only add new characters at the end of s,
it is optimal to match as many characters in s as possible.
Use two pointers to keep track of how many characters have been matched in s and t.
We then need to append the last (t.length - j) characters of t onto s.

n = length of s
Time Complexity: O(n)
Space Complexity: O(1)

var appendCharacters = function(s, t) {
  let j = 0;
  for (let i = 0; i < s.length && j < t.length; i++) {
    if (s[i] === t[j]) j++;
  }
  return t.length - j;
};
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