Count Complete Substrings - Sliding Window [JS]

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Description 

Solution: Sliding Window

There can be up to 26 unique characters in a substring, and each character occurs exactly k times.
For each number of unique characters, maintain a sliding window of characters where

  1. The difference between two adjacent characters is at most 2.
  2. The length of the window doesn't exceed the size the substring is meant to be: unique characters * k.
    Count the number of substrings with exactly unique unique characters which occur k times each.

Time Complexity: O(26n)
Space Complexity: O(1)

var countCompleteSubstrings = function(word, k) {
  let n = word.length, complete = 0;
  for (let unique = 1; unique <= 26; unique++) {
    let substrSize = unique * k;
    let charsWithKFreq = 0;
    let count = Array(26).fill(0);
    for (let j = 0, i = 0; j < n; j++) {
      // diff between adjacent chars > 2, reset the window completely
      if (j > 0 && diff(word[j], word[j - 1]) > 2) {
        count = Array(26).fill(0);
        charsWithKFreq = 0;
        i = j;
      }
      count[word.charCodeAt(j) - 97]++;
      if (count[word.charCodeAt(j) - 97] === k) charsWithKFreq++;
      // move left pointer up while window is too big for the substring size
      while (j - i + 1 > substrSize) {
        if (count[word.charCodeAt(i) - 97] === k) charsWithKFreq--;
        count[word.charCodeAt(i) - 97]--;
        i++;
      }
      if (charsWithKFreq === unique) complete++;
    }
  }
  return complete;
};

function diff(a, b) {
  return Math.abs(a.charCodeAt() - b.charCodeAt());
}
Location: Auckland, New Zealand

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