Add solution #828

Author: JoshCrozier

README.md

@@ -636,6 +636,7 @@
 825|[Friends Of Appropriate Ages](./0825-friends-of-appropriate-ages.js)|Medium|
 826|[Most Profit Assigning Work](./0826-most-profit-assigning-work.js)|Medium|
 827|[Making A Large Island](./0827-making-a-large-island.js)|Hard|
+828|[Count Unique Characters of All Substrings of a Given String](./0828-count-unique-characters-of-all-substrings-of-a-given-string.js)|Hard|
 830|[Positions of Large Groups](./0830-positions-of-large-groups.js)|Easy|
 831|[Masking Personal Information](./0831-masking-personal-information.js)|Medium|
 841|[Keys and Rooms](./0841-keys-and-rooms.js)|Medium|

solutions/0828-count-unique-characters-of-all-substrings-of-a-given-string.js

@@ -0,0 +1,41 @@
+/**
+ * 828. Count Unique Characters of All Substrings of a Given String
+ * https://leetcode.com/problems/count-unique-characters-of-all-substrings-of-a-given-string/
+ * Difficulty: Hard
+ *
+ * Let's define a function countUniqueChars(s) that returns the number of unique characters in s.
+ * - For example, calling countUniqueChars(s) if s = "LEETCODE" then "L", "T", "C", "O", "D" are
+ *   the unique characters since they appear only once in s, therefore countUniqueChars(s) = 5.
+ *
+ * Given a string s, return the sum of countUniqueChars(t) where t is a substring of s. The test
+ * cases are generated such that the answer fits in a 32-bit integer.
+ *
+ * Notice that some substrings can be repeated so in this case you have to count the repeated
+ * ones too.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var uniqueLetterString = function(s) {
+  const n = s.length;
+  const lastPositions = {};
+  let result = 0;
+
+  for (let i = 0; i < n; i++) {
+    const char = s[i];
+    if (!lastPositions[char]) lastPositions[char] = [-1, -1];
+
+    const [prevPrev, prev] = lastPositions[char];
+    result += (i - prev) * (prev - prevPrev);
+    lastPositions[char] = [prev, i];
+  }
+
+  for (const char of Object.keys(lastPositions)) {
+    const [prev, pos] = lastPositions[char];
+    if (pos >= 0) result += (n - pos) * (pos - prev);
+  }
+
+  return result;
+};