Add solution #443

Author: JoshCrozier

README.md

@@ -176,6 +176,7 @@
 414|[Third Maximum Number](./0414-third-maximum-number.js)|Easy|
 419|[Battleships in a Board](./0419-battleships-in-a-board.js)|Medium|
 442|[Find All Duplicates in an Array](./0442-find-all-duplicates-in-an-array.js)|Medium|
+443|[String Compression](./0443-string-compression.js)|Medium|
 448|[Find All Numbers Disappeared in an Array](./0448-find-all-numbers-disappeared-in-an-array.js)|Easy|
 451|[Sort Characters By Frequency](./0451-sort-characters-by-frequency.js)|Medium|
 452|[Minimum Number of Arrows to Burst Balloons](./0452-minimum-number-of-arrows-to-burst-balloons.js)|Medium|

solutions/0443-string-compression.js

@@ -0,0 +1,42 @@
+/**
+ * 443. String Compression
+ * https://leetcode.com/problems/string-compression/
+ * Difficulty: Medium
+ *
+ * Given an array of characters chars, compress it using the following algorithm:
+ *
+ * Begin with an empty string s. For each group of consecutive repeating characters in chars:
+ * - If the group's length is 1, append the character to s.
+ * - Otherwise, append the character followed by the group's length.
+ *
+ * The compressed string s should not be returned separately, but instead, be stored in the
+ * input character array chars. Note that group lengths that are 10 or longer will be split
+ * into multiple characters in chars.
+ *
+ * After you are done modifying the input array, return the new length of the array.
+ *
+ * You must write an algorithm that uses only constant extra space.
+ */
+
+/**
+ * @param {character[]} chars
+ * @return {number}
+ */
+var compress = function(chars) {
+  let pointer = 0;
+  for (let i = 0; i < chars.length; i++) {
+    const char = chars[i];
+    let count = 0;
+    while (i < chars.length && chars[i] === char) {
+      count++;
+      i++;
+    }
+    chars[pointer++] = char;
+    if (count !== 1) {
+      String(count).split('').forEach(n => chars[pointer++] = n);
+    }
+    i--;
+  }
+  chars.length = pointer;
+  return pointer;
+};