Add solution #1593

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,406 LeetCode solutions in JavaScript
+# 1,407 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1230,6 +1230,7 @@
 1590|[Make Sum Divisible by P](./solutions/1590-make-sum-divisible-by-p.js)|Medium|
 1591|[Strange Printer II](./solutions/1591-strange-printer-ii.js)|Hard|
 1592|[Rearrange Spaces Between Words](./solutions/1592-rearrange-spaces-between-words.js)|Easy|
+1593|[Split a String Into the Max Number of Unique Substrings](./solutions/1593-split-a-string-into-the-max-number-of-unique-substrings.js)|Medium|
 1598|[Crawler Log Folder](./solutions/1598-crawler-log-folder.js)|Easy|
 1657|[Determine if Two Strings Are Close](./solutions/1657-determine-if-two-strings-are-close.js)|Medium|
 1668|[Maximum Repeating Substring](./solutions/1668-maximum-repeating-substring.js)|Easy|

solutions/1593-split-a-string-into-the-max-number-of-unique-substrings.js

@@ -0,0 +1,38 @@
+/**
+ * 1593. Split a String Into the Max Number of Unique Substrings
+ * https://leetcode.com/problems/split-a-string-into-the-max-number-of-unique-substrings/
+ * Difficulty: Medium
+ *
+ * Given a string s, return the maximum number of unique substrings that the given string can
+ * be split into.
+ *
+ * You can split string s into any list of non-empty substrings, where the concatenation of
+ * the substrings forms the original string. However, you must split the substrings such that
+ * all of them are unique.
+ *
+ * A substring is a contiguous sequence of characters within a string.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var maxUniqueSplit = function(s) {
+  return backtrack(0, new Set());
+
+  function backtrack(start, seen) {
+    if (start === s.length) return seen.size;
+
+    let maxSplits = 0;
+    for (let end = start + 1; end <= s.length; end++) {
+      const substring = s.slice(start, end);
+      if (!seen.has(substring)) {
+        seen.add(substring);
+        maxSplits = Math.max(maxSplits, backtrack(end, seen));
+        seen.delete(substring);
+      }
+    }
+
+    return maxSplits;
+  }
+};