Add solution #745

Author: JoshCrozier

README.md

@@ -273,6 +273,7 @@
 733|[Flood Fill](./0733-flood-fill.js)|Easy|
 735|[Asteroid Collision](./0735-asteroid-collision.js)|Medium|
 739|[Daily Temperatures](./0739-daily-temperatures.js)|Medium|
+745|[Prefix and Suffix Search](./0745-prefix-and-suffix-search.js)|Hard|
 746|[Min Cost Climbing Stairs](./0746-min-cost-climbing-stairs.js)|Easy|
 747|[Largest Number At Least Twice of Others](./0747-largest-number-at-least-twice-of-others.js)|Easy|
 748|[Shortest Completing Word](./0748-shortest-completing-word.js)|Easy|

solutions/0745-prefix-and-suffix-search.js

@@ -0,0 +1,45 @@
+/**
+ * 745. Prefix and Suffix Search
+ * https://leetcode.com/problems/prefix-and-suffix-search/
+ * Difficulty: Hard
+ *
+ * Design a special dictionary that searches the words in it by a prefix and a suffix.
+ *
+ * Implement the WordFilter class:
+ * - WordFilter(string[] words) Initializes the object with the words in the dictionary.
+ * - f(string pref, string suff) Returns the index of the word in the dictionary, which
+ *   has the prefix pref and the suffix suff. If there is more than one valid index,
+ *   return the largest of them. If there is no such word in the dictionary, return -1.
+ */
+
+/**
+ * @param {string[]} words
+ */
+var WordFilter = function(words) {
+  this.map = new Map();
+
+  for (let i = 0; i < words.length; i++) {
+    let prefix = '';
+
+    for (let j = 0; j <= words[i].length; j++) {
+      let suffix = '';
+
+      for (let k = 0; k <= words[i].length; k++) {
+        suffix = words[i].slice(k);
+        this.map.set(prefix + '#' + suffix, i);
+      }
+
+      prefix += words[i][j];
+    }
+  }
+};
+
+/**
+ * @param {string} pref
+ * @param {string} suff
+ * @return {number}
+ */
+WordFilter.prototype.f = function(pref, suff) {
+  const target = `${pref}#${suff}`;
+  return this.map.has(target) ? this.map.get(target) : -1;
+};