Add solution #1487

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,325 LeetCode solutions in JavaScript
+# 1,326 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1136,6 +1136,7 @@
 1482|[Minimum Number of Days to Make m Bouquets](./solutions/1482-minimum-number-of-days-to-make-m-bouquets.js)|Medium|
 1483|[Kth Ancestor of a Tree Node](./solutions/1483-kth-ancestor-of-a-tree-node.js)|Hard|
 1486|[XOR Operation in an Array](./solutions/1486-xor-operation-in-an-array.js)|Easy|
+1487|[Making File Names Unique](./solutions/1487-making-file-names-unique.js)|Medium|
 1491|[Average Salary Excluding the Minimum and Maximum Salary](./solutions/1491-average-salary-excluding-the-minimum-and-maximum-salary.js)|Easy|
 1492|[The kth Factor of n](./solutions/1492-the-kth-factor-of-n.js)|Medium|
 1493|[Longest Subarray of 1's After Deleting One Element](./solutions/1493-longest-subarray-of-1s-after-deleting-one-element.js)|Medium|

solutions/1487-making-file-names-unique.js

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+/**
+ * 1487. Making File Names Unique
+ * https://leetcode.com/problems/making-file-names-unique/
+ * Difficulty: Medium
+ *
+ * Given an array of strings names of size n. You will create n folders in your file system such
+ * that, at the ith minute, you will create a folder with the name names[i].
+ *
+ * Since two files cannot have the same name, if you enter a folder name that was previously used,
+ * the system will have a suffix addition to its name in the form of (k), where, k is the smallest
+ * positive integer such that the obtained name remains unique.
+ *
+ * Return an array of strings of length n where ans[i] is the actual name the system will assign
+ * to the ith folder when you create it.
+ */
+
+/**
+ * @param {string[]} names
+ * @return {string[]}
+ */
+var getFolderNames = function(names) {
+  const map = new Map();
+  const result = [];
+
+  for (const name of names) {
+    if (!map.has(name)) {
+      result.push(name);
+      map.set(name, 1);
+    } else {
+      let suffix = map.get(name);
+      let newName = `${name}(${suffix})`;
+      while (map.has(newName)) {
+        suffix++;
+        newName = `${name}(${suffix})`;
+      }
+      result.push(newName);
+      map.set(name, suffix + 1);
+      map.set(newName, 1);
+    }
+  }
+
+  return result;
+};