Add solution #1647

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,446 LeetCode solutions in JavaScript
+# 1,447 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1269,6 +1269,7 @@
 1642|[Furthest Building You Can Reach](./solutions/1642-furthest-building-you-can-reach.js)|Medium|
 1643|[Kth Smallest Instructions](./solutions/1643-kth-smallest-instructions.js)|Hard|
 1646|[Get Maximum in Generated Array](./solutions/1646-get-maximum-in-generated-array.js)|Easy|
+1647|[Minimum Deletions to Make Character Frequencies Unique](./solutions/1647-minimum-deletions-to-make-character-frequencies-unique.js)|Medium|
 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|
 1669|[Merge In Between Linked Lists](./solutions/1669-merge-in-between-linked-lists.js)|Medium|

solutions/1647-minimum-deletions-to-make-character-frequencies-unique.js

@@ -0,0 +1,40 @@
+/**
+ * 1647. Minimum Deletions to Make Character Frequencies Unique
+ * https://leetcode.com/problems/minimum-deletions-to-make-character-frequencies-unique/
+ * Difficulty: Medium
+ *
+ * A string s is called good if there are no two different characters in s that have the
+ * same frequency.
+ *
+ * Given a string s, return the minimum number of characters you need to delete to make s good.
+ *
+ * The frequency of a character in a string is the number of times it appears in the string.
+ * For example, in the string "aab", the frequency of 'a' is 2, while the frequency of 'b' is 1.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var minDeletions = function(s) {
+  const charFrequencies = new Array(26).fill(0);
+  for (const char of s) {
+    charFrequencies[char.charCodeAt(0) - 97]++;
+  }
+
+  charFrequencies.sort((a, b) => b - a);
+  let result = 0;
+  const usedFrequencies = new Set();
+
+  for (const freq of charFrequencies) {
+    if (freq === 0) break;
+    let currentFreq = freq;
+    while (usedFrequencies.has(currentFreq) && currentFreq > 0) {
+      currentFreq--;
+      result++;
+    }
+    usedFrequencies.add(currentFreq);
+  }
+
+  return result;
+};