Add solution #3085

Author: JoshCrozier

README.md

@@ -2163,6 +2163,7 @@
 3079|[Find the Sum of Encrypted Integers](./solutions/3079-find-the-sum-of-encrypted-integers.js)|Easy|
 3083|[Existence of a Substring in a String and Its Reverse](./solutions/3083-existence-of-a-substring-in-a-string-and-its-reverse.js)|Easy|
 3084|[Count Substrings Starting and Ending with Given Character](./solutions/3084-count-substrings-starting-and-ending-with-given-character.js)|Medium|
+3085|[Minimum Deletions to Make String K-Special](./solutions/3085-minimum-deletions-to-make-string-k-special.js)|Medium|
 3105|[Longest Strictly Increasing or Strictly Decreasing Subarray](./solutions/3105-longest-strictly-increasing-or-strictly-decreasing-subarray.js)|Easy|
 3108|[Minimum Cost Walk in Weighted Graph](./solutions/3108-minimum-cost-walk-in-weighted-graph.js)|Hard|
 3110|[Score of a String](./solutions/3110-score-of-a-string.js)|Easy|

solutions/3085-minimum-deletions-to-make-string-k-special.js

@@ -0,0 +1,45 @@
+/**
+ * 3085. Minimum Deletions to Make String K-Special
+ * https://leetcode.com/problems/minimum-deletions-to-make-string-k-special/
+ * Difficulty: Medium
+ *
+ * You are given a string word and an integer k.
+ *
+ * We consider word to be k-special if |freq(word[i]) - freq(word[j])| <= k for all indices i
+ * and j in the string.
+ *
+ * Here, freq(x) denotes the frequency of the character x in word, and |y| denotes the absolute
+ * value of y.
+ *
+ * Return the minimum number of characters you need to delete to make word k-special.
+ */
+
+/**
+ * @param {string} word
+ * @param {number} k
+ * @return {number}
+ */
+var minimumDeletions = function(word, k) {
+  const freq = new Array(26).fill(0);
+
+  for (const char of word) {
+    freq[char.charCodeAt(0) - 97]++;
+  }
+
+  const counts = freq.filter(x => x > 0).sort((a, b) => a - b);
+  let minDeletions = Infinity;
+  for (let i = 0; i < counts.length; i++) {
+    let deletions = 0;
+    for (let j = 0; j < i; j++) {
+      deletions += counts[j];
+    }
+    for (let j = i; j < counts.length; j++) {
+      if (counts[j] - counts[i] > k) {
+        deletions += counts[j] - (counts[i] + k);
+      }
+    }
+    minDeletions = Math.min(minDeletions, deletions);
+  }
+
+  return minDeletions;
+};