Add solution #3002

Author: JoshCrozier

README.md

@@ -2125,6 +2125,7 @@
 2997|[Minimum Number of Operations to Make Array XOR Equal to K](./solutions/2997-minimum-number-of-operations-to-make-array-xor-equal-to-k.js)|Medium|
 2998|[Minimum Number of Operations to Make X and Y Equal](./solutions/2998-minimum-number-of-operations-to-make-x-and-y-equal.js)|Medium|
 2999|[Count the Number of Powerful Integers](./solutions/2999-count-the-number-of-powerful-integers.js)|Hard|
+3002|[Maximum Size of a Set After Removals](./solutions/3002-maximum-size-of-a-set-after-removals.js)|Medium|
 3024|[Type of Triangle](./solutions/3024-type-of-triangle.js)|Easy|
 3042|[Count Prefix and Suffix Pairs I](./solutions/3042-count-prefix-and-suffix-pairs-i.js)|Easy|
 3066|[Minimum Operations to Exceed Threshold Value II](./solutions/3066-minimum-operations-to-exceed-threshold-value-ii.js)|Medium|

solutions/3002-maximum-size-of-a-set-after-removals.js

@@ -0,0 +1,41 @@
+/**
+ * 3002. Maximum Size of a Set After Removals
+ * https://leetcode.com/problems/maximum-size-of-a-set-after-removals/
+ * Difficulty: Medium
+ *
+ * You are given two 0-indexed integer arrays nums1 and nums2 of even length n.
+ *
+ * You must remove n / 2 elements from nums1 and n / 2 elements from nums2. After the removals,
+ * you insert the remaining elements of nums1 and nums2 into a set s.
+ *
+ * Return the maximum possible size of the set s.
+ */
+
+/**
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @return {number}
+ */
+var maximumSetSize = function(nums1, nums2) {
+  const n = nums1.length;
+  const set1 = new Set(nums1);
+  const set2 = new Set(nums2);
+
+  let unique1 = set1.size;
+  let unique2 = set2.size;
+  let common = 0;
+
+  for (const num of set1) {
+    if (set2.has(num)) common++;
+  }
+
+  unique1 -= common;
+  unique2 -= common;
+
+  const max1 = Math.min(unique1, n / 2);
+  const max2 = Math.min(unique2, n / 2);
+  const remaining = n / 2 - max1 + n / 2 - max2;
+  const maxCommon = Math.min(common, remaining);
+
+  return max1 + max2 + maxCommon;
+};