Add solution #3347

Author: JoshCrozier

README.md

@@ -2712,6 +2712,7 @@
 3342|[Find Minimum Time to Reach Last Room II](./solutions/3342-find-minimum-time-to-reach-last-room-ii.js)|Medium|
 3343|[Count Number of Balanced Permutations](./solutions/3343-count-number-of-balanced-permutations.js)|Hard|
 3344|[Maximum Sized Array](./solutions/3344-maximum-sized-array.js)|Medium|
+3347|[Maximum Frequency of an Element After Performing Operations II](./solutions/3347-maximum-frequency-of-an-element-after-performing-operations-ii.js)|Hard|
 3349|[Adjacent Increasing Subarrays Detection I](./solutions/3349-adjacent-increasing-subarrays-detection-i.js)|Easy|
 3350|[Adjacent Increasing Subarrays Detection II](./solutions/3350-adjacent-increasing-subarrays-detection-ii.js)|Medium|
 3353|[Minimum Total Operations](./solutions/3353-minimum-total-operations.js)|Easy|

solutions/3347-maximum-frequency-of-an-element-after-performing-operations-ii.js

@@ -0,0 +1,62 @@
+/**
+ * 3347. Maximum Frequency of an Element After Performing Operations II
+ * https://leetcode.com/problems/maximum-frequency-of-an-element-after-performing-operations-ii
+ * Difficulty: Hard
+ *
+ * You are given an integer array nums and two integers k and numOperations.
+ *
+ * You must perform an operation numOperations times on nums, where in each operation you:
+ * - Select an index i that was not selected in any previous operations.
+ * - Add an integer in the range [-k, k] to nums[i].
+ *
+ * Return the maximum possible frequency of any element in nums after performing the operations.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @param {number} numOperations
+ * @return {number}
+ */
+var maxFrequency = function(nums, k, numOperations) {
+  const n = nums.length;
+  nums.sort((a, b) => a - b);
+
+  const count = new Map();
+  for (const num of nums) {
+    count.set(num, (count.get(num) || 0) + 1);
+  }
+
+  let result = 0;
+  let left = 0;
+  let right = 0;
+  for (let mid = 0; mid < n; mid++) {
+    while (nums[mid] - nums[left] > k) {
+      left++;
+    }
+
+    while (right < n - 1 && nums[right + 1] - nums[mid] <= k) {
+      right++;
+    }
+
+    const total = right - left + 1;
+    result = Math.max(
+      result,
+      Math.min(total - count.get(nums[mid]), numOperations) + count.get(nums[mid])
+    );
+  }
+
+  left = 0;
+  for (right = 0; right < n; right++) {
+    let mid = Math.floor((nums[left] + nums[right]) / 2);
+
+    while (mid - nums[left] > k || nums[right] - mid > k) {
+      left++;
+      mid = Math.floor((nums[left] + nums[right]) / 2);
+    }
+
+    result = Math.max(result, Math.min(right - left + 1, numOperations));
+  }
+
+  return result;
+};