Add solution #220

Author: JoshCrozier

README.md

@@ -154,6 +154,7 @@
 216|[Combination Sum III](./0216-combination-sum-iii.js)|Medium|
 217|[Contains Duplicate](./0217-contains-duplicate.js)|Easy|
 219|[Contains Duplicate II](./0219-contains-duplicate-ii.js)|Easy|
+220|[Contains Duplicate III](./0220-contains-duplicate-iii.js)|Hard|
 225|[Implement Stack using Queues](./0225-implement-stack-using-queues.js)|Easy|
 226|[Invert Binary Tree](./0226-invert-binary-tree.js)|Easy|
 229|[Majority Element II](./0229-majority-element-ii.js)|Medium|

solutions/0220-contains-duplicate-iii.js

@@ -0,0 +1,45 @@
+/**
+ * 220. Contains Duplicate III
+ * https://leetcode.com/problems/contains-duplicate-iii/
+ * Difficulty: Hard
+ *
+ * You are given an integer array nums and two integers indexDiff and valueDiff.
+ *
+ * Find a pair of indices (i, j) such that:
+ * - i != j,
+ * - abs(i - j) <= indexDiff.
+ * - abs(nums[i] - nums[j]) <= valueDiff, and
+ *
+ * Return true if such pair exists or false otherwise.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} indexDiff
+ * @param {number} valueDiff
+ * @return {boolean}
+ */
+var containsNearbyAlmostDuplicate = function(nums, indexDiff, valueDiff) {
+  if (valueDiff < 0) return false;
+
+  const map = new Map();
+
+  for (let i = 0; i < nums.length; i++) {
+    const limit = valueDiff + 1;
+    const item = Math.floor(nums[i] / limit);
+
+    if (map.has(item)
+      || map.has(item - 1) && Math.abs(nums[i] - map.get(item - 1)) < limit
+      || map.has(item + 1) && Math.abs(nums[i] - map.get(item + 1)) < limit) {
+      return true;
+    }
+
+    map.set(item, nums[i]);
+
+    if (i >= indexDiff) {
+      map.delete(Math.floor(nums[i - indexDiff] / limit));
+    }
+  }
+
+  return false;
+};