Add solution #632

Author: JoshCrozier

README.md

@@ -477,6 +477,7 @@
 628|[Maximum Product of Three Numbers](./0628-maximum-product-of-three-numbers.js)|Easy|
 629|[K Inverse Pairs Array](./0629-k-inverse-pairs-array.js)|Hard|
 630|[Course Schedule III](./0630-course-schedule-iii.js)|Hard|
+632|[Smallest Range Covering Elements from K Lists](./0632-smallest-range-covering-elements-from-k-lists.js)|Hard|
 633|[Sum of Square Numbers](./0633-sum-of-square-numbers.js)|Medium|
 636|[Exclusive Time of Functions](./0636-exclusive-time-of-functions.js)|Medium|
 637|[Average of Levels in Binary Tree](./0637-average-of-levels-in-binary-tree.js)|Easy|

solutions/0632-smallest-range-covering-elements-from-k-lists.js

@@ -0,0 +1,56 @@
+/**
+ * 632. Smallest Range Covering Elements from K Lists
+ * https://leetcode.com/problems/smallest-range-covering-elements-from-k-lists/
+ * Difficulty: Hard
+ *
+ * You have k lists of sorted integers in non-decreasing order. Find the smallest range that
+ * includes at least one number from each of the k lists.
+ *
+ * We define the range [a, b] is smaller than range [c, d] if b - a < d - c or a < c
+ * if b - a == d - c.
+ */
+
+/**
+ * @param {number[][]} nums
+ * @return {number[]}
+ */
+var smallestRange = function(nums) {
+  const k = nums.length;
+  const listCountMap = new Map();
+  let coveredLists = 0;
+  let minDiff = Infinity;
+  let minStart;
+  let minEnd;
+
+  const allElements = nums.flatMap((list, listIndex) =>
+    list.map(num => ({ num, index: listIndex }))
+  ).sort((a, b) => a.num - b.num);
+
+  let windowStart = 0;
+  for (let windowEnd = 0; windowEnd < allElements.length; windowEnd++) {
+    const currentElement = allElements[windowEnd];
+    const listIndex = currentElement.index;
+    const count = listCountMap.get(listIndex) ?? 0;
+
+    if (count === 0) coveredLists += 1;
+    listCountMap.set(listIndex, count + 1);
+
+    while (coveredLists === k) {
+      const leftElement = allElements[windowStart];
+      const range = currentElement.num - leftElement.num;
+
+      if (range < minDiff) {
+        minDiff = range;
+        minStart = leftElement.num;
+        minEnd = currentElement.num;
+      }
+      const leftListIndex = leftElement.index;
+      const leftCount = listCountMap.get(leftListIndex);
+      listCountMap.set(leftListIndex, leftCount - 1);
+      if (leftCount - 1 === 0) coveredLists -= 1;
+      windowStart += 1;
+    }
+  }
+
+  return [minStart, minEnd];
+};