Add solution #1040

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,109 LeetCode solutions in JavaScript
+# 1,110 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -847,6 +847,7 @@
 1037|[Valid Boomerang](./solutions/1037-valid-boomerang.js)|Easy|
 1038|[Binary Search Tree to Greater Sum Tree](./solutions/1038-binary-search-tree-to-greater-sum-tree.js)|Medium|
 1039|[Minimum Score Triangulation of Polygon](./solutions/1039-minimum-score-triangulation-of-polygon.js)|Medium|
+1040|[Moving Stones Until Consecutive II](./solutions/1040-moving-stones-until-consecutive-ii.js)|Medium|
 1041|[Robot Bounded In Circle](./solutions/1041-robot-bounded-in-circle.js)|Medium|
 1047|[Remove All Adjacent Duplicates In String](./solutions/1047-remove-all-adjacent-duplicates-in-string.js)|Easy|
 1051|[Height Checker](./solutions/1051-height-checker.js)|Easy|

solutions/1040-moving-stones-until-consecutive-ii.js

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+/**
+ * 1040. Moving Stones Until Consecutive II
+ * https://leetcode.com/problems/moving-stones-until-consecutive-ii/
+ * Difficulty: Medium
+ *
+ * There are some stones in different positions on the X-axis. You are given an integer array
+ * stones, the positions of the stones.
+ *
+ * Call a stone an endpoint stone if it has the smallest or largest position. In one move, you
+ * pick up an endpoint stone and move it to an unoccupied position so that it is no longer an
+ * endpoint stone.
+ * - In particular, if the stones are at say, stones = [1,2,5], you cannot move the endpoint
+ *   stone at position 5, since moving it to any position (such as 0, or 3) will still keep that
+ *   stone as an endpoint stone.
+ *
+ * The game ends when you cannot make any more moves (i.e., the stones are in three consecutive
+ * positions).
+ *
+ * Return an integer array answer of length 2 where:
+ * - answer[0] is the minimum number of moves you can play, and
+ * - answer[1] is the maximum number of moves you can play.
+ */
+
+/**
+ * @param {number[]} stones
+ * @return {number[]}
+ */
+var numMovesStonesII = function(stones) {
+  stones.sort((a, b) => a - b);
+  const n = stones.length;
+  const maxMoves = Math.max(stones[n - 1] - stones[1] - n + 2, stones[n - 2] - stones[0] - n + 2);
+  let minMoves = n;
+
+  let left = 0;
+  for (let right = 0; right < n; right++) {
+    while (stones[right] - stones[left] + 1 > n) {
+      left++;
+    }
+    const windowSize = right - left + 1;
+    if (windowSize === n - 1 && stones[right] - stones[left] + 1 === n - 1) {
+      minMoves = Math.min(minMoves, 2);
+    } else {
+      minMoves = Math.min(minMoves, n - windowSize);
+    }
+  }
+
+  return [minMoves, maxMoves];
+};