Add solution #957

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,042 LeetCode solutions in JavaScript
+# 1,043 LeetCode solutions in JavaScript
 
 [https://leetcode.com/](https://leetcode.com/)
 
@@ -766,6 +766,7 @@
 954|[Array of Doubled Pairs](./solutions/0954-array-of-doubled-pairs.js)|Medium|
 955|[Delete Columns to Make Sorted II](./solutions/0955-delete-columns-to-make-sorted-ii.js)|Medium|
 956|[Tallest Billboard](./solutions/0956-tallest-billboard.js)|Hard|
+957|[Prison Cells After N Days](./solutions/0957-prison-cells-after-n-days.js)|Medium|
 966|[Vowel Spellchecker](./solutions/0966-vowel-spellchecker.js)|Medium|
 970|[Powerful Integers](./solutions/0970-powerful-integers.js)|Easy|
 976|[Largest Perimeter Triangle](./solutions/0976-largest-perimeter-triangle.js)|Easy|

solutions/0957-prison-cells-after-n-days.js

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+/**
+ * 957. Prison Cells After N Days
+ * https://leetcode.com/problems/prison-cells-after-n-days/
+ * Difficulty: Medium
+ *
+ * There are 8 prison cells in a row and each cell is either occupied or vacant.
+ *
+ * Each day, whether the cell is occupied or vacant changes according to the following rules:
+ * - If a cell has two adjacent neighbors that are both occupied or both vacant, then the cell
+ *   becomes occupied.
+ * - Otherwise, it becomes vacant.
+ *
+ * Note that because the prison is a row, the first and the last cells in the row can't have two
+ * adjacent neighbors.
+ *
+ * You are given an integer array cells where cells[i] == 1 if the ith cell is occupied and
+ * cells[i] == 0 if the ith cell is vacant, and you are given an integer n.
+ *
+ * Return the state of the prison after n days (i.e., n such changes described above).
+ */
+
+/**
+ * @param {number[]} cells
+ * @param {number} n
+ * @return {number[]}
+ */
+var prisonAfterNDays = function(cells, n) {
+  const nextState = cells => [
+    0,
+    ...Array.from({ length: 6 }, (_, i) => cells[i] === cells[i + 2] ? 1 : 0),
+    0
+  ];
+
+  let current = [...cells];
+  const seen = new Map();
+  let cycleLength = 0;
+
+  while (n > 0) {
+    const stateKey = current.join('');
+    if (seen.has(stateKey)) {
+      const cycleStart = seen.get(stateKey);
+      cycleLength = cycleStart - n;
+      n %= cycleLength;
+    }
+    seen.set(stateKey, n);
+    if (n > 0) {
+      current = nextState(current);
+      n--;
+    }
+  }
+
+  return current;
+};