Add solution #2022

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,691 LeetCode solutions in JavaScript
+# 1,692 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1550,6 +1550,7 @@
 2017|[Grid Game](./solutions/2017-grid-game.js)|Medium|
 2018|[Check if Word Can Be Placed In Crossword](./solutions/2018-check-if-word-can-be-placed-in-crossword.js)|Medium|
 2019|[The Score of Students Solving Math Expression](./solutions/2019-the-score-of-students-solving-math-expression.js)|Hard|
+2022|[Convert 1D Array Into 2D Array](./solutions/2022-convert-1d-array-into-2d-array.js)|Easy|
 2027|[Minimum Moves to Convert String](./solutions/2027-minimum-moves-to-convert-string.js)|Easy|
 2033|[Minimum Operations to Make a Uni-Value Grid](./solutions/2033-minimum-operations-to-make-a-uni-value-grid.js)|Medium|
 2037|[Minimum Number of Moves to Seat Everyone](./solutions/2037-minimum-number-of-moves-to-seat-everyone.js)|Easy|

solutions/2022-convert-1d-array-into-2d-array.js

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+/**
+ * 2022. Convert 1D Array Into 2D Array
+ * https://leetcode.com/problems/convert-1d-array-into-2d-array/
+ * Difficulty: Easy
+ *
+ * You are given a 0-indexed 1-dimensional (1D) integer array original, and two integers,
+ * m and n. You are tasked with creating a 2-dimensional (2D) array with  m rows and n
+ * columns using all the elements from original.
+ *
+ * The elements from indices 0 to n - 1 (inclusive) of original should form the first row
+ * of the constructed 2D array, the elements from indices n to 2 * n - 1 (inclusive) should
+ * form the second row of the constructed 2D array, and so on.
+ *
+ * Return an m x n 2D array constructed according to the above procedure, or an empty 2D array
+ * if it is impossible.
+ */
+
+/**
+ * @param {number[]} original
+ * @param {number} m
+ * @param {number} n
+ * @return {number[][]}
+ */
+var construct2DArray = function(original, m, n) {
+  if (original.length !== m * n) return [];
+
+  const result = [];
+  for (let i = 0; i < m; i++) {
+    result.push(original.slice(i * n, (i + 1) * n));
+  }
+
+  return result;
+};