Add solution #1289

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,211 LeetCode solutions in JavaScript
+# 1,212 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -973,6 +973,7 @@
 1286|[Iterator for Combination](./solutions/1286-iterator-for-combination.js)|Medium|
 1287|[Element Appearing More Than 25% In Sorted Array](./solutions/1287-element-appearing-more-than-25-in-sorted-array.js)|Easy|
 1288|[Remove Covered Intervals](./solutions/1288-remove-covered-intervals.js)|Medium|
+1289|[Minimum Falling Path Sum II](./solutions/1289-minimum-falling-path-sum-ii.js)|Hard|
 1290|[Convert Binary Number in a Linked List to Integer](./solutions/1290-convert-binary-number-in-a-linked-list-to-integer.js)|Easy|
 1291|[Sequential Digits](./solutions/1291-sequential-digits.js)|Medium|
 1292|[Maximum Side Length of a Square with Sum Less than or Equal to Threshold](./solutions/1292-maximum-side-length-of-a-square-with-sum-less-than-or-equal-to-threshold.js)|Medium|

solutions/1289-minimum-falling-path-sum-ii.js

@@ -0,0 +1,36 @@
+/**
+ * 1289. Minimum Falling Path Sum II
+ * https://leetcode.com/problems/minimum-falling-path-sum-ii/
+ * Difficulty: Hard
+ *
+ * Given an n x n integer matrix grid, return the minimum sum of a falling path with non-zero
+ * shifts.
+ *
+ * A falling path with non-zero shifts is a choice of exactly one element from each row of grid
+ * such that no two elements chosen in adjacent rows are in the same column.
+ */
+
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var minFallingPathSum = function(grid) {
+  const n = grid.length;
+  let prevRow = [...grid[0]];
+
+  for (let row = 1; row < n; row++) {
+    const currRow = new Array(n);
+    for (let col = 0; col < n; col++) {
+      let minPrev = Infinity;
+      for (let prevCol = 0; prevCol < n; prevCol++) {
+        if (prevCol !== col) {
+          minPrev = Math.min(minPrev, prevRow[prevCol]);
+        }
+      }
+      currRow[col] = grid[row][col] + minPrev;
+    }
+    prevRow = currRow;
+  }
+
+  return Math.min(...prevRow);
+};