Add solution #1139

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,148 LeetCode solutions in JavaScript
+# 1,149 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -895,6 +895,7 @@
 1131|[Maximum of Absolute Value Expression](./solutions/1131-maximum-of-absolute-value-expression.js)|Medium|
 1137|[N-th Tribonacci Number](./solutions/1137-n-th-tribonacci-number.js)|Easy|
 1138|[Alphabet Board Path](./solutions/1138-alphabet-board-path.js)|Medium|
+1139|[Largest 1-Bordered Square](./solutions/1139-largest-1-bordered-square.js)|Medium|
 1143|[Longest Common Subsequence](./solutions/1143-longest-common-subsequence.js)|Medium|
 1161|[Maximum Level Sum of a Binary Tree](./solutions/1161-maximum-level-sum-of-a-binary-tree.js)|Medium|
 1189|[Maximum Number of Balloons](./solutions/1189-maximum-number-of-balloons.js)|Easy|

solutions/1139-largest-1-bordered-square.js

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+/**
+ * 1139. Largest 1-Bordered Square
+ * https://leetcode.com/problems/largest-1-bordered-square/
+ * Difficulty: Medium
+ *
+ * Given a 2D grid of 0s and 1s, return the number of elements in the largest square subgrid
+ * that has all 1s on its border, or 0 if such a subgrid doesn't exist in the grid.
+ */
+
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var largest1BorderedSquare = function(grid) {
+  const rows = grid.length;
+  const cols = grid[0].length;
+  const left = Array(rows).fill().map(() => Array(cols).fill(0));
+  const top = Array(rows).fill().map(() => Array(cols).fill(0));
+  let maxSide = 0;
+
+  for (let i = 0; i < rows; i++) {
+    for (let j = 0; j < cols; j++) {
+      if (grid[i][j]) {
+        left[i][j] = j > 0 ? left[i][j - 1] + 1 : 1;
+        top[i][j] = i > 0 ? top[i - 1][j] + 1 : 1;
+
+        let side = Math.min(left[i][j], top[i][j]);
+        while (side > maxSide) {
+          if (left[i - side + 1][j] >= side && top[i][j - side + 1] >= side) {
+            maxSide = side;
+            break;
+          }
+          side--;
+        }
+      }
+    }
+  }
+
+  return maxSide * maxSide;
+};