Add solution #302

Author: JoshCrozier

README.md

@@ -290,6 +290,7 @@
 299|[Bulls and Cows](./solutions/0299-bulls-and-cows.js)|Medium|
 300|[Longest Increasing Subsequence](./solutions/0300-longest-increasing-subsequence.js)|Medium|
 301|[Remove Invalid Parentheses](./solutions/0301-remove-invalid-parentheses.js)|Hard|
+302|[Smallest Rectangle Enclosing Black Pixels](./solutions/0302-smallest-rectangle-enclosing-black-pixels.js)|Hard|
 303|[Range Sum Query - Immutable](./solutions/0303-range-sum-query-immutable.js)|Easy|
 304|[Range Sum Query 2D - Immutable](./solutions/0304-range-sum-query-2d-immutable.js)|Medium|
 306|[Additive Number](./solutions/0306-additive-number.js)|Medium|

solutions/0302-smallest-rectangle-enclosing-black-pixels.js

@@ -0,0 +1,63 @@
+/**
+ * 302. Smallest Rectangle Enclosing Black Pixels
+ * https://leetcode.com/problems/smallest-rectangle-enclosing-black-pixels/
+ * Difficulty: Hard
+ *
+ * You are given an m x n binary matrix image where 0 represents a white pixel and 1 represents
+ * a black pixel.
+ *
+ * The black pixels are connected (i.e., there is only one black region). Pixels are connected
+ * horizontally and vertically.
+ *
+ * Given two integers x and y that represents the location of one of the black pixels, return
+ * the area of the smallest (axis-aligned) rectangle that encloses all black pixels.
+ *
+ * You must write an algorithm with less than O(mn) runtime complexity
+ */
+
+/**
+ * @param {character[][]} image
+ * @param {number} x
+ * @param {number} y
+ * @return {number}
+ */
+var minArea = function(image, x, y) {
+  const rows = image.length;
+  const cols = image[0].length;
+  const top = binarySearch(0, x, true, true);
+  const bottom = binarySearch(x, rows - 1, true, false);
+  const left = binarySearch(0, y, false, true);
+  const right = binarySearch(y, cols - 1, false, false);
+
+  return (bottom - top + 1) * (right - left + 1);
+
+  function binarySearch(start, end, isRow, isMin) {
+    let result = isMin ? end : start;
+    while (start <= end) {
+      const mid = Math.floor((start + end) / 2);
+      let hasBlack = false;
+      for (let i = 0; i < (isRow ? cols : rows); i++) {
+        const pixel = isRow ? image[mid][i] : image[i][mid];
+        if (pixel === '1') {
+          hasBlack = true;
+          break;
+        }
+      }
+      if (hasBlack) {
+        result = mid;
+        if (isMin) {
+          end = mid - 1;
+        } else {
+          start = mid + 1;
+        }
+      } else {
+        if (isMin) {
+          start = mid + 1;
+        } else {
+          end = mid - 1;
+        }
+      }
+    }
+    return result;
+  }
+};