Add solution #835

Author: JoshCrozier

README.md

@@ -642,6 +642,7 @@
 831|[Masking Personal Information](./0831-masking-personal-information.js)|Medium|
 833|[Find And Replace in String](./0833-find-and-replace-in-string.js)|Medium|
 834|[Sum of Distances in Tree](./0834-sum-of-distances-in-tree.js)|Hard|
+835|[Image Overlap](./0835-image-overlap.js)|Medium|
 841|[Keys and Rooms](./0841-keys-and-rooms.js)|Medium|
 844|[Backspace String Compare](./0844-backspace-string-compare.js)|Easy|
 846|[Hand of Straights](./0846-hand-of-straights.js)|Medium|

solutions/0835-image-overlap.js

@@ -0,0 +1,60 @@
+/**
+ * 835. Image Overlap
+ * https://leetcode.com/problems/image-overlap/
+ * Difficulty: Medium
+ *
+ * You are given two images, img1 and img2, represented as binary, square matrices of size n x n.
+ * A binary matrix has only 0s and 1s as values.
+ *
+ * We translate one image however we choose by sliding all the 1 bits left, right, up, and/or down
+ * any number of units. We then place it on top of the other image. We can then calculate the
+ * overlap by counting the number of positions that have a 1 in both images.
+ *
+ * Note also that a translation does not include any kind of rotation. Any 1 bits that are
+ * translated outside of the matrix borders are erased.
+ *
+ * Return the largest possible overlap.
+ */
+
+/**
+ * @param {number[][]} img1
+ * @param {number[][]} img2
+ * @return {number}
+ */
+var largestOverlap = function(img1, img2) {
+  const positions1 = [];
+  const positions2 = [];
+
+  for (let i = 0; i < img1.length; i++) {
+    for (let j = 0; j < img1.length; j++) {
+      if (img1[i][j] === 1) {
+        positions1.push([i, j]);
+      }
+      if (img2[i][j] === 1) {
+        positions2.push([i, j]);
+      }
+    }
+  }
+
+  if (positions1.length === 0 || positions2.length === 0) {
+    return 0;
+  }
+
+  const translations = new Map();
+  let maxOverlap = 0;
+
+  for (const [r1, c1] of positions1) {
+    for (const [r2, c2] of positions2) {
+      const translation = `${r2 - r1},${c2 - c1}`;
+
+      if (!translations.has(translation)) {
+        translations.set(translation, 0);
+      }
+
+      translations.set(translation, translations.get(translation) + 1);
+      maxOverlap = Math.max(maxOverlap, translations.get(translation));
+    }
+  }
+
+  return maxOverlap;
+};