Add solution #839

Author: JoshCrozier

README.md

@@ -646,6 +646,7 @@
 836|[Rectangle Overlap](./0836-rectangle-overlap.js)|Easy|
 837|[New 21 Game](./0837-new-21-game.js)|Medium|
 838|[Push Dominoes](./0838-push-dominoes.js)|Medium|
+839|[Similar String Groups](./0839-similar-string-groups.js)|Hard|
 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/0839-similar-string-groups.js

@@ -0,0 +1,71 @@
+/**
+ * 839. Similar String Groups
+ * https://leetcode.com/problems/similar-string-groups/
+ * Difficulty: Hard
+ *
+ * Two strings, X and Y, are considered similar if either they are identical or we can make them
+ * equivalent by swapping at most two letters (in distinct positions) within the string X.
+ *
+ * For example, "tars" and "rats" are similar (swapping at positions 0 and 2), and "rats" and
+ * "arts" are similar, but "star" is not similar to "tars", "rats", or "arts".
+ *
+ * Together, these form two connected groups by similarity: {"tars", "rats", "arts"} and {"star"}.
+ * Notice that "tars" and "arts" are in the same group even though they are not similar. Formally,
+ * each group is such that a word is in the group if and only if it is similar to at least one
+ * other word in the group.
+ *
+ * We are given a list strs of strings where every string in strs is an anagram of every other
+ * string in strs. How many groups are there?
+ */
+
+/**
+ * @param {string[]} strs
+ * @return {number}
+ */
+var numSimilarGroups = function(strs) {
+  const parent = Array.from({ length: strs.length }, (_, i) => i);
+  const rank = new Array(strs.length).fill(0);
+
+  for (let i = 0; i < strs.length; i++) {
+    for (let j = i + 1; j < strs.length; j++) {
+      if (areSimilar(strs[i], strs[j])) union(i, j);
+    }
+  }
+
+  const groups = new Set();
+  for (let i = 0; i < strs.length; i++) groups.add(find(i));
+
+  return groups.size;
+
+  function find(x) {
+    if (parent[x] !== x) {
+      parent[x] = find(parent[x]);
+    }
+    return parent[x];
+  }
+
+  function union(x, y) {
+    const rootX = find(x);
+    const rootY = find(y);
+
+    if (rootX === rootY) return;
+
+    if (rank[rootX] < rank[rootY]) {
+      parent[rootX] = rootY;
+    } else {
+      parent[rootY] = rootX;
+      if (rank[rootX] === rank[rootY]) rank[rootX]++;
+    }
+  }
+
+  function areSimilar(a, b) {
+    if (a === b) return true;
+
+    let diff = 0;
+    for (let i = 0; i < a.length; i++) {
+      if (a[i] !== b[i] && ++diff > 2) return false;
+    }
+
+    return diff === 2;
+  }
+};