Add solution #1092

Author: JoshCrozier

README.md

@@ -497,6 +497,7 @@
 1071|[Greatest Common Divisor of Strings](./1071-greatest-common-divisor-of-strings.js)|Easy|
 1079|[Letter Tile Possibilities](./1079-letter-tile-possibilities.js)|Medium|
 1081|[Smallest Subsequence of Distinct Characters](./1081-smallest-subsequence-of-distinct-characters.js)|Medium|
+1092|[Shortest Common Supersequence](./1092-shortest-common-supersequence.js)|Hard|
 1103|[Distribute Candies to People](./1103-distribute-candies-to-people.js)|Easy|
 1108|[Defanging an IP Address](./1108-defanging-an-ip-address.js)|Easy|
 1122|[Relative Sort Array](./1122-relative-sort-array.js)|Easy|

solutions/1092-shortest-common-supersequence.js

@@ -0,0 +1,50 @@
+/**
+ * 1092. Shortest Common Supersequence
+ * https://leetcode.com/problems/shortest-common-supersequence/
+ * Difficulty: Hard
+ *
+ * Given two strings str1 and str2, return the shortest string that has both str1 and str2 as
+ * subsequences. If there are multiple valid strings, return any of them.
+ *
+ * A string s is a subsequence of string t if deleting some number of characters from t (possibly
+ * 0) results in the string s.
+ */
+
+/**
+ * @param {string} str1
+ * @param {string} str2
+ * @return {string}
+ */
+var shortestCommonSupersequence = function(str1, str2) {
+  const dp = new Array(str1.length + 1).fill().map(() => new Array(str2.length + 1).fill(0));
+  let result = '';
+
+  for (let i = 0; i <= str1.length; i++) {
+    dp[i][0] = i;
+  }
+  for (let j = 0; j <= str2.length; j++) {
+    dp[0][j] = j;
+  }
+
+  for (let i = 1; i <= str1.length; i++) {
+    for (let j = 1; j <= str2.length; j++) {
+      dp[i][j] = str1[i-1] === str2[j-1] ? dp[i-1][j-1] + 1 : Math.min(dp[i-1][j], dp[i][j-1]) + 1;
+    }
+  }
+
+  for (let i = str1.length, j = str2.length; i > 0 || j > 0;) {
+    if (!i) {
+      result = str2[--j] + result;
+    } else if (!j) {
+      result = str1[--i] + result;
+    } else if (str1[i-1] === str2[j-1]) {
+      result = str1[--i] + (str2[--j], result);
+    } else {
+      dp[i][j] === dp[i-1][j] + 1
+        ? result = str1[--i] + result
+        : result = str2[--j] + result;
+    }
+  }
+
+  return result;
+};