Add SCS.

Author: trekhleb

README.md

@@ -39,6 +39,7 @@
   * [Fisher–Yates Shuffle](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/fisher-yates) - random permutation of a finite sequence
   * [Longest Common Subsequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/longest-common-subsequnce) (LCS) 
   * [Longest Increasing subsequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/longest-increasing-subsequence)
+  * [Shortest Common Supersequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/shortest-common-supersequence) (SCS)
 * **String**
   * [Levenshtein Distance](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/levenshtein-distance) - minimum edit distance between two sequences
   * [Hamming Distance](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/hamming-distance) - number of positions at which the symbols are different
@@ -92,7 +93,7 @@
   * [Longest Common Subsequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/longest-common-subsequnce) (LCS)
   * [Longest Common Substring](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/longest-common-substring)
   * [Longest Increasing subsequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/longest-increasing-subsequence)
-  * Shortest common supersequence
+  * [Shortest Common Supersequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/shortest-common-supersequence)
   * Knapsack problem
   * Maximum subarray
   * Maximum sum path

src/algorithms/sets/shortest-common-supersequence/README.md

@@ -0,0 +1,24 @@
+# Shortest Common Supersequence
+
+The shortest common supersequence (SCS) of two sequences `X` and `Y` 
+is the shortest sequence which has `X` and `Y` as subsequences.
+
+In other words assume we're given two strings str1 and str2, find 
+the shortest string that has both str1 and str2 as subsequences.
+
+This is a problem closely related to the longest common 
+subsequence problem.
+
+## Example
+
+```
+Input:   str1 = "geek",  str2 = "eke"
+Output: "geeke"
+
+Input:   str1 = "AGGTAB",  str2 = "GXTXAYB"
+Output:  "AGXGTXAYB"
+```
+
+## References
+
+- [GeeksForGeeks](https://www.geeksforgeeks.org/shortest-common-supersequence/)

src/algorithms/sets/shortest-common-supersequence/__test__/shortestCommonSupersequence.test.js

@@ -0,0 +1,35 @@
+import shortestCommonSupersequence from '../shortestCommonSupersequence';
+
+describe('shortestCommonSupersequence', () => {
+  it('should find shortest common supersequence of two sequences', () => {
+    // LCS (longest common subsequence) is empty
+    expect(shortestCommonSupersequence(
+      ['A', 'B', 'C'],
+      ['D', 'E', 'F'],
+    )).toEqual(['A', 'B', 'C', 'D', 'E', 'F']);
+
+    // LCS (longest common subsequence) is "EE"
+    expect(shortestCommonSupersequence(
+      ['G', 'E', 'E', 'K'],
+      ['E', 'K', 'E'],
+    )).toEqual(['G', 'E', 'K', 'E', 'K']);
+
+    // LCS (longest common subsequence) is "GTAB"
+    expect(shortestCommonSupersequence(
+      ['A', 'G', 'G', 'T', 'A', 'B'],
+      ['G', 'X', 'T', 'X', 'A', 'Y', 'B'],
+    )).toEqual(['A', 'G', 'G', 'X', 'T', 'X', 'A', 'Y', 'B']);
+
+    // LCS (longest common subsequence) is "BCBA".
+    expect(shortestCommonSupersequence(
+      ['A', 'B', 'C', 'B', 'D', 'A', 'B'],
+      ['B', 'D', 'C', 'A', 'B', 'A'],
+    )).toEqual(['A', 'B', 'D', 'C', 'A', 'B', 'D', 'A', 'B']);
+
+    // LCS (longest common subsequence) is "BDABA".
+    expect(shortestCommonSupersequence(
+      ['B', 'D', 'C', 'A', 'B', 'A'],
+      ['A', 'B', 'C', 'B', 'D', 'A', 'B', 'A', 'C'],
+    )).toEqual(['A', 'B', 'C', 'B', 'D', 'C', 'A', 'B', 'A', 'C']);
+  });
+});

src/algorithms/sets/shortest-common-supersequence/shortestCommonSupersequence.js

@@ -0,0 +1,71 @@
+import longestCommonSubsequnce from '../longest-common-subsequnce/longestCommonSubsequnce';
+
+/**
+ * @param {string[]} set1
+ * @param {string[]} set2
+ * @return {string[]}
+ */
+
+export default function shortestCommonSupersequence(set1, set2) {
+  // Let's first find the longest common subsequence of two sets.
+  const lcs = longestCommonSubsequnce(set1, set2);
+
+  // If LCS is empty then the shortest common supersequnce would be just
+  // concatenation of two sequences.
+  if (lcs.length === 1 && lcs[0] === '') {
+    return set1.concat(set2);
+  }
+
+  // Now let's add elements of set1 and set2 in order before/inside/after the LCS.
+  let supersequence = [];
+
+  let setIndex1 = 0;
+  let setIndex2 = 0;
+  let lcsIndex = 0;
+  let setOnHold1 = false;
+  let setOnHold2 = false;
+
+  while (lcsIndex < lcs.length) {
+    // Add elements of the first set to supersequence in correct order.
+    if (setIndex1 < set1.length) {
+      if (!setOnHold1 && set1[setIndex1] !== lcs[lcsIndex]) {
+        supersequence.push(set1[setIndex1]);
+        setIndex1 += 1;
+      } else {
+        setOnHold1 = true;
+      }
+    }
+
+    // Add elements of the second set to supersequence in correct order.
+    if (setIndex2 < set2.length) {
+      if (!setOnHold2 && set2[setIndex2] !== lcs[lcsIndex]) {
+        supersequence.push(set2[setIndex2]);
+        setIndex2 += 1;
+      } else {
+        setOnHold2 = true;
+      }
+    }
+
+    // Add LCS element to the supersequence in correct order.
+    if (setOnHold1 && setOnHold2) {
+      supersequence.push(lcs[lcsIndex]);
+      lcsIndex += 1;
+      setIndex1 += 1;
+      setIndex2 += 1;
+      setOnHold1 = false;
+      setOnHold2 = false;
+    }
+  }
+
+  // Attach set1 leftovers.
+  if (setIndex1 < set1.length) {
+    supersequence = supersequence.concat(set1.slice(setIndex1));
+  }
+
+  // Attach set2 leftovers.
+  if (setIndex2 < set2.length) {
+    supersequence = supersequence.concat(set2.slice(setIndex2));
+  }
+
+  return supersequence;
+}