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Oleksii Trekhleb 2018-04-27 08:19:11 +03:00
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* [FisherYates Shuffle](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/fisher-yates) - random permutation of a finite sequence * [FisherYates 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 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) * [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** * **String**
* [Levenshtein Distance](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/levenshtein-distance) - minimum edit distance between two sequences * [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 * [Hamming Distance](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/hamming-distance) - number of positions at which the symbols are different
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* [Longest Common Subsequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/longest-common-subsequnce) (LCS) * [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 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) * [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 * Knapsack problem
* Maximum subarray * Maximum subarray
* Maximum sum path * Maximum sum path

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# 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/)

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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']);
});
});

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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;
}