Shortest Common Supersequence

Problem statement

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.

Example 1:

Input: str1 = "abac", str2 = "cab"Output: "cabac"Explanation: str1 = "abac" is a subsequence of "cabac" because we can delete the first "c".str2 = "cab" is a subsequence of "cabac" because we can delete the last "ac".The answer provided is the shortest such string that satisfies these properties.

Example 2:

Input: str1 = "aaaaaaaa", str2 = "aaaaaaaa"Output: "aaaaaaaa"

Constraints:

• 1 <= str1.length, str2.length <= 1000
• str1 and str2 consist of lowercase English letters.

My solution

/**
 * @param {string} str1
 * @param {string} str2
 * @return {string}
 */
var shortestCommonSupersequence = function(str1, str2) {
    const lcs = findLongestSubSequence(str1, str2)
    
    if (lcs.length === 1 && lcs[0] === "") {
        return set1.concat(set2)
    }
    
    let subSequence = [];
    
    let str1Index = 0;
    let str2Index = 0;
    let lcsIndex = 0;
    let str1OnHold = false;
    let str2OnHold = false;
    
    while (lcsIndex < lcs.length) {
        if (str1Index < str1.length) {
            if (!str1OnHold && lcs[lcsIndex] !== str1[str1Index]) {
                subSequence.push(str1[str1Index]);
                str1Index++
            } else {
                str1OnHold = true;
            }
        }
        
        if (str2Index < str2.length) {
            if (!str2OnHold && lcs[lcsIndex] !== str2[str2Index]) {
                subSequence.push(str2[str2Index]);
                str2Index++
            } else {
                str2OnHold = true;
            }
        }
        
        if (str1OnHold && str2OnHold) {
            subSequence.push(lcs[lcsIndex]);
            lcsIndex++
            str1Index++
            str2Index++
            str1OnHold = false;
            str2OnHold = false
        }
    }
    
    // console.log(str1Index, str2Index)
    
    if (str1Index < str1.length) {
        subSequence = subSequence.concat(str1.slice(str1Index))
    }
    
    if (str2Index < str2.length) {
        subSequence = subSequence.concat(str2.slice(str2Index))
    }
    
    return subSequence.join("")
};

function findLongestSubSequence(str1, str2) {
    const lcsMatrix = Array.from({
        length: str2.length + 1
    }, () => Array.from({
        length: str1.length + 1
    }, () => null))
    
    for (let i = 0; i <= str1.length; i++) {
        lcsMatrix[0][i] = 0
    }
    
    for (let i = 0; i <= str2.length; i++) {
        lcsMatrix[i][0] = 0
    }
    
    for (let rowIdx = 1; rowIdx <= str2.length; rowIdx++) {
        for (let colIdx = 1; colIdx <= str1.length; colIdx++) {
            if (str1[colIdx - 1] === str2[rowIdx - 1]) {
                lcsMatrix[rowIdx][colIdx] = lcsMatrix[rowIdx - 1][colIdx - 1] + 1
            } else {
                lcsMatrix[rowIdx][colIdx] = Math.max(
                    lcsMatrix[rowIdx - 1][colIdx],
                    lcsMatrix[rowIdx][colIdx - 1]
                )
            }
        }
    }
    
    if (lcsMatrix[str2.length][str1.length] === 0) {
        return ""
    }
    
    const longestSequence = [];
    let colIdx = str1.length;
    let rowIdx = str2.length;
    
    while (colIdx > 0 || rowIdx > 0) {
        if (str1[colIdx - 1] === str2[rowIdx - 1]) {
            longestSequence.unshift(str1[colIdx - 1])
            colIdx--
            rowIdx--
        } else if (lcsMatrix[rowIdx][colIdx] === lcsMatrix[rowIdx][colIdx - 1]) {
            colIdx--
        } else {
            rowIdx--
        }
    }
    
    return longestSequence.join("")
}