28_Shortest_Common_Supersequence.cpp

// 1092. Shortest Common Supersequence

// 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.

class Solution
{
public:
    string shortestCommonSupersequence(string str1, string str2)
    {
        int m = str1.length(), n = str2.length();
        vector<vector<int>> lcs(m + 1, vector<int>(n + 1, 0));

        // Compute LCS table
        for (int i = 1; i <= m; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                if (str1[i - 1] == str2[j - 1])
                {
                    lcs[i][j] = 1 + lcs[i - 1][j - 1];
                }
                else
                {
                    lcs[i][j] = max(lcs[i - 1][j], lcs[i][j - 1]);
                }
            }
        }

        // Build the shortest supersequence
        int i = m, j = n;
        string result = "";

        while (i > 0 && j > 0)
        {
            if (str1[i - 1] == str2[j - 1])
            {
                result += str1[i - 1];
                i--, j--;
            }
            else if (lcs[i - 1][j] > lcs[i][j - 1])
            {
                result += str1[i - 1];
                i--;
            }
            else
            {
                result += str2[j - 1];
                j--;
            }
        }

        while (i > 0)
            result += str1[--i];
        while (j > 0)
            result += str2[--j];

        reverse(result.begin(), result.end());
        return result;
    }
};

/*
This code finds the shortest common supersequence of two strings using dynamic programming:
1. First creates a LCS (Longest Common Subsequence) table using 2D vector
2. Fills the table using dynamic programming where each cell represents length of LCS up to that point
3. After table is built, constructs the supersequence by:
   - Starting from bottom-right of table
   - If characters match, add once and move diagonally
   - If different, take character from string with larger LCS value and move in that direction
   - If remaining characters in either string, add them
4. Finally reverses the result since characters were added from end to beginning
The algorithm ensures the result contains both input strings as subsequences while being as short as possible.
*/