# 49. Shortest Common Supersequence

Given two strings `str1` and `str2`, return the shortest string that has both `str1` and `str2` as subsequences.  If multiple answers exist, you may return any of them.

*(A string S is a subsequence of string T if deleting some number of characters from T (possibly 0, and the characters are chosen anywhere from T) 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.
```

## Solution: (Find the longest common subsequence)

```cpp
class Solution
{
public:
    string findLCS(string &s1, string &s2)
    {
        int n = s1.length();
        int m = s2.length();

        if (n == 0 || m == 0)
        {
            return "";
        }

        vector<vector<string>> dp(n + 1, vector<string>(m + 1, ""));

        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= m; j++)
            {
                if (s1[i - 1] == s2[j - 1])
                {
                    dp[i][j] = dp[i - 1][j - 1] + s1[i - 1];
                }
                else
                {
                    string exci = dp[i - 1][j];
                    string excj = dp[i][j - 1];

                    dp[i][j] = (exci.length()> excj.length()) ? exci : excj;
                }
            }
        }

        return dp[n][m];
    }

    string shortestCommonSupersequence(string &str1, string &str2)
    {
        string lcs = findLCS(str1, str2);

        int k = 0;

        int i = 0, j = 0;

        string res = "";

        while (k < lcs.length())
        {
            while (str1[i] != lcs[k])
            {
                res += str1[i];
                i++;
            }

            while (str2[j] != lcs[k])
            {
                res += str2[j];
                j++;
            }

            res += lcs[k];

            i++;
            j++;
            k++;
        }

        res += str1.substr(i);
        res += str2.substr(j);

        return res;
    }
};
```

**Time Complexity: O(n\* m)**


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