<p>Given two strings <code>str1</code> and <code>str2</code>, return <em>the shortest string that has both </em><code>str1</code><em> and </em><code>str2</code><em> as <strong>subsequences</strong></em>. If there are multiple valid strings, return <strong>any</strong> of them.</p>

<p>A string <code>s</code> is a <strong>subsequence</strong> of string <code>t</code> if deleting some number of characters from <code>t</code> (possibly <code>0</code>) results in the string <code>s</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> str1 = &quot;abac&quot;, str2 = &quot;cab&quot;
<strong>Output:</strong> &quot;cabac&quot;
<strong>Explanation:</strong> 
str1 = &quot;abac&quot; is a subsequence of &quot;cabac&quot; because we can delete the first &quot;c&quot;.
str2 = &quot;cab&quot; is a subsequence of &quot;cabac&quot; because we can delete the last &quot;ac&quot;.
The answer provided is the shortest such string that satisfies these properties.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> str1 = &quot;aaaaaaaa&quot;, str2 = &quot;aaaaaaaa&quot;
<strong>Output:</strong> &quot;aaaaaaaa&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= str1.length, str2.length &lt;= 1000</code></li>
	<li><code>str1</code> and <code>str2</code> consist of lowercase English letters.</li>
</ul>

class Solution:
    def shortestCommonSupersequence(self, str1: str, str2: str) -> str:
        dp = [[""] * (len(str2)+1) for _ in range(len(str1)+1)]
        
        for i in range(len(str1)-1, -1, -1):
            for j in range(len(str2)-1, -1, -1):
                if str1[i] == str2[j]:
                    dp[i][j] = str1[i] + dp[i+1][j+1]
                else:
                    if len(dp[i+1][j]) > len(dp[i][j+1]):
                        dp[i][j] = dp[i+1][j]
                    else:
                        dp[i][j] = dp[i][j+1]
        x = dp[0][0]
        ans = ""
        i, j = 0, 0
        for c in x:
            while str1[i] != c:
                ans += str1[i]
                i += 1
            while str2[j] != c:
                ans += str2[j]
                j += 1
            ans += c
            i += 1
            j += 1
        return ans + str1[i:] + str2[j:]






































        # dp = [[""] * (len(str2)+1) for _ in range(len(str1) +1)]

        # for i in range(len(str1)-1, -1, -1):
        #     for j in range(len(str2)-1, -1, -1):
        #         if str1[i] == str2[j]:
        #             dp[i][j] = str1[i] + dp[i+1][j+1]
        #         else:
        #             if len(dp[i+1][j]) > len(dp[i][j+1]):
        #                 dp[i][j] = dp[i+1][j]
        #             else:
        #                 dp[i][j] = dp[i][j+1]
        
        # x = dp[0][0]
        # # print(x)
        # ans, i, j = "", 0, 0
        # for c in x:
        #     while c != str1[i]:
        #         ans += str1[i]
        #         i += 1
        #     while c != str2[j]:
        #         ans += str2[j]
        #         j += 1
        #     ans += c
        #     i += 1
        #     j += 1
        # return ans + str1[i:] + str2[j:]











































        # #find longest substring X and len(str1) + len(str2) - X

        # dp = [[""] * (len(str2)+1) for _ in range((len(str1)+1))]
        
        # for i in range(len(str1)-1, -1, -1):
        #     for j in range(len(str2)-1, -1, -1):
        #         if str1[i] == str2[j]:
        #             dp[i][j] = str1[i] + dp[i+1][j+1]
        #         else:
        #             if len(dp[i+1][j]) >= len(dp[i][j+1]):
        #                 dp[i][j] = dp[i+1][j]
        #             else:
        #                 dp[i][j] = dp[i][j+1]
        # #longest common string
        # LCS = dp[0][0]
        # # print(dp)
        # ans, i, j = "", 0, 0
        # for c in LCS:
        #     while str1[i] != c:
        #         ans += str1[i]
        #         i += 1
        #     while str2[j] != c:
        #         ans += str2[j]
        #         j += 1
        #     ans += c
        #     i, j = i+1, j+1

        # # print(ans, LCS, x, str1[i:], x, str2[j:])
        # #ans = 같은 부분까지는 합치고, str1[i:] + str2[j:] 남은 부분은 더한다.
        # return ans + str1[i:] + str2[j:]

        
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