<p>Given two strings <code>s1</code> and&nbsp;<code>s2</code>, return <em>the lowest <strong>ASCII</strong> sum of deleted characters to make two strings equal</em>.</p>

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

<pre>
<strong>Input:</strong> s1 = &quot;sea&quot;, s2 = &quot;eat&quot;
<strong>Output:</strong> 231
<strong>Explanation:</strong> Deleting &quot;s&quot; from &quot;sea&quot; adds the ASCII value of &quot;s&quot; (115) to the sum.
Deleting &quot;t&quot; from &quot;eat&quot; adds 116 to the sum.
At the end, both strings are equal, and 115 + 116 = 231 is the minimum sum possible to achieve this.
</pre>

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

<pre>
<strong>Input:</strong> s1 = &quot;delete&quot;, s2 = &quot;leet&quot;
<strong>Output:</strong> 403
<strong>Explanation:</strong> Deleting &quot;dee&quot; from &quot;delete&quot; to turn the string into &quot;let&quot;,
adds 100[d] + 101[e] + 101[e] to the sum.
Deleting &quot;e&quot; from &quot;leet&quot; adds 101[e] to the sum.
At the end, both strings are equal to &quot;let&quot;, and the answer is 100+101+101+101 = 403.
If instead we turned both strings into &quot;lee&quot; or &quot;eet&quot;, we would get answers of 433 or 417, which are higher.
</pre>

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

<ul>
	<li><code>1 &lt;= s1.length, s2.length &lt;= 1000</code></li>
	<li><code>s1</code> and <code>s2</code> consist of lowercase English letters.</li>
</ul>

class Solution {
public:
    int minimumDeleteSum(string s1, string s2) {
        std::vector dp(s1.size(), std::vector(s2.size(), INT_MAX));
        std::function<int(int, int)> dpf = [&s1, &s2, &dp, &dpf](int x1, int x2) -> int{
            if(x1 == s1.size() && x2 == s2.size()){
                return 0;
            }
            if(x1 == s1.size()){
                return s2[x2] + dpf(x1, x2+1);
            }
            if(x2 == s2.size()){
                return s1[x1] + dpf(x1+1, x2);
            }
            if(dp[x1][x2] != INT_MAX){
                return dp[x1][x2];
            }
            if(s1[x1] == s2[x2]){
                dp[x1][x2] = dpf(x1+1, x2+1);
            }
            return dp[x1][x2] = std::min({
                dp[x1][x2],
                s1[x1] + dpf(x1+1, x2),
                s2[x2] + dpf(x1, x2+1)
            });
        };
        return dpf(0, 0);
    }
};
18000