<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_view s1, string_view s2) {
        
        int cache[s1.size()+1][s2.size()+1];
        cache[0][0] = 0;

        for (int s1i = 1; s1i <= s1.size(); ++s1i)
            cache[s1i][0] = cache[s1i-1][0] + s1[s1i-1];

        for (int s2i = 1; s2i <= s2.size(); ++s2i)
            cache[0][s2i] = cache[0][s2i-1] + s2[s2i-1];

        for (int s1i = 1; s1i <= s1.size(); ++s1i)
        {
            const char currS1 = s1[s1i-1];
            for (int s2i = 1; s2i <= s2.size(); ++s2i)
            {
                const char currS2 = s2[s2i-1];
                if (currS1 == currS2)
                {
                    cache[s1i][s2i] = cache[s1i-1][s2i-1];
                }
                else
                {
                    const int top = cache[s1i-1][s2i];
                    const int left = cache[s1i][s2i-1];

                    cache[s1i][s2i] = std::min( top + currS1, left + currS2 );
                }
            }
        }
        return cache[s1.size()][s2.size()];
    }
};
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