<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 minSum[1001][1001]; 
            // minSum[prefixLen1][prefixLen2] - answer for s1.substr(prefixLen1) and s2.substr(prefixLen2)
    
        for(int p1 = 0; p1 <= s1.size(); ++p1){
            for(int p2 = 0; p2 <= s2.size(); ++p2){
                if(p1 == 0 && p2 == 0){
                    minSum[p1][p2] = 0;
                } else if(p1 == 0){
                    minSum[p1][p2] = minSum[p1][p2-1] + s2[p2 -1];
                } else if (p2 == 0){
                    minSum[p1][p2] = minSum[p1 - 1][p2] + s1[p1- 1];
                } else{
                    if(s1[p1 - 1] == s2[p2 - 1]){
                        minSum[p1][p2] = minSum[p1 - 1][p2 - 1];
                    } else{
                        minSum[p1][p2] = min(
                        minSum[p1][p2-1] + s2[p2 -1],
                        minSum[p1 - 1][p2] + s1[p1- 1]
                        );
                    }
                }
            }
        }
        return minSum[s1.size()][s2.size()];
    }
};
10200