<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 {
    string s1, s2;
    int INF = static_cast<int>(1e9)+7;
    int dp[1001][1001];
    int f(int i, int j) {
        if (dp[i][j] != -1) return dp[i][j];
        // base case
        if (i >= s1.size() && j >= s2.size()) {
            return 0;
        }
        if (i >= s1.size()) return s2[j] + f(i, j+1);
        if (j >= s2.size()) return s1[i] + f(i + 1, j);
        int cur = INF;
        if (s1[i] == s2[j]) {
            cur = min(cur, f(i+1, j+1));
        }
        cur = min(cur, f(i+1, j) + s1[i]);
        cur = min(cur, f(i, j+1) + s2[j]);
        return dp[i][j] = cur;
    }
public:
    int minimumDeleteSum(string s1, string s2) {
        this->s1 = s1;
        this->s2 = s2;
        for (int i = 0; i < 1001; ++i) {
            for (int j = 0; j < 1001; ++j) {
                dp[i][j] = -1;
            }
        }
        return f(0, 0);
    }
};
12600