Sync LeetCode submission Runtime - 12 ms (81.39%), Memory - 18.2 MB (65.94%)

Author: koralkulacoglu

LeetCode/0712-minimum-ascii-delete-sum-for-two-strings/README.md

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+<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>

LeetCode/0712-minimum-ascii-delete-sum-for-two-strings/solution.cpp

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+class Solution {
+public:
+    int minimumDeleteSum(string s1, string s2) {
+        int n = s1.size();
+        int m = s2.size();
+
+        vector<vector<int>> dp(n+1, vector<int>(m+1, 0));
+        for (int i=1; i<=n; i++) dp[i][0] = dp[i-1][0] + s1[i-1];
+        for (int i=1; i<=m; i++) dp[0][i] = dp[0][i-1] + s2[i-1];
+
+        for (int i=1; i<=n; i++) {
+            for (int j=1; j<=m; j++) {
+                if (s1[i-1] == s2[j-1]) dp[i][j] = dp[i-1][j-1];
+                else dp[i][j] = min(dp[i-1][j] + s1[i-1], dp[i][j-1] + s2[j-1]);
+            }
+        }
+
+        return dp[n][m];
+    }
+};
+