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Author: NamVr

1744-number-of-ways-to-form-a-target-string-given-a-dictionary/README.md

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+<p>You are given a list of strings of the <strong>same length</strong> <code>words</code> and a string <code>target</code>.</p>
+
+<p>Your task is to form <code>target</code> using the given <code>words</code> under the following rules:</p>
+
+<ul>
+	<li><code>target</code> should be formed from left to right.</li>
+	<li>To form the <code>i<sup>th</sup></code> character (<strong>0-indexed</strong>) of <code>target</code>, you can choose the <code>k<sup>th</sup></code> character of the <code>j<sup>th</sup></code> string in <code>words</code> if <code>target[i] = words[j][k]</code>.</li>
+	<li>Once you use the <code>k<sup>th</sup></code> character of the <code>j<sup>th</sup></code> string of <code>words</code>, you <strong>can no longer</strong> use the <code>x<sup>th</sup></code> character of any string in <code>words</code> where <code>x &lt;= k</code>. In other words, all characters to the left of or at index <code>k</code> become unusuable for every string.</li>
+	<li>Repeat the process until you form the string <code>target</code>.</li>
+</ul>
+
+<p><strong>Notice</strong> that you can use <strong>multiple characters</strong> from the <strong>same string</strong> in <code>words</code> provided the conditions above are met.</p>
+
+<p>Return <em>the number of ways to form <code>target</code> from <code>words</code></em>. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;acca&quot;,&quot;bbbb&quot;,&quot;caca&quot;], target = &quot;aba&quot;
+<strong>Output:</strong> 6
+<strong>Explanation:</strong> There are 6 ways to form target.
+&quot;aba&quot; -&gt; index 0 (&quot;<u>a</u>cca&quot;), index 1 (&quot;b<u>b</u>bb&quot;), index 3 (&quot;cac<u>a</u>&quot;)
+&quot;aba&quot; -&gt; index 0 (&quot;<u>a</u>cca&quot;), index 2 (&quot;bb<u>b</u>b&quot;), index 3 (&quot;cac<u>a</u>&quot;)
+&quot;aba&quot; -&gt; index 0 (&quot;<u>a</u>cca&quot;), index 1 (&quot;b<u>b</u>bb&quot;), index 3 (&quot;acc<u>a</u>&quot;)
+&quot;aba&quot; -&gt; index 0 (&quot;<u>a</u>cca&quot;), index 2 (&quot;bb<u>b</u>b&quot;), index 3 (&quot;acc<u>a</u>&quot;)
+&quot;aba&quot; -&gt; index 1 (&quot;c<u>a</u>ca&quot;), index 2 (&quot;bb<u>b</u>b&quot;), index 3 (&quot;acc<u>a</u>&quot;)
+&quot;aba&quot; -&gt; index 1 (&quot;c<u>a</u>ca&quot;), index 2 (&quot;bb<u>b</u>b&quot;), index 3 (&quot;cac<u>a</u>&quot;)
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;abba&quot;,&quot;baab&quot;], target = &quot;bab&quot;
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> There are 4 ways to form target.
+&quot;bab&quot; -&gt; index 0 (&quot;<u>b</u>aab&quot;), index 1 (&quot;b<u>a</u>ab&quot;), index 2 (&quot;ab<u>b</u>a&quot;)
+&quot;bab&quot; -&gt; index 0 (&quot;<u>b</u>aab&quot;), index 1 (&quot;b<u>a</u>ab&quot;), index 3 (&quot;baa<u>b</u>&quot;)
+&quot;bab&quot; -&gt; index 0 (&quot;<u>b</u>aab&quot;), index 2 (&quot;ba<u>a</u>b&quot;), index 3 (&quot;baa<u>b</u>&quot;)
+&quot;bab&quot; -&gt; index 1 (&quot;a<u>b</u>ba&quot;), index 2 (&quot;ba<u>a</u>b&quot;), index 3 (&quot;baa<u>b</u>&quot;)
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= words.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= words[i].length &lt;= 1000</code></li>
+	<li>All strings in <code>words</code> have the same length.</li>
+	<li><code>1 &lt;= target.length &lt;= 1000</code></li>
+	<li><code>words[i]</code> and <code>target</code> contain only lowercase English letters.</li>
+</ul>

1744-number-of-ways-to-form-a-target-string-given-a-dictionary/solution.py

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+class Solution:
+    def numWays(self, words: List[str], target: str) -> int:
+        MOD = 1000000007
+        word_length = len(words[0])
+        target_length = len(target)
+        char_frequency = [[0] * 26 for _ in range(word_length)]
+
+        # Step 1: Calculate frequency of each character at every index in "words".
+        for word in words:
+            for j in range(word_length):
+                char_frequency[j][ord(word[j]) - ord("a")] += 1
+
+        # Step 2: Initialize two DP arrays: prev_count and curr_count.
+        prev_count = [0] * (target_length + 1)
+        curr_count = [0] * (target_length + 1)
+
+        # Base case: There is one way to form an empty target string.
+        prev_count[0] = 1
+
+        # Step 3: Fill the DP arrays.
+        for curr_word in range(1, word_length + 1):
+            curr_count = prev_count.copy()
+            for curr_target in range(1, target_length + 1):
+                cur_pos = ord(target[curr_target - 1]) - ord("a")
+
+                # If characters match, add the number of ways.
+                curr_count[curr_target] += (
+                    char_frequency[curr_word - 1][cur_pos]
+                    * prev_count[curr_target - 1]
+                ) % MOD
+                curr_count[curr_target] %= MOD
+
+            # Move current row to previous row for the next iteration.
+            prev_count = curr_count.copy()
+
+        # Step 4: The result is in prev[target_length].
+        return curr_count[target_length]