mirror of
				https://gitee.com/coder-xiaomo/leetcode-problemset
				synced 2025-10-26 07:18:56 +08:00 
			
		
		
		
	
		
			
				
	
	
		
			53 lines
		
	
	
		
			3.5 KiB
		
	
	
	
		
			HTML
		
	
	
	
	
	
			
		
		
	
	
			53 lines
		
	
	
		
			3.5 KiB
		
	
	
	
		
			HTML
		
	
	
	
	
	
| <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 <= 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> </p>
 | |
| <p><strong>Example 1:</strong></p>
 | |
| 
 | |
| <pre>
 | |
| <strong>Input:</strong> words = ["acca","bbbb","caca"], target = "aba"
 | |
| <strong>Output:</strong> 6
 | |
| <strong>Explanation:</strong> There are 6 ways to form target.
 | |
| "aba" -> index 0 ("<u>a</u>cca"), index 1 ("b<u>b</u>bb"), index 3 ("cac<u>a</u>")
 | |
| "aba" -> index 0 ("<u>a</u>cca"), index 2 ("bb<u>b</u>b"), index 3 ("cac<u>a</u>")
 | |
| "aba" -> index 0 ("<u>a</u>cca"), index 1 ("b<u>b</u>bb"), index 3 ("acc<u>a</u>")
 | |
| "aba" -> index 0 ("<u>a</u>cca"), index 2 ("bb<u>b</u>b"), index 3 ("acc<u>a</u>")
 | |
| "aba" -> index 1 ("c<u>a</u>ca"), index 2 ("bb<u>b</u>b"), index 3 ("acc<u>a</u>")
 | |
| "aba" -> index 1 ("c<u>a</u>ca"), index 2 ("bb<u>b</u>b"), index 3 ("cac<u>a</u>")
 | |
| </pre>
 | |
| 
 | |
| <p><strong>Example 2:</strong></p>
 | |
| 
 | |
| <pre>
 | |
| <strong>Input:</strong> words = ["abba","baab"], target = "bab"
 | |
| <strong>Output:</strong> 4
 | |
| <strong>Explanation:</strong> There are 4 ways to form target.
 | |
| "bab" -> index 0 ("<u>b</u>aab"), index 1 ("b<u>a</u>ab"), index 2 ("ab<u>b</u>a")
 | |
| "bab" -> index 0 ("<u>b</u>aab"), index 1 ("b<u>a</u>ab"), index 3 ("baa<u>b</u>")
 | |
| "bab" -> index 0 ("<u>b</u>aab"), index 2 ("ba<u>a</u>b"), index 3 ("baa<u>b</u>")
 | |
| "bab" -> index 1 ("a<u>b</u>ba"), index 2 ("ba<u>a</u>b"), index 3 ("baa<u>b</u>")
 | |
| </pre>
 | |
| 
 | |
| <p> </p>
 | |
| <p><strong>Constraints:</strong></p>
 | |
| 
 | |
| <ul>
 | |
| 	<li><code>1 <= words.length <= 1000</code></li>
 | |
| 	<li><code>1 <= words[i].length <= 1000</code></li>
 | |
| 	<li>All strings in <code>words</code> have the same length.</li>
 | |
| 	<li><code>1 <= target.length <= 1000</code></li>
 | |
| 	<li><code>words[i]</code> and <code>target</code> contain only lowercase English letters.</li>
 | |
| </ul>
 |