Class Solution
java.lang.Object
g2301_2400.s2301_match_substring_after_replacement.Solution
2301 - Match Substring After Replacement.<p>Hard</p>
<p>You are given two strings <code>s</code> and <code>sub</code>. You are also given a 2D character array <code>mappings</code> where <code>mappings[i] = [old<sub>i</sub>, new<sub>i</sub>]</code> indicates that you may <strong>replace</strong> any number of <code>old<sub>i</sub></code> characters of <code>sub</code> with <code>new<sub>i</sub></code>. Each character in <code>sub</code> <strong>cannot</strong> be replaced more than once.</p>
<p>Return <code>true</code> <em>if it is possible to make</em> <code>sub</code> <em>a substring of</em> <code>s</code> <em>by replacing zero or more characters according to</em> <code>mappings</code>. Otherwise, return <code>false</code>.</p>
<p>A <strong>substring</strong> is a contiguous non-empty sequence of characters within a string.</p>
<p><strong>Example 1:</strong></p>
<p><strong>Input:</strong> s = “fool3e7bar”, sub = “leet”, mappings = [[“e”,“3”],[“t”,“7”],[“t”,“8”]]</p>
<p><strong>Output:</strong> true</p>
<p><strong>Explanation:</strong> Replace the first ‘e’ in sub with ‘3’ and ‘t’ in sub with ‘7’.</p>
<p>Now sub = “l3e7” is a substring of s, so we return true.</p>
<p><strong>Example 2:</strong></p>
<p><strong>Input:</strong> s = “fooleetbar”, sub = “f00l”, mappings = [[“o”,“0”]]</p>
<p><strong>Output:</strong> false</p>
<p><strong>Explanation:</strong> The string “f00l” is not a substring of s and no replacements can be made.</p>
<p>Note that we cannot replace ‘0’ with ‘o’.</p>
<p><strong>Example 3:</strong></p>
<p><strong>Input:</strong> s = “Fool33tbaR”, sub = “leetd”, mappings = [[“e”,“3”],[“t”,“7”],[“t”,“8”],[“d”,“b”],[“p”,“b”]]</p>
<p><strong>Output:</strong> true</p>
<p><strong>Explanation:</strong> Replace the first and second ‘e’ in sub with ‘3’ and ‘d’ in sub with ‘b’.</p>
<p>Now sub = “l33tb” is a substring of s, so we return true.</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 <= sub.length <= s.length <= 5000</code></li>
<li><code>0 <= mappings.length <= 1000</code></li>
<li><code>mappings[i].length == 2</code></li>
<li><code>old<sub>i</sub> != new<sub>i</sub></code></li>
<li><code>s</code> and <code>sub</code> consist of uppercase and lowercase English letters and digits.</li>
<li><code>old<sub>i</sub></code> and <code>new<sub>i</sub></code> are either uppercase or lowercase English letters or digits.</li>
</ul>
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Constructor Summary
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Method Summary
Modifier and TypeMethodDescriptionbooleanmatchReplacement(String s, String sub, char[][] mappings)
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Constructor Details
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Solution
public Solution()
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Method Details
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matchReplacement
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