62 lines
1.8 KiB
Markdown
62 lines
1.8 KiB
Markdown
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---
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id: 587d7db6367417b2b2512b9b
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title: Find Characters with Lazy Matching
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challengeType: 1
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forumTopicId: 301341
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dashedName: find-characters-with-lazy-matching
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---
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# --description--
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In regular expressions, a <dfn>greedy</dfn> match finds the longest possible part of a string that fits the regex pattern and returns it as a match. The alternative is called a <dfn>lazy</dfn> match, which finds the smallest possible part of the string that satisfies the regex pattern.
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You can apply the regex `/t[a-z]*i/` to the string `"titanic"`. This regex is basically a pattern that starts with `t`, ends with `i`, and has some letters in between.
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Regular expressions are by default greedy, so the match would return `["titani"]`. It finds the largest sub-string possible to fit the pattern.
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However, you can use the `?` character to change it to lazy matching. `"titanic"` matched against the adjusted regex of `/t[a-z]*?i/` returns `["ti"]`.
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**Note:** Parsing HTML with regular expressions should be avoided, but pattern matching an HTML string with regular expressions is completely fine.
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# --instructions--
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Fix the regex `/<.*>/` to return the HTML tag `<h1>` and not the text `"<h1>Winter is coming</h1>"`. Remember the wildcard `.` in a regular expression matches any character.
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# --hints--
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The `result` variable should be an array with `<h1>` in it
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```js
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assert(result[0] == '<h1>');
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```
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`myRegex` should use lazy matching
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```js
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assert(/\?/g.test(myRegex));
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```
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`myRegex` should not include the string `h1`
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```js
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assert(!myRegex.source.match('h1'));
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```
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# --seed--
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## --seed-contents--
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```js
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let text = "<h1>Winter is coming</h1>";
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let myRegex = /<.*>/; // Change this line
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let result = text.match(myRegex);
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```
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# --solutions--
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```js
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let text = "<h1>Winter is coming</h1>";
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let myRegex = /<.*?>/; // Change this line
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let result = text.match(myRegex);
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```
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