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Fixes typo in anagrams.js #102

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Fixes typo in anagrams.js #102

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2 changes: 1 addition & 1 deletion book/analysis/anagrams.js
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Expand Up @@ -91,7 +91,7 @@ assert.equal(false, anagramSortAndCompare('abcde', 'abcd'))
At first glance you may be tempted to think that this algorithm is
$$O(n)$$, since there is one simple iteration to compare the *n*
characters after the sorting process. However, the two calls to the
Python `sorted` method are not without their own cost. Sorting is
Javascript `sort` method are not without their own cost. Sorting is
typically either $$O(n^{2})$$ or $$O(n\log n)$$, so the sorting
operations dominate the iteration. In the end, this algorithm will have
the same order of magnitude as that of the sorting process.
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