Class Solution
java.lang.Object
g1101_1200.s1170_compare_strings_by_frequency_of_the_smallest_character.Solution
1170 - Compare Strings by Frequency of the Smallest Character.<p>Medium</p>
<p>Let the function <code>f(s)</code> be the <strong>frequency of the lexicographically smallest character</strong> in a non-empty string <code>s</code>. For example, if <code>s = "dcce"</code> then <code>f(s) = 2</code> because the lexicographically smallest character is <code>'c'</code>, which has a frequency of 2.</p>
<p>You are given an array of strings <code>words</code> and another array of query strings <code>queries</code>. For each query <code>queries[i]</code>, count the <strong>number of words</strong> in <code>words</code> such that <code>f(queries[i])</code> < <code>f(W)</code> for each <code>W</code> in <code>words</code>.</p>
<p>Return <em>an integer array</em> <code>answer</code><em>, where each</em> <code>answer[i]</code> <em>is the answer to the</em> <code>i<sup>th</sup></code> <em>query</em>.</p>
<p><strong>Example 1:</strong></p>
<p><strong>Input:</strong> queries = [“cbd”], words = [“zaaaz”]</p>
<p><strong>Output:</strong> [1]</p>
<p><strong>Explanation:</strong> On the first query we have f(“cbd”) = 1, f(“zaaaz”) = 3 so f(“cbd”) < f(“zaaaz”).</p>
<p><strong>Example 2:</strong></p>
<p><strong>Input:</strong> queries = [“bbb”,“cc”], words = [“a”,“aa”,“aaa”,“aaaa”]</p>
<p><strong>Output:</strong> [1,2]</p>
<p><strong>Explanation:</strong> On the first query only f(“bbb”) < f(“aaaa”). On the second query both f(“aaa”) and f(“aaaa”) are both > f(“cc”).</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 <= queries.length <= 2000</code></li>
<li><code>1 <= words.length <= 2000</code></li>
<li><code>1 <= queries[i].length, words[i].length <= 10</code></li>
<li><code>queries[i][j]</code>, <code>words[i][j]</code> consist of lowercase English letters.</li>
</ul>
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Solution
public Solution()
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Method Details
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numSmallerByFrequency
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