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    <p>Here's my solution in C++:</p> <pre><code>int min_width(const string&amp; x, const set&lt;char&gt;&amp; y) { vector&lt;int&gt; at; for (int i = 0; i &lt; x.length(); i++) if (y.count(x[i]) &gt; 0) at.push_back(i); int ret = x.size(); int start = 0; map&lt;char, int&gt; count; for (int end = 0; end &lt; at.size(); end++) { count[x[at[end]]]++; while (count[x[at[start]]] &gt; 1) count[x[at[start++]]]--; if (count.size() == y.size() &amp;&amp; ret &gt; at[end] - at[start] + 1) ret = at[end] - at[start] + 1; } return ret; } </code></pre> <p><strong>Edit</strong>: Here's an implementation of Jack's idea. It's the same time complexity as mine, but without the inner loop that confuses you.</p> <pre><code>int min_width(const string&amp; x, const set&lt;char&gt;&amp; y) { int ret = x.size(); map&lt;char, int&gt; index; set&lt;int&gt; index_set; for (int j = 0; j &lt; x.size(); j++) { if (y.count(x[j]) &gt; 0) { if (index.count(x[j]) &gt; 0) index_set.erase(index[x[j]]); index_set.insert(j); index[x[j]] = j; if (index.size() == y.size()) { int i = *index_set.begin(); if (ret &gt; j-i+1) ret = j-i+1; } } } return ret; } </code></pre> <p>In Java it can be implemented nicely with LinkedHashMap:</p> <pre><code>static int minWidth(String x, HashSet&lt;Character&gt; y) { int ret = x.length(); Map&lt;Character, Integer&gt; index = new LinkedHashMap&lt;Character, Integer&gt;(); for (int j = 0; j &lt; x.length(); j++) { char ch = x.charAt(j); if (y.contains(ch)) { index.remove(ch); index.put(ch, j); if (index.size() == y.size()) { int i = index.values().iterator().next(); if (ret &gt; j - i + 1) ret = j - i + 1; } } } return ret; } </code></pre> <p>All operations inside the loop take constant time (assuming hashed elements disperse properly). </p>
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    1. POMinimum window width in string x that contains all characters of string y
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    1. USSheldon L. Cooper
    1. USSheldon L. Cooper
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    1. COThe time complexity is linear in the size of the string "x" (assuming a constant size alphabet.) This implementation assumes there's always a solution.
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    2. CO@Sheldon, I tested your code and it does produce correct result, however you have a while loop inside a for loop, so the complexity looks to be O(n^2) to me.
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    3. CONo, it's O(n). The while loop executes at most n iterations OVERALL, since "start" is incremented at each iteration.
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