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    <p>The original answer is correct for: is a string <code>a</code> a <em>substring</em> of <code>b</code> (misread).</p> <p>Using a trie, you can simply add all strings to it in a first iteration, and in the 2nd iteration, start reading each word, let it be <code>w</code>. If you find a word that you finished your read, but did not reach the string terminator (<code>$</code> usually), you reach some node <code>v</code> in the trie. <br>By doing a <a href="http://en.wikipedia.org/wiki/Depth-first_search" rel="nofollow">DFS</a> from <code>v</code>, you can get all strings which <code>w</code> is prefix of them.</p> <p>high level pseudo code:</p> <pre><code>t &lt;- new trie for each word w: t.add(w) for each word w: node &lt;- t.getLastNode(w) if node.val != $ collection&lt;- DFS(node) (excluding w itself) w is a prefix of each word in collection </code></pre> <p>Note: in order to optimize it, you might need to do some extra work: if <code>a</code> is prefix of <code>b</code>, and <code>b</code> is prefix of <code>c</code>, then <code>a</code> is prefix of <code>c</code>, so - when you do the DFS, if you reach some node that was already searched - just append its strings to the current prefix. <br>Still, since there could be quadric number of possibilities (<code>"a", "aa", "aaa", ....</code> ), getting all of them requires quadric time.</p> <hr> <p>Original answer: finding if <code>a</code> is a <em>substring</em> of <code>b</code>:</p> <p>The suggested solution runs in a quadric complexity, you will need to check each two pairs, giving you <code>O(n* (n-1) * |S|)</code>.</p> <p>You can build a <a href="http://en.wikipedia.org/wiki/Suffix_tree" rel="nofollow">suffix tree</a> from the strings in the first iteration, and in the 2nd iteration check if each string is a non trivial entry (not itself) of another string. <br>This solution is <code>O(n*|S|)</code></p>
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    1. POFind strings which are prefixes of other strings
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    1. COI guess the radix tree also can be used here with the same amount of operations.
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    2. COlike Archeg said, a Radix Tree can be used to save space in the Trie.
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