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copied!<p>You can avoid recursion here. Just walk the array. Compare the current array element to the string you are accumulating. When you see a new "main" element for the first time, append it to the string you are building. When you see a closing instance of a main element that matches the suffix of the string you have so far, pop it. Build up the resulting array as you walk.</p> <p>For example, given</p> <pre><code>["one", "foo", "two", "baz", "two", "one", "three", "lulz", "wtf", "three"] </code></pre> <p>You first see "one" so emit that</p> <pre><code>result = ["one"] working = "one" </code></pre> <p>Then you see <code>"foo"</code> so emit, but keep only <code>"one"</code> as the string to compare against.</p> <pre><code>result = ["one", "one.foo"] working = "one" </code></pre> <p>Next you have <code>"two"</code> so emit and keep <code>"one.two"</code>. </p> <pre><code>result = ["one", "one.foo", "one.two"] working = "one.two" </code></pre> <p>Next is <code>"baz"</code> emit but do not keep.</p> <pre><code>result = ["one", "one.foo", "one.two", "one.two.baz"] working = "one.two" </code></pre> <p>Next is a <code>"two"</code> so you can pop!</p> <pre><code>result = ["one", "one.foo", "one.two", "one.two.baz"] working = "one" </code></pre> <p>Next, a <code>"one"</code> so pop again!</p> <pre><code>result = ["one", "one.foo", "one.two", "one.two.baz"] working = "" </code></pre> <p>Now a three, which is a main tag, so emit and keep:</p> <pre><code>result = ["one", "one.foo", "one.two", "one.two.baz", "three"] working = "three" </code></pre> <p>Straightforward! I think you can take it from here. No fancy recursion is needed. Of course you may have to check for errors in nesting and balancing.</p>

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