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    <p>Both of these approaches are correct. It is indeed legal to have multiple recursive calls from a function, and the meaning is what you'd think - just do one call, then the next, then the next, etc.</p> <p>Interestingly, I don't think that the recursive version does make exponentially many calls. It makes at most two recursive calls, but each are on a problem whose size is (approximately) half as large as the original call. Essentially, the recurrence looks like this:</p> <pre><code>T(1) = 1 T(2) = 1 T(n) &lt;= T(n / 2) + T(n / 2 + 1) + 1 </code></pre> <p>I use "less than or equal to here" to say that in the best case you might just make one call, but in the worst case you make at most two.</p> <p>I want to prove that this function T(n) &lt;= max{cn + d, a} for some constants c, d, and a. This would prove that T(n) = O(n) and thus makes at most linearly many calls. As our base case, we have that</p> <pre><code>T(1) = 1 T(2) = 1 </code></pre> <p>so let's set a = 1. For the inductive step, we consider three cases. First, let's consider when floor(n / 2) &lt;= 2 and floor(n / 2 + 1) &lt;= 2:</p> <pre><code>T(n) &lt;= T(n / 2) + T(n / 2 + 1) + 1 &lt;= 1 + 1 + 1 &lt;= 3 </code></pre> <p>If we assume that cn + d >= 3, when n = 3 or n = 4, then this works out correctly. In particular, this means that 3c + d >= 3 and 4c + d >= 3.</p> <p>In the next case, let's see what happens if floor(n / 2) &lt;= 2 and floor(n / 2 + 1) >= 2. Then we have that</p> <pre><code>T(n) &lt;= T(n / 2) + T(n / 2 + 1) + 1 &lt;= 1 + max{c(n / 2 + 1) + d, 1} + 1 &lt;= 2 + max{c(n / 2 + 1) + d, 1} &lt;= 3 + c(n / 2 + 1) + d </code></pre> <p>So if we have that 3 + c(n / 2 + 1) + d &lt;= cn + d, this claim still holds. Note that we're only in this case if n = 5, and so this means that we must have that</p> <pre><code>3 + c(n / 2 + 1) + d &lt;= cn + d 3 + c(n / 2 + 1) &lt;= cn 3 + c(5 / 2 + 1) &lt;= 5c 3 + 5c/2 + c &lt;= 5c 3 + 7c/2 &lt;= 5c 4 &lt;= 3c / 2 8 / 3 &lt;= c </code></pre> <p>So we must have that c >= 8 / 3.</p> <p>And finally, the case where neither n / 2 nor n / 2 + 1 are less than three:</p> <pre><code>T(n) &lt;= T(n / 2) + T(n / 2 + 1) + 1 &lt;= c(n / 2) + d + c(n / 2 + 1) + d + 1 &lt;= cn / 2 + cn / 2 + c + 2d + 1 = cn + c + 2d + 1 </code></pre> <p>This is less than cn + d if</p> <pre><code> cn + c + 2d + 1 &lt;= cn + d c + 2d + 1 &lt;= d c + d + 1 &lt;= 0 </code></pre> <p>This works if d = -c - 1.</p> <p>From earlier, we know that 3c + d >= 3, which works if 2c - 1 >= 3, or 2c >= 4, so c >= 2. We also have that 4c + d >= 3, which also works if c >= 2. Letting c = 8 / 3, we get that d = -11 / 3, so</p> <pre><code>T(n) &lt;= max{8n/3 - 11/3, 1} </code></pre> <p>So T(n) = O(n) and the recursion makes only linearly many calls.</p> <hr> <p>The short version of this is that both the recursive and iterative versions take linear time. Don't fear recursion's exponential-time blowup without being sure it's exponential. :-) Though admittedtly, in this case, I really like the iterative version more. It's clearer, more intuitive, and more immediately O(n).</p>
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    1. COWhat an explanation , I will read it for sure :) I am noob in this .
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    2. CO@Nishant- Here's hoping I didn't botch the math. :-)
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    3. CO@Paul Hankin- The value of F(n) might be no greater than n, but the time to compute it might be if we charge each recursive call a constant unit of time. This recursive formulation (T(1) = 1, T(2n) = T(n) + 1, T(2n + 1) = T(n) + T(n + 1) + 1) can't easily be bounded by n.
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