maxorand's blog

By maxorand, 3 years ago, In English

If the given array's size is N then what should I declare the size of the array for Segment Tree?

If I set the Segment Tree's array size to 2 × 2k (or maybe something like this 2 × 2k + 5) where k = ⌈ log2N, won't that be enough? If not, then why?

 
 
 
 
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3 years ago, # |
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2*N is enough.

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    3 years ago, # ^ |
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    No, 2 * N will not hold for any N that is not a power of 2.

    For example if N = 105 then you will need roughly 262143 sized segment tree, which is greater than 2 * N. Actually 4 * N is the safest way.

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3 years ago, # |
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Actually if you can write N as 2x where x is an integer then you can see that you need 2x + 1 - 1 nodes in your segment tree. So what can you do for other N's? Well, you can find the next power of 2 after N, let's say it is 2x and then you can declare 2x + 1 - 1 sized segment tree. So yes it will work as you said.

But if you don't want to do that much calculation, you can always declare 4 * N with your eyes closed!

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3 years ago, # |
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Here is a nice Quora answer which proves that 4 is in fact the smallest k such that kN is enough space for a segment tree on an array of size N.

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3 years ago, # |
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I usually adopt the implementation of segment tree mentioned in this book Competitive Programmer's Handbook — a new book on competitive programming.

According to my own understanding, if there are N elements, then I will find the minimum M = 2m that is no less than N. Then, the total number of nodes in the segment tree should be 2 * M.

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15 months ago, # |
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You should change a little bit k=(int)log2(N)+1; Rest of them are ok .

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15 months ago, # |
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Your solution fail for N=7 . According to you Size=2* (2^2) = 8 . But this is wrong because in tree , the leftmost node of lower level is 8 and the rightmost node of lower level is 13 . That's why it fails . Your fault is that your logarithm just chopped some fractional number . FIrst you have to calculate the leftmost node of the lower level , which is 2^k , where k=log2(N)+1 ; . We have to multiply 2^k by 2 . Because we know that in each level there has 2^i nodes . So in the lowest level there has 2^k nodes . So the rightmost node is 2^k + 2^k = 2* 2^k . That's why it works for 2 times 2^k