sdnr1's blog

By sdnr1, history, 11 days ago, In English,

There is hardly any need for improving the time complexity of initializing a Binary Indexed Tree (BIT). Anyway I want to share this neat little trick to initialize a BIT in O(N) time.

Let bit[N] be BIT array and a[N] be the array we need to initialize our BIT with. Notice that for all i we need to add a[i] to positions i, i + (i & -i) and so on in the bit array. Also, a[i + (i & -i)] will also be added to all these positions except i. We can add these 2 in the required places together! Take a look at the following code:

int bit[N], a[N];

void initialize(int n)
    for(int i=1; i<=n; i++)
        bit[i] += a[i];
        if(i + (i & -i) <= n)
            bit[i + (i & -i)] += bit[i];

Really easy and elegant! Although I have not come across any problems that needed this trick, but there might be a situation where N is too large to initialize the array in O(NlogN) while the number of queries are such that O(logN) per query is feasible (maybe in a 2D BIT problem).

The same technique can be for bulk updates (put update value for position i at a[i] for all N) and even for bulk queries. By bulk I mean number of updates/queries are O(N). Leaving bulk queries as an exercise for the reader :P (its really easy if you understood the above).

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11 days ago, # |
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3/4 of your blogs are about BITs, you must be very fond of them :)

11 days ago, # |
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This strategy only works for offline updates.

10 days ago, # |
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Can anyone explain me to easy way how Binary Index tree works. Thanks :)

9 days ago, # |
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This is really cool trick, but it rarely improves solution significantly.

Some time ago I've describes another way of constructing BITs in O(N) here. And Jakube pointed to this StackOverflow question with answer that describes your approach.

Some thoughs over your exercise:

General approach to perform bulk operations of single type