It's sqrt decomposition. It didn't rime. Bummer...

Cutting to the chase — to be modified

“Why not logarithm my problems?” you might ask. My answer will be: you can't always do that.

It has been written about sqrt decomposition many times. Still, many people see as a poor cousin of segment trees or as the crippled child of data structures. There are a lot of tutorials on the topic, but I'll try my best to offer a different perspective and bust the segment tree supremacy by giving this little kid a flamethrower.

Basics

You are given an array of length n. Answer to m query of types:

1. “What is the maximum element in interval [x,y] ? ”.
2. The elements on position x take the value y.

The brute force goes like that: we shall iterate from x to y for each query and compute the answer. The update will be O(1) and query O(N). That is O(N*M).

The key element in sqrt decomposition is skipping. Let's define a k-segment as a continuous subsequence from an array that has the length less or equal to k. We shall divide our array in as little k-segments as possible. For example array [1,2,3,4,5,6,7,8,9,10,11,12,13] divided in 3-segments looks like: [1,2,3] [4,5,6] [7,8,9] [10,11,12] [13]. Note that this greedy split of adding elements to a segment if possible ensures us with a maximum number of k-segments.

Now let's apply the k-segment concept to our queries. Let's say we want to find the maximum on interval [3,11]. That is find the maximum out of elements 3] [4,5,6] [7,8,9] [10,11. Let's observe we have “full” and “broken” 3-segments in our example. What is the maximum number of “broken” segments ? It's two ( the left and right border segments ).

Finally , having all this stuff sorted out , I'll present you the final solution.

We record for each segment it's elements and the maximum element. An update is made by replacing the element on the x-th position and updating the maximum on the k-segment. A query can be responded by calculating the maximum on “broken” k-segments and the maximum of the values of the “full” segments.

The complexity is O(K) for an update , O(K+N/K) for a query and O(N) for building the structure. If we take K = sqrt(N) , then the complexity will be O(N+M sqrt N).

Here is an (object oriented) approach to implement the problem above.

You can also read more about the basics (and other stuff) here .

Replacing lazy deletion

You are given an array of length n. Answer to m query of types:

1. “What is the maximum element in interval [x,y] ? ”.
2. The elements on positions [x,y] take the value z.

The only thing we must keep in addition is a variable ff that works as follows:

• if ff != 0 then the maximum will be fixed
• if ff == 0 then we have to compute the maximum

It works just as lazy deletion on segment trees. That is, if we need to look for a value inside a k-segment then is the segment is fixed we update all it's elements with ff. If we need to compute the maximum on a fixed segment is not necessary to go inside it, but rather just respond with value ff. Complexity will be O(N+M sqrt N).

Why the k-segment notation ?

Sometimes taking K=sqrt(N) isn't the best idea. It was the case the following problem form National Olympics in Informatics Romania.

We have a vector v with n elements and a number m. The vector has to be split in exactly m continuous subsequences such that each element has to be in one subsequence. We define the control value of some subsequence as [(maximum-minimum+1)/2]. We have to split the array such that the maximum control value of each subsequence will be as small as possible. ( n <= 1,000,000 and m <= 1000 )

For example if we split the array [13,4,6,19,4,10] in [13,4,6] , [19] , [4,10] we have the control values 5,0,3 so the answer will be 5.

Firstly , let's solve the problem in a not-so-good complexity. We binary search the result, making a greedy verification function. As long as one can expand the current continuous sequence to the right side ( i.e. [(maximum-minimum+1)/2] <= binary_searched_value ) , we move to the right. If we can split the array in m or less sequences, the verification function will return true. The complexity will be O(N log max_ans).

Now we have to improve the complexity of some verification. We split the array in k-sequences ( of length k ). For each sequence keep maximum and minimum. Now, instead of trying to add only one value to the sequence we can try to add k values to it. This will give us the complexity O(N/K+M*K) per verification as we have maximum N/K “full” k-segments to be added and we start to add a new continuous segment to the solution at most M times.

The only thing lest to do it to choose K. If we choose K = sqrt(N) ( <= M ), this will give no improvement. But if we choose K as 100 , this will make our algorithm significantly faster. Final complexity will be O(N + (N/K+M*K) log max_ans).

Give that kid a flamethrower

Now we're talking business, right ?

1. Conex graph

You are given a conex graph with n nodes and m edges. You are given q queries of the type k , a[1] , a[2] … a[k] ( 1 <= a[i] <= m ) and you have to find out for each query if the graph is conex after eliminating edges with indexes a[1] , a[2] … a[k]. Note that each query is independent. ( q*k , m <= 100,000 )

The solution that uses sqrt decomposition splits the queries in groups of k = sqrt(q). Now let's solve the problem for each group of queries.

    nbr = int(sqrt(q));
    for (int i=1;i<=q;++i)
    {
        bunch.push_back(i);
        if ( int(bunch.size()) == nbr )
        {
            solve(bunch);
            bunch.clear();
        }
    }
    if ( bunch.size() )
        solve(bunch); 

First of all , let's compress all the nodes that we are not interested in. That is we keep all edges instead of edges in all queries in the current bunch and make a new node for each conex component in this graph. The new graph will have only nodes and no edges.

Let's compute the answer for each query in our bunch. For a query we will add edges in the new graph. Each added edge has a corresponding edge in the old graph. The edges in the old graph that I referred to are all groups of edges for all other k-1 queries.

The complexity will be O(N+sqrt(Q)*K) per bunch. In total the complexity will be O(sqrt(Q) * (N+sqrt(Q)*K)) = O(N*sqrt(Q)+Q*K).

To be sure you understood, let's take a quick example. In the picture the edges marked with red are the first query and the ones with blue are the second one. Suppose we make a bunch out of queries 1 and 2.

The new graph will be made out of nodes 0' = [0,1,4] , 1' = [3] , 2' = [2,5,6,7]. When we solve query one, edges (1,4) , (0,5) , (0,6) will be added, so in the new graph we will have an edge added between 0' and 2' , therefore the new graph will not be conex. When we solve query two, edges (1,3) , (3,5) are added, so (0',1') and (1',2') are added in the new graph, therefore the new graph will be conex.

You can find a similar problem here .