PROBLEM LINK:
Author: Fedor Korobeinikov
Tester: Hiroto Sekido
Editorialist: Kevin Atienza
DIFFICULTY:
MEDIUM
PREREQUISITES:
sqrt decomposition, preprocessing
PROBLEM:
Given a sequence of N integers A_1, A_2, \ldots, A_N, where each A_i is between 1 to M, you are to answer Q queries of the following kind:
- Given L and R, where 1 \le L \le R \le N, what is the maximum |x - y| such that L \le x, y \le R and A_x = A_y?
Note that in the problem, Q is actually K.
QUICK EXPLANATION:
For each i, 1 \le i \le N, precompute the following in O(N) time:
- \text{next}[i], the smallest j > i such that A_i = A_j
- \text{prev}[i], the largest j < i such that A_i = A_j
Let S = \lfloor \sqrt{N} \rfloor, and B = \lceil N/S \rceil. Decompose the array into B blocks, each of size S (except possibly the last). For each i, 1 \le i \le N, and 0 \le j \le B-1, precompute the following in O(N \sqrt{N}) time:
- \text{last_in_blocks}[j][i], the largest k \le jS+S such that A_k = A_i
- \text{block_ans}[j][i], the answer for the query (L,R) = (jS+1,i). For a fixed j, all the \text{block_ans}[j][i] can be computed in O(N) time.
Now, to answer a query (L,R), first find the blocks j_L and j_R where L and R belong in (0 \le j_L, j_R < B). Then the answer is at least \text{block_ans}[j_L+1][R], and the only pairs (x,y) not yet considered are those where L \le x \le j_LS+S. To consider those, one can simply try all x in that range, and find the highest y \le R such that A_x = A_y. Finding that y can be done by using \text{last_in_blocks}[j_R-1][x] and a series of \text{next} calls. To make that last part run in O(S) time, consider only the x such that \text{prev}[x] < L.
EXPLANATION:
Weâ€™ll explain the solution for subtask 1 first, because our solution for subtask 2 will build upon it. However, we will first make the assumption that M \le N, otherwise we can simply replace the values A_1, \ldots A_N with numbers from 1 to N, and should only take O(N) time with a set. However, we donâ€™t recommended that you actually do it; this is only to make the analysis clearer.
O(N^2) per query
First, a simple brute-force O(N^2)-time per query is very simple to implement, so getting the first subtask is not an issue at all. Iâ€™m even providing you with a pseudocode on how to do it
def answer_query(L, R):
for d in R-L...1 by -1
for x in L...R-d
y = x+d
if A[x] == A[y]
return d
return 0
Weâ€™re simply checking every possible answer from [0,R-L] in decreasing order. Note that the whole algorithm runs in O(QN^2) time, which could get TLE if the test cases were stronger. But in case you canâ€™t get your solution accepted, then itâ€™s time to optimize your query time toâ€¦
O(N) per query
To obtain a faster running time, we have to use the fact that we are finding the maximum |x-y|. What this means is that for every value v, we are only concerned with the first and last time it occurs in [L,R].
We first consider the following alternative O(N^2)-time per query solution:
def answer_query(L, R):
answer = 0
for y in L...R
for x in L...y
if A[x] == A[y]
answer = max(answer, y - x)
return answer
The idea here is that for every y, we are seeking A_x, which is the first occurrence of A_y in [L,y], because all the other occurrences will result in a smaller y - x value. Now, to speed it up, notice that we donâ€™t have to recompute this x every time we encounter the value A_x, because we are already reading the values A_L, \ldots, A_R in order, so we already have the information â€śwhen did A_y first appearâ€ť before we ever need it! Hereâ€™s an implementation (in pseudocode):
def answer_query(L, R):
index = new map/dictionary
answer = 0
for y in L...R
if not index.has_key(A[y])
index[A[y]] = y
answer = max(answer, y - index[A[y]])
return answer
Now, notice that this runs in O(N) time if one uses a hash map for example!
We mention here that itâ€™s possible to drop the use of a hash map by using the fact that the values A_y are in [1,M]. This means that we can simply allocate an array of length M, instead of creating a hash map from scratch or clearing it. However, we must be careful when we reinitialize this array, because it is long! There are two ways of â€śinitializingâ€ť it:
- We clear the array every time weâ€™re done using it, but we only clear those we just encountered. This required listing all the indices we accessed.
- We maintain a parallel array that contains when array was last accessed for each index. To clear the array, we simply update the current time.
Weâ€™ll show how to do the second one:
class LazyMap:
index[1..M]
found[1..M] # all initialized to zero
time = 0
def clear():
this.time++
def has_key(i):
return this.found[i] == this.time
def set(i, value): # called on the statement x[i] = value for example
this.found[i] = this.time
this.index[i] = value
def get(i): # called on the expression x[i] for example
return this.index[i]
index = new LazyMap()
def answer_query(L, R):
index.clear()
answer = 0
for y in L...R
if not index.has_key(A[y])
index[A[y]] = y
answer = max(answer, y - index[A[y]])
return answer
Using this, the algorithm still runs in O(N) time (remember that we assume M \le N), but most likely with a lower constant.
The overall algorithm runs in O(QN) time.
sqrt decomposition
When one encounters an array with queries in it, there are usually two ways to preprocess the array so that the queries can be done in sublinear time:
- sqrt decomposition, which splits up the array into \lceil N/S \rceil blocks of size S each. S is usually taken to be \lfloor \sqrt{N} \rfloor (hence the term â€śsqrt decompositionâ€ť). Usually, one can reduce the running time to O((N+Q)\sqrt{N}) or O((N+Q)\sqrt{N \log N}). Sometimes, depending on the problem, it may also yield O((N + Q)N^{2/3}) time.
- build some tree structure on top of the array. This usually yields an O(N + Q \log N) or O((N + Q) \log N) time algorithm.
There are other less common ways, such as lazy updates or combinations of the above, but first weâ€™ll try out whether the above work.
Suppose we have selected the parameter S, and we have split the array into B = \lceil N/S \rceil blocks of size S, except possibly the last block which may contain fewer than S elements. Suppose we want to answer a particular query (L,R). Note that L and R will belong to some block. For simplicity, we assume that they belong to different blocks, because if they are on the same block, then R - L \le S, so we can use the O(S) time query above.
Thus, the general picture will be:
|...........|...........|...........|...........|...........|...........|...........|
^ ^ ^ ^
L E_L E_R R
We have marked two additional points, E_L and E_R, which are the boundaries of the blocks completely inside [L,R]. Now, it would be nice if we have already precomputed the answer for the query pair (E_L,E_R), because then we will only have to deal with at most 2(S-1) remaining values: [L,E_L) and [E_R,R]. We can indeed precompute the answers at the boundaries, but we can do even better: we can precompute the answers for all pairs (E,R), where E is a boundary point and R is any point in the array! There are only O(BN) pairs, and we can compute the answers in O(BN) time also:
class LazyMap:
...
S = floor(sqrt(N))
B = ceil(N/S)
index = new LazyMap()
block_ans[1..B][1..N]
def precompute():
answer = 0
for b in 1...B
index.clear()
E = b*S-S+1 # left endpoint of the b'th block
answer = 0
for R in E...N
if not index.has_key(A[R])
index[A[R]] = R
answer = max(answer, R - index[A[R]])
block_ans[b][R] = answer
(if you read the â€śquick explanationâ€ť, note that there is a slight difference here: weâ€™re indexing the blocks from 1 to B instead of 0 to B-1)
This means that, in the query, the only remaining values we havenâ€™t considered yet are those in [L,E_L). To consider those, we have to know, for each x in [L,E_L), the last occurrence of A_x in [L,R]. To do so, we will need the following information:
- \text{next}[i], the smallest j > i such that A_i = A_j
- \text{prev}[i], the largest j < i such that A_i = A_j
- \text{last_in_blocks}[j][i], the largest k within the first j blocks such that A_k = A_i
How will this help us? Well, we want to find A_x's last occurrences in [L,R]. So first, we find its last occurrence in the blocks up to E_R (itâ€™s just \text{last_in_blocks}[\text{floor}(R/S)][x]). However, itâ€™s possible that A_x appears in [E_R,R], so we need to use its \text{next} pointers, until we find the last one. Since there are at most S-1 elements in [E_R,R], this seems fast, but it could easily take O(S^2) time for example when most of the values in [L,E_L) and [E_R,R] are equal. Thankfully, this is easily fixed: we only care about the first occurrence of A_x, so if it has been encountered before, then we donâ€™t have to process it again! This ensures that for distinct value in [E_R,R], its set of indices is iterated only once. This therefore guarantees an O(S) running time!
Checking whether an A_x has been encountered before can also be done using the index
approach, or alternatively as \text{prev}[x] \ge L:
def answer_query(L, R):
b_L = ((L+S-1)/S)
b_R = R/S
if b_L >= b_R
# old query here
else
E_L = b_L*S
answer = block_ans[b_L+1][R]
for x in L...E_L
if prev[x] < L # i.e. x hasn't been encountered before
y = last_in_blocks[floor(R/S)][x]
while next[y] <= R
y = next[y]
answer = max(answer, y - x)
return answer
One can now see that the query time is now O(S) Note that b_L \le b_R means that L and R are within O(S) elements away, so we can do the old query instead.
Letâ€™s now see how to precompute \text{next}, \text{prev} and \text{last_in_blocks}. First, \text{next}[i] and \text{prev}[i] can easily be computed in O(N) time with the following code:
...
next[1..N]
prev[1..N]
last[1..M] # initialized to 0
...
def precompute():
...
for i in 1...N
next[i] = N+1
prev[i] = 0
for i in 1...N
j = last[A[i]]
if j != 0
next[j] = i
prev[i] = j
last[A[i]] = i
The last
array stores the last index encountered for every value, and is updated as we traverse the array.
And then \text{last_in_blocks} can be compute in O(BN) time:
...
last_in_blocks[1..B][1..N] # initialized to 0
...
def precompute():
...
for b in 1...B
L = b*S-S+1
R = min(b*S,N)
for y in L...R
if next[y] > R
x = y
while x > 0
last_in_blocks[b][x] = y
x = prev[x]
for x in 1...N
for b in 2...B
if last_in_blocks[b][x] == 0
last_in_blocks[b][x] = last_in_blocks[b-1][x]
The first loop finds the last value encountered at each block (with the check next[y] > R
), and proceeds setting the last_in_blocks
of all the indices until that position with equal value, using the prev
pointer. The second loop fills out the remaining entries, because some values do not have representatives in some blocks.
Running time
Now, what is the total running time then? The precomputation runs in O(BN) time, and each query takes O(S) time, so overall it is O(NB + QS). But remember that B = \Theta(N/S), so the algorithm is just O(N^2/S + QS). But we still have the freedom to choose the value of S. Now, most will simply choose S = \Theta(\sqrt{N}), so that the running time is O((N+Q)\sqrt{N}), but we are special, so we will be more pedantic.
Note that N^2/S is a decreasing function while QS is an increasing function. Also, remember that O(f(x)+g(x)) = O(\max(f(x),g(x))) (why?). Therefore, the best choice for S is one that makes N^2/S and QS equal (at least asymptotically). Thus, we want the choice S = \Theta(N/\sqrt{Q}) instead, and the running time is O(N\sqrt{Q}+Q) (the +Q is there to account for when Q > N^2). For this problem, thereâ€™s not much difference between this and O((N+Q)\sqrt{N}), but the running time O(N\sqrt{Q}+Q) is mostly for theoretical interest, and when Q is much less than N (or much more), youâ€™ll feel the difference.
Optimization side note: there is another way to do the old query without using our LazyMap
, or at least calling has_key
: traverse the array backwards. Here is an example:
...
_index[1..M]
...
def answer_query(L, R):
...
if b_L >= b_R
# old query
answer = 0
for y in R...L by -1
_index[A[y]] = y
for y in L...R
answer = max(answer, y - _index[A[y]])
else
...
return answer
I found that this is a teeny teeny bit faster than the original O(S) old query
Also, when choosing S, one does not have to choose \lfloor \sqrt{N} \rfloor, or even \lfloor N/\sqrt{Q} \rfloor, because there is still a constant hidden in the \Theta notation. This means that you still have the freedom to choose a multiplicative constant for S, which in practice essentially amounts to the freedom to select S however you want. To get the best value for S, try generating a large input (with varying values of M !), and finding the best choice for S via ternary search. The goal is to get the precomputation part and the query part roughly equal in running time. This technique of tweaking the parameters is incredibly useful in long contests where the time limit is usually tight.
Time Complexity:
O(M + (N + Q)\sqrt{N}) but a theoretically it is O(N \sqrt{Q} + Q)
Note that in the problem, Q is actually K.