_We're starting a series of articles describing tricks useful in programming contests. Please keep the comments in English._
Being flexible and easy to code, the square root trick is a pretty popular in the Romanian programming contests community. It even has a name: "jmenul lu Batog" which means Batog's trick :). Bogdan Batog introduced it to a few high school students more than 10 years ago and the trick entered romanian coding contest folklore.
Being flexible and easy to code, the square root trick is pretty popular in the Romanian programming contests community. It even has a name: "jmenul lu Batog" which means Batog's trick :). Bogdan Batog introduced it to a few high school students more than ten years ago and the trick entered romanian coding contest folklore.
Enough introduction, let’s check out a few problems!
The idea is that we can use bucketing or a two level tree as some people call it to improve naive data structures or algorithms. The square root part appears when we minimize the function $n/x + x$, we'll see more about that later on.
Let’s check out a few problems that explain how the trick works.
h2. Range Sum
bq.. Given A, an n elements array, implement a data structure for point updates and range sum queries:
- set(i, x): A[i] := k,
- sum(lo, hi) returns A[lo] + A[lo+1] + .. + A[hi].
- update(i, x): A[i] := x,
- query(lo, hi) returns A[lo] + A[lo+1] + .. + A[hi].
p. The naive solution uses an array. It takes $O(1)$ time for an update and $O(r-l) = O(n)$ for the range sum.
p. The naive solution uses an array. It takes $O(1)$ time for an update and $O(hi - lo) = O(n)$ for the range sum.
A more efficient solution splits the array into length $k$ slices and stores the slice sums in an array $S$.
p. A more efficient solution splits the array into length $k$ slices and stores the slice sums in an array $S$.
The update takes constant time.
!<{margin-right: 20px; auto;display:block;}blog/square-root-trick?image01.png!
p. The update takes constant time, because we have to update the value for A and the value for the corresponding $S$.
The query is interesting. The elements of the first and last slice (partially contained in the queried range) have to be traversed one by one, but for slices completely contained in our range we can use the values in $S$ directly and get a performance boost.
The code looks like this:
== code(c) |
S[i/k] = S[i/k] - A[i] + x
A[i] = x
==
p. Here is an update example:
!{margin-right: 20px; auto;display:block;}blog/square-root-trick?image01.png!
The query is interesting. Slices completely contained in our range are summed up fast. The elements of the first and last slice (partially contained in the queried range) have to be traversed one by one.
p. In $update(6, 5)$ we have to change $A[6]$ to 5 which results in changing the value of $S[1]$ to keep $S$ up to date.
!<{margin-right: 20px; auto;display:block;}blog/square-root-trick?image00.png!
!{margin-right: 20px; auto;display:block;}blog/square-root-trick?image00.png!
In $query(2, 14)$ we get
== code(c) |
query(2, 14) = A[2] + A[3]+
(A[4] + A[5] + A[6] + A[7]) +
(A[8] + A[9] + A[10] + A[11]) +
A[12] + A[13] + A[14]
= A[2] + A[3] +
S[1] + S[2] +
A[12] + A[13] + A[14]
= 0 + 7 + 11 + 9 + 5 + 2 + 0
= 34
==
Here's how the code looks:
The code looks like this:
== code(c) |
i = lo
while (i + 1) % k != 0 and i <= hi:
sum += A[i]
i += 1
while i + k <= hi:
sum += S[i/k]
i += k
while i <= hi:
sum += A[i]
i += 1
def update(S, A, i, k, x):
S[i/k] = S[i/k] - A[i] + x
A[i] = x
def query(S, A, lo, hi, k):
s = 0
i = lo
while (i + 1) % k != 0 and i <= hi:
s += A[i]
i += 1
while i + k <= hi:
s += S[i/k]
i += k
while i <= hi:
s += A[i]
i += 1
return s
==
The query takes less than $k + n/k + k = 2k + n/k$ time. $2k + n/k$ is minimized when $k$ is $O(sqrt(n))$. For $k = sqrt(n)$ the query takes $O(sqrt(n))$ time.
Each query takes on average $k/2 + n/k + k/2 = k + n/k$ time. This is minimized for $k = sqrt(n)$. So we get a $O(sqrt(n))$ time complexity query.
This trick also works for other associative operations, like: min, gcd, product etc.
bq. Given a set $S$ of $n$ points and a query point $p$, find the point in $S$ closest to $p$.
For uniformly distributed points, a good strategy is to represent the space as a grid and maintain a list of inner points for each cell. For a given query point, we can check the cell the point falls into and its neighbouring cells. For a $sqrt(n) * sqrt(n)$ grid we’ll have one point per cell, on average. So, on average, finding the point in $S$ closest to $p$, requires traversing a constant number of cells.
For uniformly distributed points, a good strategy is to represent the space as a grid and maintain a list of inner points for each cell. For a given query point, we can check the cell the point falls into and its neighbouring cells. For a $sqrt(n) * sqrt(n)$ grid we’ll have one point per cell, on average. On average, finding the point in $S$ closest to $p$, requires traversing a constant number of cells.
h2. Longest common subsequence solution
h2. Longest common subsequence
bq. Given two strings A (n characters) and B (m characters), find their longest common subsequence. (eg. The longest common sub sequence for abcabc and abcbcca is abcbc.)
h2. Additional problems
# *(Josephus Problem)* $n$ people numbered from 1 to $n$ sit in a circle and play a game. Starting from the first person and every $k$th person is eliminated. Write an algorithm that prints out the order in which people are eliminated.
# *(Level Ancestor)* You are given an tree of size $n$. $ancestor(node, levelsUp)$ finds the node’s ancestor that is $levelsUp$ steps up. For example, $ancestor(node, 1)$ returns the father and ancestor(node, 2) returns the grandfather. Implement $ancestor(node, levelsUp)$ efficiently. ($O(sqrt(n))$ per query)
# *(Range Median)* You are given an array of size $n$. Implement a data structure to perform update operations $a[i] = k$ and range median operations efficiently. The range median query, $median(l, r)$ returns the median element of the sorted subsequence $a[l..r]$. $O(log{n})$ per update and $O(sqrt(n)log(n))$ per query
# *(Josephus Problem)* $n$ people numbered from 1 to $n$ sit in a circle and play a game. Starting from the first person and every $kth$ person is eliminated. Write an algorithm that prints out the order in which people are eliminated.
# *(Level Ancestor)* You are given an tree of size $n$. $ancestor(node, levelsUp)$ finds the node’s ancestor that is $levelsUp$ steps up. For example, $ancestor(node, 1)$ returns the father and ancestor(node, 2) returns the grandfather. Implement $ancestor(node, levelsUp)$ efficiently. ( $O(sqrt(n))$ per query)
# *(Range Median)* You are given an array of size $n$. Implement a data structure to perform update operations $a[i] = k$ and range median operations efficiently. The range median query, $median(l, r)$ returns the median element of the sorted subsequence $a[l..r]$. $O(log(n))$ per update and -$O(sqrt(n)log(n))$- $O(sqrt(n)log(n)log(U))$ per query
h2. Hope you've enjoyed it! Try to solve Range Median and the other problems in the comments section using this trick.
h2. Hope you've enjoyed it!
Try using the trick to solve Range Median and the other problems in the comments section.
