Sqrt (or Square Root) Decomposition Technique

Software Development
#1

Hello Everyone,

The key concept of this technique is to decompose given array into small chunks specifically of size sqrt(n).

Let’s say we have an array of n elements and we decompose this array into small chunks of size sqrt(n). We will be having exactly sqrt(n) such chunks provided that n is a perfect square. Therefore, now our array on n elements is decomposed into sqrt(n) blocks, where each block contains sqrt(n) elements (assuming size of array is perfect square).

Let’s consider these chunks or blocks as an individual array each of which contains sqrt(n) elements and you have computed your desired answer(according to your problem) individually for all the chunks. Now, you need to answer certain queries asking you the answer for the elements in range l to r(l and r are starting and ending indices of the array) in the original n sized array.

The naive approach is simply to iterate over each element in range l to r and calculate its corresponding answer. Therefore, the Time Complexity per query will be O(n).

Time Complexity: Let’s consider a query [l = 1 and r = n-2] (n is the size of the array and has a 0 based indexing). Therefore, for this query exactly ( sqrt(n) – 2 ) blocks will be completely overlapped where as the first and last block will be partially overlapped with just one element left outside the overlapping range. So, the completely overlapped blocks can be summed up in ( sqrt(n) – 2 ) ~ sqrt(n) iterations, whereas first and last block are needed to be traversed one by one separately. But as we know that the number of elements in each block is at max sqrt(n), to sum up last block individually we need to make,
(sqrt(n)-1) ~ sqrt(n) iterations and same for the last block.
So, the overall Complexity = O(sqrt(n)) + O(sqrt(n)) + O(sqrt(n)) = O(3*sqrt(N)) = O(sqrt(n))

Update Queries(Point update) :

In this query we simply find the block in which the given index lies, then subtract its previous value and add the new updated value as per the point update query.
Time Complexity : O(1)

Implementation :

The implementation of the above trick is given below

// C++ program to demonstrate working of Square Root

// Decomposition.

#include "iostream"

#include "math.h"

using namespace std;

#define MAXN 10000

#define SQRSIZE 100

int arr[MAXN]; // original array

int block[SQRSIZE]; // decomposed array

int blk_sz; // block size

// Time Complexity : O(1)

void update( int idx, int val)

{

int blockNumber = idx / blk_sz;

block[blockNumber] += val - arr[idx];

arr[idx] = val;

}

// Time Complexity : O(sqrt(n))

int query( int l, int r)

{

int sum = 0;

while (l<r and l%blk_sz!=0 and l!=0)

{

// traversing first block in range

sum += arr[l];

l++;

}

while (l+blk_sz <= r)

{

// traversing completely overlapped blocks in range

sum += block[l/blk_sz];

l += blk_sz;

}

while (l<=r)

{

// traversing last block in range

sum += arr[l];

l++;

}

return sum;

}

// Fills values in input[]

void preprocess( int input[], int n)

{

// initiating block pointer

int blk_idx = -1;

// calculating size of block

blk_sz = sqrt (n);

// building the decomposed array

for ( int i=0; i<n; i++)

{

arr[i] = input[i];

if (i%blk_sz == 0)

{

// entering next block

// incementing block pointer

blk_idx++;

}

block[blk_idx] += arr[i];

}

}

// Driver code

int main()

{

// We have used separate array for input because

// the purpose of this code is to explain SQRT

// decomposition in competitive programming where

// we have multiple inputs.

int input[] = {1, 5, 2, 4, 6, 1, 3, 5, 7, 10};

int n = sizeof (input)/ sizeof (input[0]);

preprocess(input, n);

cout << "query(3,8) : " << query(3, 8) << endl;

cout << "query(1,6) : " << query(1, 6) << endl;

update(8, 0);

cout << "query(8,8) : " << query(8, 8) << endl;

return 0;

}

Output:

query(3,8) : 26 query(1,6) : 21 query(8,8) : 0

Note : The above code works even if n is not perfect square. In the case, the last block will contain even less number of elements than sqrt(n), thus reducing the number of iterations.
Let’s say n = 10. In this case we will have 4 blocks first three block of size 3 and last block of size 1.