Square Root Decomposition

Author: Benjamin Qi

Splitting up data into smaller chunks to speed up processing.

Focus Problem – read through this problem before continuing!

You should already have done this problem in Point Update Range Sum, but here we'll present two more approaches. Both run in time.


See the first example in CPH.

int n,q;
vi x;
ll block[450];
int BLOCK;
ll sum(int a) { // sum of first a elements of array in O(sqrt(n)) time
ll ans = 0;
F0R(i,a/BLOCK) ans += block[i];
FOR(i,a/BLOCK*BLOCK,a) ans += x[i];
return ans;


See the CPH section on "batch processing."

Maintain a "buffer" of the latest updates (up to ). The answer for each sum query can be calculated with prefix sums and by examining each update within the buffer. When the buffer gets too large (), clear it and recalculate prefix sums.

int n,q;
vi x;
vl cum;
void build() {
cum = {0};
trav(t,x) cum.pb(cum.bk+t);
int main() {

Mo's Algorithm

See CPH 27.3 and the CF link above.

CFvery brief description

Additional Notes

Low constraints (ex. ) and/or high time limits (greater than 2s) can be signs that sqrt decomp is intended.

CPH 262:

In practice, it is not necessary to use the exact value of as a parameter, and instead we may use parameters and where is different from . The optimal parameter depends on the problem and input. For example, if an algorithm often goes through the blocks but rarely inspects single elements inside the blocks, it may be a good idea to divide the array into blocks, each of which contains elements.

As another example, if an update takes time proportional to the size of one block () while a query takes time proportional to the number of blocks times () then we can set to make both updates and queries take time .

Solutions with worse complexities are not necessarily slower (at least for problems with reasonable input sizes, ex. ). I recall an instance where a fast solution passed (where came from a BIT) while an solution did not ... So constant factors are important!

On Trees

The techniques mentioned in the blogs below are extremely rare but worth a mention.

Some more discussion about how sqrt decomp can be used:

CFformat isn't great but tree example is ok



Problems where the best solution I know of involves sqrt decomp.

StatusSourceProblem NameDifficultyTagsSolutionURL
JOIEasyView Solution
Show Tags


View Solution
PlatVery HardExternal Sol
DMOJVery Hard
DMOJVery HardCheck DMOJ


Problems that can be solved without it. But you might as well try to use it!!

StatusSourceProblem NameDifficultyTagsSolutionURL
Show Tags

Mo's algorithm, 2D SRQ

View Solution
IOINormalExternal Sol
PlatHardExternal Sol
CFHardCheck CF
TLXHardCheck TLX
CSAHardCheck CSA
Old GoldHard
Show Tags

Convex Hull

External Sol
CFVery Hard
Show Tags

Convex Hull

Check CF
IOIVery HardExternal Sol
PlatVery HardExternal Sol
IOIVery Hard
Show Tags


External Sol

Module Progress:

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