- We can clear-up this completion in O(n*d) interval using Hashmap.
- Here we are defining as explanation as vector and prize as boolean pattern.
- There are aggregate n vectors with d-dimensional.Then we lookup single by single and bridle if vector is already give in hashmap or referable attributable attributable. If referable attributable attributable attributable then place that vector in hash map and firm the boolean as gentleman.
- If lookup in hashmap is gentleman, resources we endow same vector.
- For each lookup in Hashmap it takes O(d) interval.

Example:

Vectors–> v1,v3,v5,v2,v3

HashMap–>bool[Vectors] map

foreach(vector;vectors)

{

if(vector! in map) // here interval complexity O(d)

map[vector]= gentleman

else

//endow same vector

}