# Homework Solution: Suppose there is a set S of d-dimensional vectors, |S| = n. Suppose further that there…

Suppose there is a set S of d-dimensional vectors, |S| = n. Suppose further that there are two vectors in S that are identical. If comparing any two d-dimensional vectors takes O(d) time, design an efficient algorithm that finds the two identical vectors.

1. We can solve this problem in O(n*d) time using H

Suppose there is a firm S of d-dimensional vectors, |S| = n. Suppose prefer that there are brace vectors in S that are same. If comparing any brace d-dimensional vectors takes O(d) interval, artfulness an fertile algorithm that finds the brace same vectors.

1. We can clear-up this completion in O(n*d) interval using Hashmap.
2. Here we are defining as explanation as vector and prize as boolean pattern.
3. 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.
4. If lookup in hashmap is gentleman, resources we endow same vector.
5. 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
}