Question

Explain the space-time tradeoff implemented in efficient hash tables? b) Use hashing to design and algorithm to identify all duplicate elements in an array, e.g., for the following array [ “r6i”, “opd”, “r6i”, “ydg”, “x5s”, “pc1”, “tni”, “594”, “ wi5”, “ip0”, “vvj”, “oad”, “ydg”, “‘opd” ] the elements [“r6i”, “opd”, “ydg”] are duplicates. The time complexity of the algorithm must be O(n)

Answer 1

In hash tables time complexity will be of o(n) but space for elements will be high for stroing them,we can acess the required element in O(1) time

Here is the algorithm code in cpp:-

Explanation:-First we create a hash table of (unordered map) of string,int ,by default values will be zero.we traverse through the entire array and increse the count for the element by 1,if we encounter same element the count will increase .At last we traverse through th enitire hash if count >1 we will print element corresponding to that element.

#include<bits/stdc++.h>
using namespace std;
#include <unordered_map>
#include<iostream>
int main(){
   string arr[]={"r6i", "opd", "r6i", "ydg", "x5s", "pc1", "tni", "594", "wi5", "ip0", "vvj", "oad", "ydg", "opd" };
   unordered_map<string,int> hash;
int len=sizeof(arr)/sizeof(arr[0]);
   for(int i=0;i<len;i++){
       hash[arr[i]]++;
   }

   cout<<"print duplicates:"<<endl;
   for(auto x:hash){
       if(x.second>1){
           cout<<x.first<<" ";
       }
   }
   cout<<endl;
   return 0;
}