Given a list of numbers ( fixed list) Now given any other list, how can you efficiently find out if there is any element in the second list that is an element of the first list (fixed list) ?



Given a list of numbers ( fixed list) Now given any other list, how can you efficiently find out if there is any element in the second list that is an element of the first list (fixed list) ?

Algorithm:

  • In this method, we store all elements of second array in a hash table (unordered_set). One by one check all elements of first array and print all those elements which are not present in the hash table.

Example:

#include<stdio.h>
 
// C++ efficient program to 
// find elements which are not
// present in second array
#include<bits/stdc++.h>
using namespace std;
 
// Function for finding elements which are there in a[] but not in b[].
void findMissing(int a[], int b[], 
                 int n, int m)
{
    // Store all elements of 
    // second array in a hash table
    unordered_set  s;
    for (int i = 0; i < m; i++)
        s.insert(b[i]);
 
    // Print all elements of first array that are not present in hash table
    for (int i = 0; i < n; i++)
        if (s.find(a[i]) == s.end())
            cout << a[i] << " ";
}
 
// Driver code
int main()
{
    int a[] = { 1, 2, 6, 3, 4, 5 };
    int b[] = { 2, 4, 3, 1, 0 };
    int n = sizeof(a) / sizeof(a[0]);
    int m = sizeof(b) / sizeof(b[1]);
    findMissing(a, b, n, m);
    return 0;
} 

Output:

5 6

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