# Between Two Sets

### Problem Statement :

```There will be two arrays of integers. Determine all integers that satisfy the following two conditions:

The elements of the first array are all factors of the integer being considered
The integer being considered is a factor of all elements of the second array
These numbers are referred to as being between the two arrays. Determine how many such numbers exist.

Example
a=[2,4]
b=[24,36]

There are two numbers between the arrays: 6 and 12.
6%2=0, 6%6=0, 24%6=0 and 36%6=0 for the first value.
12%2=0, 12%6=0 and 24%12=0, 36%12=0 for the second value. Return 2.

Function Description

Complete the getTotalX function in the editor below. It should return the number of integers that are betwen the sets.

getTotalX has the following parameter(s):

int a[n]: an array of integers
int b[m]: an array of integers
Returns

int: the number of integers that are between the sets

Input Format

The first line contains two space-separated integers, n and m, the number of elements in arrays a and b.
The second line contains n distinct space-separated integers a[i] where 0 <= i < n.
The third line contains m distinct space-separated integers b[j] where 0 <= j < m.

Constraints
1 <= n,m <= 10
1 <= a[i] <= 100
1 <= b[j] <= 100```

### Solution :

```                            ```Solution in C :

C  :

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
int n,i,min,max,j,c=0,f;
int m;
scanf("%d %d",&n,&m);
int *a = malloc(sizeof(int) * n);
for(int a_i = 0; a_i < n; a_i++){
scanf("%d",&a[a_i]);
}
int *b = malloc(sizeof(int) * m);
for(int b_i = 0; b_i < m; b_i++){
scanf("%d",&b[b_i]);
}
max=a[0];
for(i=0;i<n;i++)
{
if(a[i]>max)
max=a[i];
}
min=b[0];
for(i=0;i<m;i++)
{
if(b[i]<min)
min=b[i];
}
for(i=max;i<=min;i++)
{
f=0;
for(j=0;j<n;j++)
{
if(i%a[j]!=0)
{
f=1;
break;
}
}
if(f==0)
{
for(j=0;j<m;j++)
{
if(b[j]%i!=0)
{
f=1;
break;
}
}
}
if(f==0)
c++;
}
printf("%d",c);
return 0;
}

C++  :

#include <iostream>
#include <vector>
#include <cmath>
#include <ctime>
#include <cassert>
#include <cstdio>
#include <queue>
#include <set>
#include <map>
#include <fstream>
#include <cstdlib>
#include <string>
#include <cstring>
#include <algorithm>
#include <numeric>

#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)

using namespace std;

typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef long long i64;
typedef vector<i64> vi64;
typedef vector<vi64> vvi64;

template<class T> bool uin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool uax(T &a, T b) { return a < b ? (a = b, true) : false; }

int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.precision(10);
cout << fixed;
#ifdef LOCAL_DEFINE
freopen("input.txt", "rt", stdin);
#endif

int n, m;
cin >> n >> m;
vi a(n); forn(i, n) cin >> a[i];
vi b(m); forn(i, m) cin >> b[i];
int ans = 0;
for1(k, 100) {
bool ok = true;
for (int x: a) ok &= k % x == 0;
for (int x: b) ok &= x % k == 0;
if (ok) ++ans;
}
cout << ans << '\n';

#ifdef LOCAL_DEFINE
cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
return 0;
}

Java  :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int m = in.nextInt();
int[] a = new int[n];
for(int a_i=0; a_i < n; a_i++){
a[a_i] = in.nextInt();
}
int[] b = new int[m];
for(int b_i=0; b_i < m; b_i++){
b[b_i] = in.nextInt();
}
Arrays.sort(a);
Arrays.sort(b);
int count=0;
for (int i=a[n-1]; i<=b[0]; i++){
boolean flag=true;
for (int j=0; j<n; j++){
if (i%a[j]!=0){
flag=false;
break;
}
}
if (flag==true){
for (int j=0; j<m; j++){
if (b[j]%i!=0){
flag=false;
break;
}
}
}
if (flag==true)
count++;
}
System.out.println (count);
}
}

python3  :

if __name__=='__main__':
n,m = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
ans = 0
for i in range(1,101):
flag = True
for j in a:
if i%j!=0:
flag = False
break
if flag:
for k in b:
if k%i!=0:
flag = False
break
if flag:
ans+=1
print(ans)```
```

## Contacts

We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co

## No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio

## Cube Summation

You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coor

## Direct Connections

Enter-View ( EV ) is a linear, street-like country. By linear, we mean all the cities of the country are placed on a single straight line - the x -axis. Thus every city's position can be defined by a single coordinate, xi, the distance from the left borderline of the country. You can treat all cities as single points. Unfortunately, the dictator of telecommunication of EV (Mr. S. Treat Jr.) do

## Subsequence Weighting

A subsequence of a sequence is a sequence which is obtained by deleting zero or more elements from the sequence. You are given a sequence A in which every element is a pair of integers i.e A = [(a1, w1), (a2, w2),..., (aN, wN)]. For a subseqence B = [(b1, v1), (b2, v2), ...., (bM, vM)] of the given sequence : We call it increasing if for every i (1 <= i < M ) , bi < bi+1. Weight(B) =