Elements of an array that are not divisible by any element of another array
Last Updated :
13 Jul, 2022
Given two arrays A[] and B[], write an efficient code to determine if every element of B[] is divisible by at least 1 element of A[]. Display those elements of B[], which are not divisible by any of the elements in A[].
Examples :
Input : A[] = {100, 200, 400, 100, 600}
B[] = {45, 90, 48, 1000, 3000}
Output : 45, 90, 48
The output elements are those that are
not divisible by any element of A[].
Method I (Naive Implementation):
- Iterate through every single element of B[].
- Check if it is divisible by at least 1 element of A[] or not. If not divisible by any, then print it.
Implementation:
C++
// C++ code for naive implementation
#include<iostream>
using namespace std;
// Function for checking the condition
// with 2 loops
void printNonDivisible(int A[], int B[],
int n, int m)
{
for (int i = 0; i < m; i++)
{
int j = 0;
for (j = 0; j < n; j++)
if( B[i] % A[j] == 0 )
break;
// If none of the elements in A[]
// divided B[i]
if (j == n)
cout << B[i] << endl;
}
}
// Driver code
int main()
{
int A[] = {100, 200, 400, 100};
int n = sizeof(A)/sizeof(A[0]);
int B[] = {190, 200, 87, 600, 800};
int m = sizeof(B)/sizeof(B[0]);
printNonDivisible(A, B, n, m);
return 0;
}
// Java code for naive implementation
import java.io.*;
public class GFG {
// Function for checking the condition
// with 2 loops
static void printNonDivisible(int []A, int []B,
int n, int m)
{
for (int i = 0; i < m; i++)
{
int j = 0;
for (j = 0; j < n; j++)
if( B[i] % A[j] == 0 )
break;
// If none of the elements
// in A[] divided B[i]
if (j == n)
System.out.println(B[i]);
}
}
// Driver code
static public void main (String[] args)
{
int []A = {100, 200, 400, 100};
int n = A.length;
int []B = {190, 200, 87, 600, 800};
int m = B.length;
printNonDivisible(A, B, n, m);
}
}
// This code is contributed by vt_m .
# Python3 code for naive implementation
import math as mt
# Function for checking the condition
# with 2 loops
def printNonDivisible(A, B, n, m):
for i in range(m):
j = 0
for j in range(n):
if(B[i] % A[j] == 0):
break
# If none of the elements in A[]
# divided B[i]
if (j == n - 1):
print(B[i])
# Driver code
A = [100, 200, 400, 100]
n = len(A)
B = [190, 200, 87, 600, 800]
m = len(B)
printNonDivisible(A, B, n, m)
# This code is contributed by#
# mohit kumar 29
// C# code for naive implementation
using System;
public class GFG {
// Function for checking the
// condition with 2 loops
static void printNonDivisible(int []A, int []B,
int n, int m)
{
for (int i = 0; i < m; i++)
{
int j = 0;
for (j = 0; j < n; j++)
if( B[i] % A[j] == 0 )
break;
// If none of the elements
// in A[] divided B[i]
if (j == n)
Console.WriteLine(B[i]);
}
}
// Driver code
static public void Main ()
{
int []A = {100, 200, 400, 100};
int n = A.Length;
int []B = {190, 200, 87, 600, 800};
int m = B.Length;
printNonDivisible(A, B, n, m);
}
}
// This code is contributed by vt_m .
<?php
// PHP code for naive implementation
// Function for checking
// the condition with 2 loops
function printNonDivisible($A, $B,
$n, $m)
{
for ($i = 0; $i < $m; $i++)
{
$j = 0;
for ($j = 0; $j < $n; $j++)
if( $B[$i] % $A[$j] == 0 )
break;
// If none of the elements
// in A[] divided B[i]
if ($j == $n)
echo $B[$i], "\n";
}
}
// Driver code
$A= array (100, 200, 400, 100);
$n = sizeof($A);
$B = array (190, 200, 87, 600, 800);
$m = sizeof($B);
printNonDivisible($A, $B, $n, $m);
// This code is contributed by ajit
?>
<script>
// Javascript code for naive implementation
// Function for checking the
// condition with 2 loops
function printNonDivisible(A, B, n, m)
{
for (let i = 0; i < m; i++)
{
let j = 0;
for (j = 0; j < n; j++)
if( B[i] % A[j] == 0 )
break;
// If none of the elements
// in A[] divided B[i]
if (j == n)
document.write(B[i] + "</br>");
}
}
let A = [100, 200, 400, 100];
let n = A.length;
let B = [190, 200, 87, 600, 800];
let m = B.length;
printNonDivisible(A, B, n, m);
</script>
Time Complexity :- O(n*m)
Auxiliary Space :- O(1)
Method 2 (Efficient when elements in are small)
- Maintain an array mark[] to mark the multiples of the numbers in A[].
- Mark all the multiples of all the elements in A[], till a max of B[].
- Check if mark[B[i]] value for every element n in B[] is not 0 and print if not marked.
Implementation:
C++
// CPP code for improved implementation
#include<bits/stdc++.h>
using namespace std;
// Function for printing all elements of B[]
// that are not divisible by any element of A[]
void printNonDivisible(int A[], int B[], int n,
int m)
{
// Find maximum element in B[]
int maxB = 0;
for (int i = 0; i < m; i++)
if (B[i] > maxB)
maxB = B[i];
// Initialize all multiples as marked
int mark[maxB];
memset(mark, 0, sizeof(mark));
// Marking the multiples of all the
// elements of the array.
for (int i = 0; i < n; i++)
for (int x = A[i]; x <= maxB; x += A[i])
mark[x]++;
// Print not marked elements
for (int i = 0; i < m; i++)
if (! mark[B[i]])
cout << B[i] << endl;
}
// Driver function
int main()
{
int A[] = {100, 200, 400, 100};
int n = sizeof(A)/sizeof(A[0]);
int B[] = {190, 200, 87, 600, 800};
int m = sizeof(B)/sizeof(B[0]);
printNonDivisible(A, B, n, m);
return 0;
}
// Java code for improved implementation
import java.io.*;
class GFG
{
// Function for printing all elements of B[]
// that are not divisible by any element of A[]
static void printNonDivisible(int []A, int []B,
int n,int m)
{
// Find maximum element in B[]
int maxB = 0;
for (int i = 0; i < m; i++)
if (B[i] > maxB)
maxB = B[i];
// Initialize all multiples as marked
int [] mark = new int[maxB + 1];
for(int i = 0; i < maxB; i++)
mark[i]=0;
// Marking the multiples of all the
// elements of the array.
for (int i = 0; i < n; i++)
for (int x = A[i]; x <= maxB; x += A[i])
mark[x]++;
// Print not marked elements
for (int i = 0; i < m; i++)
if (mark[B[i]] == 0)
System.out.println(B[i]);
}
// Driver code
static public void main(String[] args)
{
int []A= {100, 200, 400, 100};
int n = A.length;
int []B= {190, 200, 87, 600, 800};
int m = B.length;
printNonDivisible(A, B, n, m);
}
}
// This code is contributed by Mohit Kumar.
# Python 3 code for improved implementation
# Function for printing all elements of B[]
# that are not divisible by any element of A[]
def printNonDivisible(A, B, n, m):
# Find maximum element in B[]
maxB = 0
for i in range(0, m, 1):
if (B[i] > maxB):
maxB = B[i]
# Initialize all multiples as marked
mark = [0 for i in range(maxB)]
# Marking the multiples of all
# the elements of the array.
for i in range(0, n, 1):
for x in range(A[i], maxB, A[i]):
mark[x] += 1
# Print not marked elements
for i in range(0, m - 1, 1):
if (mark[B[i]] == 0):
print(B[i])
# Driver Code
if __name__ == '__main__':
A = [100, 200, 400, 100]
n = len(A)
B = [190, 200, 87, 600, 800]
m = len(B)
printNonDivisible(A, B, n, m)
# This code is contributed by
# Shashank_Sharma
// C# code for improved implementation
using System;
class GFG
{
// Function for printing all elements of []B
// that are not divisible by any element of []A
static void printNonDivisible(int []A, int []B,
int n, int m)
{
// Find maximum element in []B
int maxB = 0;
for (int i = 0; i < m; i++)
if (B[i] > maxB)
maxB = B[i];
// Initialize all multiples as marked
int [] mark = new int[maxB + 1];
for(int i = 0; i < maxB; i++)
mark[i] = 0;
// Marking the multiples of all the
// elements of the array.
for (int i = 0; i < n; i++)
for (int x = A[i]; x <= maxB; x += A[i])
mark[x]++;
// Print not marked elements
for (int i = 0; i < m; i++)
if (mark[B[i]] == 0)
Console.WriteLine(B[i]);
}
// Driver code
static public void Main(String[] args)
{
int []A= {100, 200, 400, 100};
int n = A.Length;
int []B= {190, 200, 87, 600, 800};
int m = B.Length;
printNonDivisible(A, B, n, m);
}
}
// This code is contributed by Rajput-Ji
<?php
// PHP code for improved implementation
// Function for printing all elements of B[]
// that are not divisible by any element of A[]
function printNonDivisible($A, $B, $n, $m)
{
// Find maximum element in B[]
$maxB = 0;
for ($i = 0; $i < $m; $i++)
{
if ($B[$i] > $maxB)
$maxB = $B[$i];
}
// Initialize all multiples as marked
$mark = array();
for ($i = 0; $i < $maxB; $i++)
{
$mark[] = "0";
}
// Marking the multiples of all
// the elements of the array.
for ($i = 0; $i < $n; $i++)
{
for ($x = $A[$i]; $x < $maxB;
$x += $A[$i])
{
$mark[$x] += 1;
}
}
// Print not marked elements
for ($i = 0; $i < $m - 1; $i++)
{
if ($mark[$B[$i]] == 0)
echo "$B[$i]\n";
}
}
// Driver Code
$A = array(100, 200, 400, 100);
$n = count($A);
$B = array(190, 200, 87, 600, 800);
$m = count($B);
printNonDivisible($A, $B, $n, $m);
// This code is contributed by
// Srathore
?>
<script>
// Javascript code for improved implementation
// Function for printing all elements of []B
// that are not divisible by any element of []A
function printNonDivisible(A, B, n, m)
{
// Find maximum element in []B
let maxB = 0;
for (let i = 0; i < m; i++)
if (B[i] > maxB)
maxB = B[i];
// Initialize all multiples as marked
let mark = new Array(maxB + 1);
for(let i = 0; i < maxB; i++)
mark[i] = 0;
// Marking the multiples of all the
// elements of the array.
for (let i = 0; i < n; i++)
for (let x = A[i]; x <= maxB; x += A[i])
mark[x]++;
// Print not marked elements
for (let i = 0; i < m; i++)
if (mark[B[i]] == 0)
document.write(B[i] + "</br>");
}
let A= [100, 200, 400, 100];
let n = A.length;
let B= [190, 200, 87, 600, 800];
let m = B.length;
printNonDivisible(A, B, n, m);
</script>
Time Complexity :- O(m + n*(max(B[]/min(A[])))
Auxiliary Space :- O(n) + O(m) + O(max(B[]))
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