Given two arrays A[] and B[], the task is to find the integers which are divisible by all the elements of array A[] and divide all the elements of array B[].
Examples:
Input: A[] = {1, 2, 2, 4}, B[] = {16, 32, 64}
Output: 4 8 16
4, 8 and 16 are the only numbers that
are multiples of all the elements of array A[]
and divide all the elements of array B[]Input: A[] = {2, 3, 6}, B[] = {42, 84}
Output: 6 42
Approach: If X is a multiple of all the elements of the first array then X must be a multiple of the LCM of all the elements of the first array.
Similarly, If X is a factor of all the elements of the second array then it must be a factor of the GCD of all the elements of the second array and such X will exist only if GCD of the second array is divisible by the LCM of the first array.
If it is divisible then X can be any value from the range [LCM, GCD] which is a multiple of LCM and evenly divides GCD.
Below is the implementation of above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the LCM of two numbersint lcm(int x, int y){ int temp = (x * y) / __gcd(x, y); return temp;}// Function to print the required numbersvoid findNumbers(int a[], int n, int b[], int m){ // To store the lcm of array a[] elements // and the gcd of array b[] elements int lcmA = 1, gcdB = 0; // Finding LCM of first array for (int i = 0; i < n; i++) lcmA = lcm(lcmA, a[i]); // Finding GCD of second array for (int i = 0; i < m; i++) gcdB = __gcd(gcdB, b[i]); // No such element exists if (gcdB % lcmA != 0) { cout << "-1"; return; } // All the multiples of lcmA which are // less than or equal to gcdB and evenly // divide gcdB will satisfy the conditions int num = lcmA; while (num <= gcdB) { if (gcdB % num == 0) cout << num << " "; num += lcmA; }}// Driver codeint main(){ int a[] = { 1, 2, 2, 4 }; int b[] = { 16, 32, 64 }; int n = sizeof(a) / sizeof(a[0]); int m = sizeof(b) / sizeof(b[0]); findNumbers(a, n, b, m); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{static int __gcd(int a, int b) { if (b == 0) return a; return __gcd(b, a % b); }// Function to return the LCM of two numbersstatic int lcm(int x, int y){ int temp = (x * y) / __gcd(x, y); return temp;}// Function to print the required numbersstatic void findNumbers(int a[], int n, int b[], int m){ // To store the lcm of array a[] elements // and the gcd of array b[] elements int lcmA = 1, gcdB = 0; // Finding LCM of first array for (int i = 0; i < n; i++) lcmA = lcm(lcmA, a[i]); // Finding GCD of second array for (int i = 0; i < m; i++) gcdB = __gcd(gcdB, b[i]); // No such element exists if (gcdB % lcmA != 0) { System.out.print("-1"); return; } // All the multiples of lcmA which are // less than or equal to gcdB and evenly // divide gcdB will satisfy the conditions int num = lcmA; while (num <= gcdB) { if (gcdB % num == 0) System.out.print(num + " "); num += lcmA; }}// Driver codepublic static void main(String[] args){ int a[] = { 1, 2, 2, 4 }; int b[] = { 16, 32, 64 }; int n = a.length; int m = b.length; findNumbers(a, n, b, m);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach from math import gcd# Function to return the LCM of two numbers def lcm( x, y) : temp = (x * y) // gcd(x, y); return temp; # Function to print the required numbers def findNumbers(a, n, b, m) : # To store the lcm of array a[] elements # and the gcd of array b[] elements lcmA = 1; __gcdB = 0; # Finding LCM of first array for i in range(n) : lcmA = lcm(lcmA, a[i]); # Finding GCD of second array for i in range(m) : __gcdB = gcd(__gcdB, b[i]); # No such element exists if (__gcdB % lcmA != 0) : print("-1"); return; # All the multiples of lcmA which are # less than or equal to gcdB and evenly # divide gcdB will satisfy the conditions num = lcmA; while (num <= __gcdB) : if (__gcdB % num == 0) : print(num, end = " "); num += lcmA; # Driver code if __name__ == "__main__" : a = [ 1, 2, 2, 4 ]; b = [ 16, 32, 64 ]; n = len(a); m = len(b); findNumbers(a, n, b, m); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System;class GFG{static int __gcd(int a, int b) { if (b == 0) return a; return __gcd(b, a % b); }// Function to return the LCM of two numbersstatic int lcm(int x, int y){ int temp = (x * y) / __gcd(x, y); return temp;}// Function to print the required numbersstatic void findNumbers(int []a, int n, int []b, int m){ // To store the lcm of array a[] elements // and the gcd of array b[] elements int lcmA = 1, gcdB = 0; // Finding LCM of first array for (int i = 0; i < n; i++) lcmA = lcm(lcmA, a[i]); // Finding GCD of second array for (int i = 0; i < m; i++) gcdB = __gcd(gcdB, b[i]); // No such element exists if (gcdB % lcmA != 0) { Console.Write("-1"); return; } // All the multiples of lcmA which are // less than or equal to gcdB and evenly // divide gcdB will satisfy the conditions int num = lcmA; while (num <= gcdB) { if (gcdB % num == 0) Console.Write(num + " "); num += lcmA; }}// Driver codepublic static void Main(String[] args){ int []a = { 1, 2, 2, 4 }; int []b = { 16, 32, 64 }; int n = a.Length; int m = b.Length; findNumbers(a, n, b, m);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approach// Function to find nth centered// tridecagonal numberfunction __gcd(a, b) { if (b == 0) return a; return __gcd(b, a % b); }// Function to return the LCM of two numbersfunction lcm(x, y){ var temp = (x * y) / __gcd(x, y); return temp;}// Function to print the required numbersfunction findNumbers(a, n, b, m){ // To store the lcm of array a[] elements // and the gcd of array b[] elements var lcmA = 1, gcdB = 0; // Finding LCM of first array for(var i = 0; i < n; i++) lcmA = lcm(lcmA, a[i]); // Finding GCD of second array for(var i = 0; i < m; i++) gcdB = __gcd(gcdB, b[i]); // No such element exists if (gcdB % lcmA != 0) { document.write("-1"); return; } // All the multiples of lcmA which are // less than or equal to gcdB and evenly // divide gcdB will satisfy the conditions var num = lcmA; while (num <= gcdB) { if (gcdB % num == 0) document.write(num + " "); num += lcmA; }}// Driver codevar a = [ 1, 2, 2, 4 ];var b = [ 16, 32, 64 ];var n = a.length;var m = b.length;findNumbers(a, n, b, m);// This code is contributed by Ankita saini</script> |
Output:
4 8 16
Time Complexity: O(max(n,m) * log(min(a, b)))
Auxiliary Space: O(1)
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