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Array with GCD of any of its subset belongs to the given array?
Here we will see one interesting problem. There is a set of N elements. We have to generate an array such that the GCD of any subset of that array lies in the given set of elements. And another constraint is that the generated array should not be more than thrice the length of the set of the GCD. For example, if there are 4 numbers {2, 4, 6, 12}, then one array will be {2, 2, 4, 2, 6, 2, 12}
To solve this problem, we have to sort the list at first, then if the GCD is the same as the minimum element of the given set, then create array just by putting GCD between each element. Otherwise, no array can be formed.
Algorithm
generateArray(arr, n)
Begin answer := empty array gcd := gcd of array arr if gcd is same as the min element of arr, then for each element e in arr, do append gcd into answer append e into answer done display answer else array cannot be formed end if End
Example
#include<iostream>
#include<vector>
#include<set>
using namespace std;
int gcd(int a, int b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
int getGCDofArray(vector<int> arr) {
int result = arr[0];
for (int i = 1; i < arr.size(); i++)
result = gcd(arr[i], result);
return result;
}
void generateArray(vector<int> arr) {
vector<int> answer;
int GCD_of_array = getGCDofArray(arr);
if(GCD_of_array == arr[0]) { //if gcd is same as minimum element
answer.push_back(arr[0]);
for(int i = 1; i < arr.size(); i++) { //push min before each
element
answer.push_back(arr[0]);
answer.push_back(arr[i]);
}
for (int i = 0; i < answer.size(); i++)
cout << answer[i] << " ";
}
else
cout << "No array can be build";
}
int main() {
int n = 4;
int data[]= {2, 4, 6, 12};
set<int> GCD(data, data + n);
vector<int> arr;
set<int>::iterator it;
for(it = GCD.begin(); it!= GCD.end(); ++it)
arr.push_back(*it);
generateArray(arr);
}Output
2 2 4 2 6 2 12
Updated on: 01-Aug-2019
72 Views
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