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Find an integer X which is divisor of all except exactly one element in an array in C++
Concept
With respect of a given array of integers, our task is to determine an integer B which isthe divisor of all except for exactly one element in the given array.
It should be noted that the GCD of all the elements is not 1.
Input
arr[] = {8, 16, 4, 24}
Output
8 8 is the divisor of all except 4.
Input
arr[] = {50, 15, 40, 41}
Output
5 5 is the divisor of all except 41.
Method
We create a prefix array A such that position or index i contains the GCD of all the elements from 1 to i. In the similar way, create a suffix array C such that index i contains the GCD of all the elements from i to n-1 (last index). It has been seen that if the GCD of A[i-1] and C[i+1] is not the divisor of the element at i, then it is the required answer.
Example
// C++ program to find the divisor of all // except for exactly one element in an array. #include <bits/stdc++.h> using namespace std; // Shows function that returns the divisor of all // except for exactly one element in an array. int getDivisor1(int a1[], int n1){ // If there's only one element in the array if (n1 == 1) return (a1[0] + 1); int A[n1], C[n1]; // Now creating prefix array of GCD A[0] = a1[0]; for (int i = 1; i < n1; i++) A[i] = __gcd(a1[i], A[i - 1]); // Now creating suffix array of GCD C[n1-1] = a1[n1-1]; for (int i = n1 - 2; i >= 0; i--) C[i] = __gcd(A[i + 1], a1[i]); // Used to iterate through the array for (int i = 0; i <= n1; i++) { // Shows variable to store the divisor int cur1; // now getting the divisor if (i == 0) cur1 = C[i + 1]; else if (i == n1 - 1) cur1 = A[i - 1]; else cur1 = __gcd(A[i - 1], C[i + 1]); // Used to check if it is not a divisor of a[i] if (a1[i] % cur1 != 0) return cur1; } return 0; } // Driver code int main(){ int a1[] = { 50,15,40,41 }; int n1 = sizeof(a1) / sizeof(a1[0]); cout << getDivisor1(a1, n1); return 0; }
Output
5
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