Given an array arr[], find nearest element for every element such that there is at least one common prime factor. In output, we need to print position of closest element.

Input: arr[] = {2, 9, 4, 3, 13} Output: 3 4 1 2 -1 Explanation : Closest element for 1st element is 3rd. =>Common prime factor of 1st and 3rd elements is 2. Closest element for 2nd element is 4th. =>Common prime factor of 2nd and 4th elements is 3.

**Naive approach**

Common prime factor will only exist if GCD of these two numbers will greater than 1. Simple brute force is to run the two loops one inside the another and iterate one by one from each index to both the sides simultaneously and find the gcd which is greater than 1. Whenever we found the answer then just break the loop and the print. If we reached the end of array after traversing both the sides then simply print -1.

// C++ program to print nearest element with at least // one common prime factor. #include<bits/stdc++.h> using namespace std; void nearestGcd(int arr[], int n) { // Loop covers the every element of arr[] for (int i=0; i<n; ++i) { int closest = -1; // Loop that covers from 0 to i-1 and i+1 // to n-1 indexes simultaneously for (int j=i-1, k=i+1; j>0 || k<=n; --j, ++k) { if (j>=0 && __gcd(arr[i], arr[j]) > 1) { closest = j+1; break; } if (k<n && __gcd(arr[i], arr[k])>1) { closest = k+1; break; } } // print position of closest element cout << closest << " "; } } // Drive code int main() { int arr[] = {2, 9, 4, 3, 13}; int n = sizeof(arr)/sizeof(arr[0]); nearestGcd(arr, n); return 0; }

Output:3 4 1 2 -1

**Time complexity: **O(n^{2})

**Auxiliary space: **O(1)

**Efficient Approach**

We find prime factors of all array elements. To quickly find prime factors, we use Sieve of Eratosthenes. For every element, we consider all prime factors and keep track of closest element with common factor.

// C++ program to print nearest element with at least // one common prime factor. #include <bits/stdc++.h> using namespace std; const int MAX = 100001; const int INF = INT_MAX; int primedivisor[MAX], dist[MAX], pos[MAX], divInd[MAX]; vector<int> divisors[MAX]; // Pre-computation of smallest prime divisor // of all numbers void sieveOfEratosthenes() { for (int i=2; i*i < MAX; ++i) { if (!primedivisor[i]) for (int j = i*i; j < MAX; j += i) primedivisor[j] = i; } // Prime number will have same divisor for (int i = 1; i < MAX; ++i) if (!primedivisor[i]) primedivisor[i] = i; } // Function to calculate all divisors of // input array void findDivisors(int arr[], int n) { for (int i=0; i<MAX; ++i) pos[i] = divInd[i] = -1, dist[i] = INF; for (int i=0; i<n; ++i) { int num = arr[i]; while (num > 1) { int div = primedivisor[num]; divisors[i].push_back(div); while (num % div == 0) num /= div; } } } void nearestGCD(int arr[], int n) { // Pre-compute all the divisors of array // element by using prime factors findDivisors(arr, n); // Traverse all elements, for (int i=0; i<n; ++i) { // For every divisor of current element, // find closest element. for (auto &div: divisors[i]) { // Visit divisor if not visited if (divInd[div] == -1) divInd[div] = i; else { // Fetch the index of visited divisor int ind = divInd[div]; // Update the divisor index to current index divInd[div] = i; // Set the minimum distance if (dist[i] > abs(ind-i)) { // Set the min distance of current // index 'i' to nearest one dist[i] = abs(ind-i); // Add 1 as indexing starts from 0 pos[i] = ind + 1; } if (dist[ind] > abs(ind-i)) { // Set the min distance of found index 'ind' dist[ind] = abs(ind-i); // Add 1 as indexing starts from 0 pos[ind] = i + 1; } } } } } // Driver code int main() { // Simple sieve to find smallest prime // divisor of number from 2 to MAX sieveOfEratosthenes(); int arr[] = {2, 9, 4, 3, 13}; int n = sizeof(arr)/sizeof(arr[0]); // function to calculate nearest distance // of every array elements nearestGCD(arr, n); // Print the nearest distance having GDC>1 for (int i=0; i<n; ++i) cout << pos[i] << " "; return 0; }

Output:3 4 1 2 -1

**Time complexity: **O(MAX * log(log (MAX) ) )

**Auxiliary space: **O(MAX)

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