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# Find the number of pairs such that their gcd is equals to 1

Given an array a of size N. The task is to find the number of pairs such that gcd(a[i], a[j]) is equal to 1, where 1 ≤ i < j ≤ N

Examples:

Input : a[] = {1, 2, 4, 6}
Output :
{1, 2}, {1, 4}, {1, 6} are such pairs

Input : a[] = {1, 2, 3, 4, 5, 6}
Output : 11

Approach :
The answer is to sum of μ(X) * C(D(X), 2) overall integer X. Where, μ(X) is Mobius function, C(N, K) is the selection of K things from N and D(X) is the number of integers in the given sequence that are divisible by X.
The correctness of the solution follows from the fact that we can do an inclusion-exclusion principle solution and to show that it is, in fact, equal to our answer. That means that we will add to the answer the number of pairs that are divisible by some intermediate (in the IEP) product D if D is formed by multiplication of even number of prime numbers and subtract this number of pairs otherwise.

So, we get:

• 1 for addition, because that is Möbius function for square-free numbers with even number of prime divisors.
• -1 for subtraction, that is Mobius function for square-free numbers with an odd number of prime divisors.
• 0 for square-free numbers. By the definition, they can’t occur in our IEP solution.

Below is the implementation of the above approach :

## C++

 // CPP program to find the number of pairs// such that gcd equals to 1#include using namespace std; #define N 100050 int lpf[N], mobius[N]; // Function to calculate least// prime factor of each numbervoid least_prime_factor(){    for (int i = 2; i < N; i++)         // If it is a prime number        if (!lpf[i])             for (int j = i; j < N; j += i)                 // For all multiples which are not                // visited yet.                if (!lpf[j])                    lpf[j] = i;} // Function to find the value of Mobius function// for all the numbers from 1 to nvoid Mobius(){    for (int i = 1; i < N; i++) {         // If number is one        if (i == 1)            mobius[i] = 1;        else {             // If number has a squared prime factor            if (lpf[i / lpf[i]] == lpf[i])                mobius[i] = 0;             // Multiply -1 with the previous number            else                mobius[i] = -1 * mobius[i / lpf[i]];        }    }} // Function to find the number of pairs// such that gcd equals to 1int gcd_pairs(int a[], int n){    // To store maximum number    int maxi = 0;     // To store frequency of each number    int fre[N] = { 0 };     // Find frequency and maximum number    for (int i = 0; i < n; i++) {        fre[a[i]]++;        maxi = max(a[i], maxi);    }     least_prime_factor();    Mobius();     // To store number of pairs with gcd equals to 1    int ans = 0;     // Traverse through the all possible elements    for (int i = 1; i <= maxi; i++) {        if (!mobius[i])            continue;         int temp = 0;        for (int j = i; j <= maxi; j += i)            temp += fre[j];         ans += temp * (temp - 1) / 2 * mobius[i];    }     // Return the number of pairs    return ans;} // Driver codeint main(){    int a[] = { 1, 2, 3, 4, 5, 6 };     int n = sizeof(a) / sizeof(a[0]);     // Function call    cout << gcd_pairs(a, n);     return 0;}

## Java

 // Java program to find the number of pairs// such that gcd equals to 1class GFG{ static int N = 100050; static int []lpf = new int[N];static int []mobius = new int[N]; // Function to calculate least// prime factor of each numberstatic void least_prime_factor(){    for (int i = 2; i < N; i++)         // If it is a prime number        if (lpf[i] == 0)             for (int j = i; j < N; j += i)                 // For all multiples which are not                // visited yet.                if (lpf[j] == 0)                    lpf[j] = i;} // Function to find the value of Mobius function// for all the numbers from 1 to nstatic void Mobius(){    for (int i = 1; i < N; i++)    {         // If number is one        if (i == 1)            mobius[i] = 1;        else        {             // If number has a squared prime factor            if (lpf[i / lpf[i]] == lpf[i])                mobius[i] = 0;             // Multiply -1 with the previous number            else                mobius[i] = -1 * mobius[i / lpf[i]];        }    }} // Function to find the number of pairs// such that gcd equals to 1static int gcd_pairs(int a[], int n){    // To store maximum number    int maxi = 0;     // To store frequency of each number    int []fre = new int[N];     // Find frequency and maximum number    for (int i = 0; i < n; i++)    {        fre[a[i]]++;        maxi = Math.max(a[i], maxi);    }     least_prime_factor();    Mobius();     // To store number of pairs with gcd equals to 1    int ans = 0;     // Traverse through the all possible elements    for (int i = 1; i <= maxi; i++)    {        if (mobius[i] == 0)            continue;         int temp = 0;        for (int j = i; j <= maxi; j += i)            temp += fre[j];         ans += temp * (temp - 1) / 2 * mobius[i];    }     // Return the number of pairs    return ans;} // Driver codepublic static void main (String[] args){    int a[] = { 1, 2, 3, 4, 5, 6 };     int n = a.length;     // Function call    System.out.print(gcd_pairs(a, n));}} // This code is contributed by PrinciRaj1992

## Python3

 # Python3 program to find the number of pairs# such that gcd equals to 1N = 100050 lpf = [0 for i in range(N)]mobius = [0 for i in range(N)] # Function to calculate least# prime factor of each numberdef least_prime_factor():     for i in range(2, N):         # If it is a prime number        if (lpf[i] == 0):             for j in range(i, N, i):                 # For all multiples which are not                # visited yet.                if (lpf[j] == 0):                    lpf[j] = i # Function to find the value of Mobius function# for all the numbers from 1 to ndef Mobius():     for i in range(1, N):         # If number is one        if (i == 1):            mobius[i] = 1        else:             # If number has a squared prime factor            if (lpf[ (i // lpf[i]) ] == lpf[i]):                mobius[i] = 0             # Multiply -1 with the previous number            else:                mobius[i] = -1 * mobius[i // lpf[i]] # Function to find the number of pairs# such that gcd equals to 1def gcd_pairs(a, n):     # To store maximum number    maxi = 0     # To store frequency of each number    fre = [0 for i in range(N)]     # Find frequency and maximum number    for i in range(n):        fre[a[i]] += 1        maxi = max(a[i], maxi)     least_prime_factor()    Mobius()     # To store number of pairs with gcd equals to 1    ans = 0     # Traverse through the all possible elements    for i in range(1, maxi + 1):        if (mobius[i] == 0):            continue         temp = 0        for j in range(i, maxi + 1, i):            temp += fre[j]         ans += temp * (temp - 1) // 2 * mobius[i]     # Return the number of pairs    return ans # Driver codea = [1, 2, 3, 4, 5, 6] n = len(a) # Function callprint(gcd_pairs(a, n)) # This code is contributed by Mohit Kumar

## C#

 // C# program to find the number of pairs// such that gcd equals to 1using System; class GFG{static int N = 100050; static int []lpf = new int[N];static int []mobius = new int[N]; // Function to calculate least// prime factor of each numberstatic void least_prime_factor(){    for (int i = 2; i < N; i++)         // If it is a prime number        if (lpf[i] == 0)             for (int j = i; j < N; j += i)                 // For all multiples which are not                // visited yet.                if (lpf[j] == 0)                    lpf[j] = i;} // Function to find the value of Mobius function// for all the numbers from 1 to nstatic void Mobius(){    for (int i = 1; i < N; i++)    {         // If number is one        if (i == 1)            mobius[i] = 1;        else        {             // If number has a squared prime factor            if (lpf[i / lpf[i]] == lpf[i])                mobius[i] = 0;             // Multiply -1 with the previous number            else                mobius[i] = -1 * mobius[i / lpf[i]];        }    }} // Function to find the number of pairs// such that gcd equals to 1static int gcd_pairs(int []a, int n){    // To store maximum number    int maxi = 0;     // To store frequency of each number    int []fre = new int[N];     // Find frequency and maximum number    for (int i = 0; i < n; i++)    {        fre[a[i]]++;        maxi = Math.Max(a[i], maxi);    }     least_prime_factor();    Mobius();     // To store number of pairs with gcd equals to 1    int ans = 0;     // Traverse through the all possible elements    for (int i = 1; i <= maxi; i++)    {        if (mobius[i] == 0)            continue;         int temp = 0;        for (int j = i; j <= maxi; j += i)            temp += fre[j];         ans += temp * (temp - 1) / 2 * mobius[i];    }     // Return the number of pairs    return ans;} // Driver codepublic static void Main (String[] args){    int []a = { 1, 2, 3, 4, 5, 6 };     int n = a.Length;     // Function call    Console.Write(gcd_pairs(a, n));}}     // This code is contributed by Rajput-Ji

## Javascript



Output:

11

Time Complexity: O(N2)

Auxiliary Space: O(N)

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