Number of co-prime pairs in an array
Last Updated :
14 Jul, 2025
Given an array arr[], count the number of pairs (i, j) such that i < j and gcd(arr[i], arr[j]) == 1.
Examples:
Input : arr[] = [1, 2, 3]
Output : 3
Explanation : Here, Co-prime pairs are (1, 2), (2, 3), (1, 3) .
Input : arr[] = [4, 8, 3, 9]
Output : 4
Explanation : Here, Co-prime pairs are (4, 3), (8, 3), (4, 9), (8, 9).
[Naive Approach] Using Two Loop
This approach checks all possible pairs in the array using two nested loops and counts those whose GCD is 1, indicating they are co-prime.
C++
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// Function to check if two numbers are co-prime
int gcd(int a, int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
// Function to count number of co-prime
// pairs in the array
int cntCoprime(vector<int>& arr) {
int n = arr.size(), count = 0;
// Check all unique pairs (i, j) such that i < j
for (int i = 0; i < n - 1; ++i) {
for (int j = i + 1; j < n; ++j) {
if (gcd(arr[i], arr[j]) == 1) {
count++;
}
}
}
return count;
}
int main() {
vector<int> arr = {4, 8, 3, 9};
cout << cntCoprime(arr) << endl;
return 0;
}
import java.util.*;
class GfG {
// Function to compute GCD
static int gcd(int a, int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
// Function to count co-prime pairs
public static int cntCoprime(int[] arr) {
int n = arr.length, count = 0;
for (int i = 0; i < n - 1; ++i) {
for (int j = i + 1; j < n; ++j) {
if (gcd(arr[i], arr[j]) == 1) {
count++;
}
}
}
return count;
}
public static void main(String[] args) {
int[] arr = {4, 8, 3, 9};
System.out.println(cntCoprime(arr));
}
}
import math
# Function to compute GCD
def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
# Function to count co-prime pairs
def cntCoprime(arr):
n = len(arr)
count = 0
for i in range(n - 1):
for j in range(i + 1, n):
if gcd(arr[i], arr[j]) == 1:
count += 1
return count
if __name__ == "__main__":
arr = [4, 8, 3, 9]
print(cntCoprime(arr))
using System;
class GfG {
// Function to compute GCD
static int gcd(int a, int b) {
while (b != 0) {
int temp = b;
b = a % b;
a = temp;
}
return a;
}
// Function to count co-prime pairs
public static int cntCoprime(int[] arr) {
int count = 0;
int n = arr.Length;
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
if (gcd(arr[i], arr[j]) == 1)
count++;
}
}
return count;
}
public static void Main() {
int[] arr = { 4, 8, 3, 9 };
Console.WriteLine(cntCoprime(arr));
}
}
// Function to compute GCD
function gcd(a, b) {
while (b !== 0) {
[a, b] = [b, a % b];
}
return a;
}
// Function to count co-prime pairs
function cntCoprime(arr) {
let count = 0;
const n = arr.length;
for (let i = 0; i < n - 1; i++) {
for (let j = i + 1; j < n; j++) {
if (gcd(arr[i], arr[j]) === 1) {
count++;
}
}
}
return count;
}
// Driver code
const arr = [4, 8, 3, 9];
console.log(cntCoprime(arr));
Time Complexity: O(n2 × log(m)),where m = max(arr[i])
Space Complexity: O(log(m)), where m = max(arr[i])
[Expected Approach] Using Mobius Function
Instead of directly counting coprime pairs, which can be inefficient, we count how many pairs share a common divisor k and then apply the Inclusion-Exclusion Principle using the Möbius function to isolate only those with gcd = 1.
For each k, we compute d(k), the number of elements divisible by k, and use d(k)*(d(k)-1) / 2 to count such pairs. However, this includes overlaps, so the Möbius function {\scriptstyle \mu(k)} is used to adjust these counts: if it is 1 if k is a product of an even number of distinct primes, –1 if odd, and 0 if k has any squared prime factor.
Finally, we sum the adjusted values as \scriptsize\sum_{k=1}^{\max(arr[i])} \mu(k) \cdot \binom{d(k)}{2}to get the total number of coprime pairs
Step by Step Implementations:
- Count frequency of each number in the array freq[]
- Compute d(k) for every k from 1 to max(arr[i])
- d(k) is the number of elements divisible by k
- Use a sieve-style loop: for each k, add up freq[j] for all j divisible by k.
- Precompute the Möbius function μ(k) up to max(arr[i])
- Use a modified Sieve of Eratosthenes:
• Set μ(1) = 1
• For each prime p, update multiples:
– Flip sign of μ(k) for each multiple of p
– Set μ(k) = 0 for multiples of p^2
- Apply the Möbius formula using Inclusion-Exclusion: {\scriptsize
\text{Number of coprime pairs} = \sum_{k=1}^{\max(arr[i])} \mu(k) \cdot \binom{d(k)}{2}
}
- Only square-free k (i.e., with non-zero μ) contribute.
- Sum up the contributions from all valid k to get the final answer
C++
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// Computes the Möbius function up to 'n'
void computeMobius(int n, vector<int>& mu) {
vector<int> is_prime(n + 1, 1);
mu[0] = 0;
mu[1] = 1;
for (int i = 2; i <= n; ++i) {
if (is_prime[i]) {
for (int j = i; j <= n; j += i) {
mu[j] *= -1;
is_prime[j] = 0;
}
// Not square-free
for (int j = i * i; j <= n; j += i * i) {
mu[j] = 0;
}
}
}
}
// Builds frequency array for values in 'arr'
void buildFre(vector<int>& arr, vector<int>& freq) {
for (int x : arr)
freq[x]++;
}
// For each k, computes how many numbers in arr
// are divisible by k
void computeDivCnt(int maxVal, vector<int>& freq, vector<int>& d) {
for (int k = 1; k <= maxVal; ++k)
for (int j = k; j <= maxVal; j += k)
d[k] += freq[j];
}
// logic to count coprime pairs using
// Möbius and Inclusion-Exclusion
int cntCoprime(vector<int>& arr) {
int maxVal = *max_element(arr.begin(), arr.end());
// d[i] -> number of elements divisible by 'i'
// mu[i] -> mobius sign for 'i'
// fre[i] -> frequency of element 'i'
vector<int> freq(maxVal + 1, 0), d(maxVal + 1, 0), mu(maxVal + 1, 1);
buildFre(arr, freq);
computeDivCnt(maxVal, freq, d);
computeMobius(maxVal, mu);
int result = 0;
for (int k = 1; k <= maxVal; ++k) {
if (mu[k] == 0 || d[k] < 2) continue;
// number of pairs that are divisible by k
int pairs = d[k] * (d[k] - 1) / 2;
result += mu[k] * pairs;
}
return result;
}
int main() {
vector<int> arr = {4, 8, 3, 9};
cout << cntCoprime(arr) << endl;
return 0;
}
import java.util.*;
class GfG {
// Computes the Möbius function up to 'n'
static void computeMobius(int n, int[] mu) {
int[] is_prime = new int[n + 1];
Arrays.fill(is_prime, 1);
mu[0] = 0;
mu[1] = 1;
for (int i = 2; i <= n; ++i) {
if (is_prime[i] == 1) {
for (int j = i; j <= n; j += i) {
mu[j] *= -1;
is_prime[j] = 0;
}
// Not square-free
for (int j = i * i; j <= n; j += i * i) {
mu[j] = 0;
}
}
}
}
// Builds frequency array for values in 'arr'
static void buildFre(int[] arr, int[] freq) {
for (int x : arr)
freq[x]++;
}
// For each k, computes how many numbers in arr
// are divisible by k
static void computeDivCnt(int maxVal, int[] freq, int[] d) {
for (int k = 1; k <= maxVal; ++k)
for (int j = k; j <= maxVal; j += k)
d[k] += freq[j];
}
// logic to count coprime pairs using
// Möbius and Inclusion-Exclusion
static int cntCoprime(int[] arr) {
int maxVal = Arrays.stream(arr).max().getAsInt();
// d[i] -> number of elements divisible by 'i'
// mu[i] -> mobius sign for 'i'
// fre[i] -> frequency of element 'i'
int[] freq = new int[maxVal + 1];
int[] d = new int[maxVal + 1];
int[] mu = new int[maxVal + 1];
Arrays.fill(mu, 1);
buildFre(arr, freq);
computeDivCnt(maxVal, freq, d);
computeMobius(maxVal, mu);
int result = 0;
for (int k = 1; k <= maxVal; ++k) {
if (mu[k] == 0 || d[k] < 2) continue;
// number of pairs that are divisible by k
int pairs = (int)((d[k] * (d[k] - 1)) / 2);
result += mu[k] * pairs;
}
return result;
}
public static void main(String[] args) {
int[] arr = {4, 8, 3, 9};
System.out.println(cntCoprime(arr));
}
}
# Computes the Möbius function up to 'n'
def computeMobius(n, mu):
is_prime = [1] * (n + 1)
mu[0] = 0
mu[1] = 1
for i in range(2, n + 1):
if is_prime[i]:
for j in range(i, n + 1, i):
mu[j] *= -1
is_prime[j] = 0
# Not square-free
for j in range(i * i, n + 1, i * i):
mu[j] = 0
# Builds frequency array for values in 'arr'
def buildFre(arr, freq):
for x in arr:
freq[x] += 1
# For each k, computes how many numbers in arr
# are divisible by k
def computeDivCnt(maxVal, freq, d):
for k in range(1, maxVal + 1):
for j in range(k, maxVal + 1, k):
d[k] += freq[j]
# logic to count coprime pairs using
# Möbius and Inclusion-Exclusion
def cntCoprime(arr):
maxVal = max(arr)
# d[i] -> number of elements divisible by 'i'
# mu[i] -> mobius sign for 'i'
# fre[i] -> frequency of element 'i'
freq = [0] * (maxVal + 1)
d = [0] * (maxVal + 1)
mu = [1] * (maxVal + 1)
buildFre(arr, freq)
computeDivCnt(maxVal, freq, d)
computeMobius(maxVal, mu)
result = 0
for k in range(1, maxVal + 1):
if mu[k] == 0 or d[k] < 2:
continue
# number of pairs that are divisible by k
pairs = d[k] * (d[k] - 1) // 2
result += mu[k] * pairs
return result
if __name__ == "__main__":
arr = [4, 8, 3, 9]
print(cntCoprime(arr))
using System;
using System.Linq;
class GfG
{
// Computes the Möbius function up to 'n'
static void computeMobius(int n, int[] mu)
{
int[] is_prime = new int[n + 1];
Array.Fill(is_prime, 1);
mu[0] = 0;
mu[1] = 1;
for (int i = 2; i <= n; ++i)
{
if (is_prime[i] == 1)
{
for (int j = i; j <= n; j += i)
{
mu[j] *= -1;
is_prime[j] = 0;
}
// Not square-free
for (int j = i * i; j <= n; j += i * i)
{
mu[j] = 0;
}
}
}
}
// Builds frequency array for values in 'arr'
static void buildFre(int[] arr, int[] freq)
{
foreach (int x in arr)
freq[x]++;
}
// For each k, computes how many numbers in arr
// are divisible by k
static void computeDivCnt(int maxVal, int[] freq, int[] d)
{
for (int k = 1; k <= maxVal; ++k)
for (int j = k; j <= maxVal; j += k)
d[k] += freq[j];
}
// logic to count coprime pairs using
// Möbius and Inclusion-Exclusion
static int cntCoprime(int[] arr)
{
int maxVal = arr.Max();
// d[i] -> number of elements divisible by 'i'
// mu[i] -> mobius sign for 'i'
// fre[i] -> frequency of element 'i'
int[] freq = new int[maxVal + 1];
int[] d = new int[maxVal + 1];
int[] mu = Enumerable.Repeat(1, maxVal + 1).ToArray();
buildFre(arr, freq);
computeDivCnt(maxVal, freq, d);
computeMobius(maxVal, mu);
int result = 0;
for (int k = 1; k <= maxVal; ++k)
{
if (mu[k] == 0 || d[k] < 2) continue;
// number of pairs that are divisible by k
int pairs = (int)(d[k] * (d[k] - 1) / 2);
result += mu[k] * pairs;
}
return result;
}
// Driver Code
public static void Main()
{
int[] arr = { 4, 8, 3, 9 };
Console.WriteLine(cntCoprime(arr));
}
}
// Computes the Möbius function up to 'n'
function computeMobius(n, mu) {
let is_prime = new Array(n + 1).fill(1);
mu[0] = 0;
mu[1] = 1;
for (let i = 2; i <= n; ++i) {
if (is_prime[i]) {
for (let j = i; j <= n; j += i) {
mu[j] *= -1;
is_prime[j] = 0;
}
// Not square-free
for (let j = i * i; j <= n; j += i * i) {
mu[j] = 0;
}
}
}
}
// Builds frequency array for values in 'arr'
function buildFre(arr, freq) {
for (let x of arr)
freq[x]++;
}
// For each k, computes how many numbers in arr
// are divisible by k
function computeDivCnt(maxVal, freq, d) {
for (let k = 1; k <= maxVal; ++k)
for (let j = k; j <= maxVal; j += k)
d[k] += freq[j];
}
// logic to count coprime pairs using
// Möbius and Inclusion-Exclusion
function cntCoprime(arr) {
const maxVal = Math.max(...arr);
// d[i] -> number of elements divisible by 'i'
// mu[i] -> mobius sign for 'i'
// fre[i] -> frequency of element 'i'
const freq = new Array(maxVal + 1).fill(0);
const d = new Array(maxVal + 1).fill(0);
const mu = new Array(maxVal + 1).fill(1);
buildFre(arr, freq);
computeDivCnt(maxVal, freq, d);
computeMobius(maxVal, mu);
let result = 0;
for (let k = 1; k <= maxVal; ++k) {
if (mu[k] === 0 || d[k] < 2) continue;
// number of pairs that are divisible by k
let pairs = (d[k] * (d[k] - 1)) / 2;
result += mu[k] * pairs;
}
return result;
}
// Driver Code
const arr = [4, 8, 3, 9];
console.log(cntCoprime(arr));
Time Complexity: O(n + m*log(m)), building divisor counts and computing the Möbius function both take O(m log(m)), where m = max(arr[i])
Space Complexity: O(m), where m = max(arr[i]).
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