Check if all the pairs of an array are coprime with each other

Given an array arr[], the task is to check if all the pairs of an array are coprime to each other. All pairs of an array are coprime when GCD(arr[i], arr[j]) = 1 holds for every pair (i, j), such that 1? i < j ? N.

Examples:

Input: arr[] = {1, 3, 8}
Output: Yes
Explanation:
Here, GCD(arr[0], arr[1]) = GCD(arr[0], arr[2]) = GCD(arr[1], arr[2]) = 1
Hence, all the pairs are coprime to each other.

Input: arr[] = {6, 67, 24, 1}
Output: No

Naive Approach: A simple solution is to iterate over every pair (A[i], A[j]) from the given array and check if the gcd(A[i], A[j]) = 1 or not. Therefore, the only positive integer(factor) that divides both of them is 1.

Below is the implementation of the naive approach:

C++

 `// C++ implementation of the``// above approach` `#include ``using` `namespace` `std;` `// Function to check if all the``// pairs of the array are coprime``// with each other or not``bool` `allCoprime(``int` `A[], ``int` `n)``{``    ``bool` `all_coprime = ``true``;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``for` `(``int` `j = i + 1; j < n; j++) {` `            ``// Check if GCD of the``            ``// pair is not equal to 1``            ``if` `(__gcd(A[i], A[j]) != 1) {` `                ``// All pairs are non-coprime``                ``// Return false``                ``all_coprime = ``false``;``                ``break``;``            ``}``        ``}``    ``}``    ``return` `all_coprime;``}` `// Driver Code``int` `main()``{``    ``int` `A[] = { 3, 5, 11, 7, 19 };``    ``int` `arr_size = ``sizeof``(A) / ``sizeof``(A[0]);``    ``if` `(allCoprime(A, arr_size)) {``        ``cout << ``"Yes"``;``    ``}``    ``else` `{``        ``cout << ``"No"``;``    ``}``    ``return` `0;``}`

Java

 `// Java implementation of the ``// above approach ``import` `java.util.*;``import` `java.lang.*;` `class` `GFG{` `// Function to check if all the ``// pairs of the array are coprime ``// with each other or not ``static` `boolean` `allCoprime(``int` `A[], ``int` `n) ``{ ``    ``boolean` `all_coprime = ``true``; ``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{ ``        ``for``(``int` `j = i + ``1``; j < n; j++) ``        ``{ ``            ` `            ``// Check if GCD of the ``            ``// pair is not equal to 1 ``            ``if` `(gcd(A[i], A[j]) != ``1``)``            ``{ ``                ` `                ``// All pairs are non-coprime ``                ``// Return false ``                ``all_coprime = ``false``; ``                ``break``; ``            ``} ``        ``} ``    ``} ``    ``return` `all_coprime; ``} ` `// Function return gcd of two number``static` `int` `gcd(``int` `a, ``int` `b) ``{ ``    ``if` `(b == ``0``) ``        ``return` `a; ``        ` `    ``return` `gcd(b, a % b); ``}` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ``int` `A[] = { ``3``, ``5``, ``11``, ``7``, ``19` `}; ``    ``int` `arr_size = A.length; ``    ` `    ``if` `(allCoprime(A, arr_size))``    ``{ ``        ``System.out.println(``"Yes"``); ``    ``} ``    ``else``    ``{ ``        ``System.out.println(``"No"``); ``    ``} ``}``}` `// This code is contributed by offbeat`

Python3

 `# Python3 implementation of the``# above approach``def` `gcd(a, b): ``    ` `    ``if` `(b ``=``=` `0``): ``        ``return` `a ``    ``else``: ``        ``return` `gcd(b, a ``%` `b) `` ` `# Function to check if all the``# pairs of the array are coprime``# with each other or not``def` `allCoprime(A, n):``     ` `    ``all_coprime ``=` `True``     ` `    ``for` `i ``in` `range``(n):``        ``for` `j ``in` `range``(i ``+` `1``, n):``             ` `            ``# Check if GCD of the``            ``# pair is not equal to 1``            ``if` `gcd(A[i], A[j]) !``=` `1``:``  ` `                ``# All pairs are non-coprime``                ``# Return false``                ``all_coprime ``=` `False``;``                ``break``     ` `    ``return` `all_coprime` `# Driver code   ``if` `__name__``=``=``"__main__"``:``     ` `    ``A ``=` `[ ``3``, ``5``, ``11``, ``7``, ``19` `]``    ``arr_size ``=` `len``(A)``    ` `    ``if` `(allCoprime(A, arr_size)):``        ``print``(``'Yes'``)``    ``else``:``        ``print``(``'No'``)` `# This code is contributed by rutvik_56`

C#

 `// C# implementation of the ``// above approach ``using` `System;``class` `GFG{` `// Function to check if all the ``// pairs of the array are coprime ``// with each other or not ``static` `bool` `allCoprime(``int` `[]A, ``                       ``int` `n) ``{ ``  ``bool` `all_coprime = ``true``; ``  ``for``(``int` `i = 0; i < n; i++)``  ``{ ``    ``for``(``int` `j = i + 1; j < n; j++) ``    ``{ ``      ``// Check if GCD of the ``      ``// pair is not equal to 1 ``      ``if` `(gcd(A[i], A[j]) != 1)``      ``{ ``        ``// All pairs are non-coprime ``        ``// Return false ``        ``all_coprime = ``false``; ``        ``break``; ``      ``} ``    ``} ``  ``} ``  ``return` `all_coprime; ``} ` `// Function return gcd of two number``static` `int` `gcd(``int` `a, ``int` `b) ``{ ``  ``if` `(b == 0) ``    ``return` `a; ` `  ``return` `gcd(b, a % b); ``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``  ``int` `[]A = {3, 5, 11, 7, 19}; ``  ``int` `arr_size = A.Length; ` `  ``if` `(allCoprime(A, arr_size))``  ``{ ``    ``Console.WriteLine(``"Yes"``); ``  ``} ``  ``else``  ``{ ``    ``Console.WriteLine(``"No"``); ``  ``} ``}``}` `// This code is contributed by Rajput-Ji`

Javascript

 ``

Output:
`Yes`

Time Complexity: O(N2 logN)
Auxiliary Space: O(1)

Efficient Approach: The key observation in the problem is that two numbers are said to be co-prime if only positive integer(factor) that divides both of them is 1. So, we can store the factors of each element of the array in the container(set, array, etc.) including this element, and check if this factor is already present or not.

Illustration:

For the array arr[] = {6, 5, 10, 3}
Since the pairs (6, 10), (6, 3) and (5, 10) have common factors, all pairs from the array are not coprime with each other.

Below is the implementation of the above approach:

C++

 `// C++ implementation of the``// above approach` `#include ``using` `namespace` `std;` `// Function to store and``// check the factors``bool` `findFactor(``int` `value,``                ``unordered_set<``int``>& factors)``{``    ``factors.insert(value);``    ``for` `(``int` `i = 2; i * i <= value; i++) {``        ``if` `(value % i == 0) {` `            ``// Check if factors are equal``            ``if` `(value / i == i) {` `                ``// Check if the factor is``                ``// already present``                ``if` `(factors.find(i)``                    ``!= factors.end()) {``                    ``return` `true``;``                ``}``                ``else` `{` `                    ``// Insert the factor in set``                    ``factors.insert(i);``                ``}``            ``}``            ``else` `{` `                ``// Check if the factor is``                ``// already present``                ``if` `(factors.find(i) != factors.end()``                    ``|| factors.find(value / i)``                           ``!= factors.end()) {``                    ``return` `true``;``                ``}``                ``else` `{` `                    ``// Insert the factors in set``                    ``factors.insert(i);``                    ``factors.insert(value / i);``                ``}``            ``}``        ``}``    ``}``    ``return` `false``;``}` `// Function to check if all the``// pairs of array elements``// are coprime with each other``bool` `allCoprime(``int` `A[], ``int` `n)``{``    ``bool` `all_coprime = ``true``;``    ``unordered_set<``int``> factors;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(A[i] == 1)``            ``continue``;` `        ``// Check if factors of A[i]``        ``// haven't occurred previously``        ``if` `(findFactor(A[i], factors)) {``            ``all_coprime = ``false``;``            ``break``;``        ``}``    ``}``    ``return` `all_coprime;``}` `// Driver Code``int` `main()``{``    ``int` `A[] = { 3, 5, 11, 7, 19 };``    ``int` `arr_size = ``sizeof``(A) / ``sizeof``(A[0]);``    ``if` `(allCoprime(A, arr_size)) {``        ``cout << ``"Yes"``;``    ``}``    ``else` `{``        ``cout << ``"No"``;``    ``}``    ``return` `0;``}`

Java

 `// Java implementation of ``// the above approach``import` `java.util.*;``class` `GFG{` `// Function to store and``// check the factors``static` `boolean` `findFactor(``int` `value,``                          ``HashSet factors)``{``  ``factors.add(value);``  ``for` `(``int` `i = ``2``; i * i <= value; i++) ``  ``{``    ``if` `(value % i == ``0``) ``    ``{``      ``// Check if factors are equal``      ``if` `(value / i == i) ``      ``{``        ``// Check if the factor is``        ``// already present``        ``if` `(factors.contains(i)) ``        ``{``          ``return` `true``;``        ``}``        ``else``        ``{``          ``// Insert the factor in set``          ``factors.add(i);``        ``}``      ``}``      ``else``      ``{``        ``// Check if the factor is``        ``// already present``        ``if` `(factors.contains(i) || ``            ``factors.contains(value / i)) ``        ``{``          ``return` `true``;``        ``}``        ``else``        ``{``          ``// Insert the factors in set``          ``factors.add(i);``          ``factors.add(value / i);``        ``}``      ``}``    ``}``  ``}``  ``return` `false``;``}` `// Function to check if all the``// pairs of array elements``// are coprime with each other``static` `boolean` `allCoprime(``int` `A[], ``int` `n)``{``  ``boolean` `all_coprime = ``true``;``  ``HashSet factors = ``          ``new` `HashSet();``  ``for` `(``int` `i = ``0``; i < n; i++) ``  ``{``    ``if` `(A[i] == ``1``)``      ``continue``;` `    ``// Check if factors of A[i]``    ``// haven't occurred previously``    ``if` `(findFactor(A[i], factors)) ``    ``{``      ``all_coprime = ``false``;``      ``break``;``    ``}``  ``}``  ``return` `all_coprime;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``  ``int` `A[] = {``3``, ``5``, ``11``, ``7``, ``19``};``  ``int` `arr_size = A.length;``  ``if` `(allCoprime(A, arr_size)) ``  ``{``    ``System.out.print(``"Yes"``);``  ``}``  ``else``  ``{``    ``System.out.print(``"No"``);``  ``}``}``}` `// This code is contributed by shikhasingrajput`

Python3

 `# Python3 implementation of ``# the above approach` `# Function to store and``# check the factors``def` `findFactor(value, factors):` `    ``factors.add(value)``    ``i ``=` `2``    ``while``(i ``*` `i <``=` `value):``      ` `      ``if` `value ``%` `i ``=``=` `0``:``        ` `            ``# Check if factors are equal``            ``if` `value ``/``/` `i ``=``=` `i:``              ` `              ``# Check if the factor is``              ``# already present``              ``if` `i ``in` `factors:``                  ``return` `bool``(``True``)``              ``else``:``                ` `                  ``# Insert the factor in set``                  ``factors.add(i)``            ``else``:    ``              ` `                ``# Check if the factor is``                ``# already present``                ``if` `(i ``in` `factors) ``or``                   ``((value ``/``/` `i) ``in` `factors):                ``                  ``return` `bool``(``True``)                ``                ``else` `:``                ` `                  ``# Insert the factors in set``                  ``factors.add(i)``                  ``factors.add(value ``/``/` `i)``                  ` `      ``i ``+``=` `1`     `    ``return` `bool``(``False``)` `# Function to check if all the``# pairs of array elements``# are coprime with each other``def` `allCoprime(A, n):` `  ``all_coprime ``=` `bool``(``True``)``  ``factors ``=` `set``()``  ` `  ``for` `i ``in` `range``(n):      ``    ``if` `A[i] ``=``=` `1``:``        ``continue`` ` `    ``# Check if factors of A[i]``    ``# haven't occurred previously``    ``if` `findFactor(A[i], factors):    ``      ``all_coprime ``=` `bool``(``False``)``      ``break``  ` `  ``return` `bool``(all_coprime)`` ` `# Driver code``A ``=` `[``3``, ``5``, ``11``, ``7``, ``19``]``arr_size ``=` `len``(A)` `if` `(allCoprime(A, arr_size)): ``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)` `# This code is contributed by divyeshrabadiya07`

C#

 `// C# implementation of ``// the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to store and``// check the factors``static` `bool` `findFactor(``int` `value,``               ``HashSet<``int``> factors)``{``    ``factors.Add(value);``    ``for``(``int` `i = 2; i * i <= value; i++) ``    ``{``        ``if` `(value % i == 0) ``        ``{``            ` `            ``// Check if factors are equal``            ``if` `(value / i == i) ``            ``{``                ` `                ``// Check if the factor is``                ``// already present``                ``if` `(factors.Contains(i)) ``                ``{``                    ``return` `true``;``                ``}``                ``else``                ``{``                    ` `                    ``// Insert the factor in set``                    ``factors.Add(i);``                ``}``            ``}``            ``else``            ``{``                ` `                ``// Check if the factor is``                ``// already present``                ``if` `(factors.Contains(i) || ``                    ``factors.Contains(value / i)) ``                ``{``                    ``return` `true``;``                ``}``                ``else``                ``{``                    ``// Insert the factors in set``                    ``factors.Add(i);``                    ``factors.Add(value / i);``                ``}``            ``}``        ``}``    ``}``    ``return` `false``;``}` `// Function to check if all the``// pairs of array elements``// are coprime with each other``static` `bool` `allCoprime(``int` `[]A, ``int` `n)``{``    ``bool` `all_coprime = ``true``;``    ``HashSet<``int``> factors = ``new` `HashSet<``int``>();``    ` `    ``for``(``int` `i = 0; i < n; i++) ``    ``{``        ``if` `(A[i] == 1)``            ``continue``;``            ` `        ``// Check if factors of A[i]``        ``// haven't occurred previously``        ``if` `(findFactor(A[i], factors)) ``        ``{``            ``all_coprime = ``false``;``            ``break``;``        ``}``    ``}``    ``return` `all_coprime;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]A = { 3, 5, 11, 7, 19 };``    ``int` `arr_size = A.Length;``    ` `    ``if` `(allCoprime(A, arr_size)) ``    ``{``        ``Console.Write(``"Yes"``);``    ``}``    ``else``    ``{``        ``Console.Write(``"No"``);``    ``}``}``}` `// This code is contributed by Amit Katiyar`

Javascript

 `// JS implementation of the``// above approach` `// Function to store and``// check the factors``function` `findFactor(value, factors)``{``    ``factors.add(value);``    ``for` `(``var` `i = 2; i * i <= value; i++) {``        ``if` `(value % i == 0) {` `            ``// Check if factors are equal``            ``if` `(value == i * i) {` `                ``// Check if the factor is``                ``// already present``                ``if` `(factors.has(i)) {``                    ``return` `true``;``                ``}``                ``else` `{` `                    ``// Insert the factor in set``                    ``factors.add(i);``                ``}``            ``}``            ``else` `{` `                ``// Check if the factor is``                ``// already present``                ``if` `(factors.has(i)``                    ``|| factors.has(Math.floor(value / i))){``                    ``return` `true``;``                ``}``                ``else` `{` `                    ``// Insert the factors in set``                    ``factors.add(i);``                    ``factors.add(Math.floor(value / i));``                ``}``            ``}``        ``}``    ``}``    ``return` `false``;``}` `// Function to check if all the``// pairs of array elements``// are coprime with each other``function` `allCoprime(A, n)``{``    ``let all_coprime = ``true``;``    ``let factors = ``new` `Set();``    ``for` `(``var` `i = 0; i < n; i++) {``        ``if` `(A[i] == 1)``            ``continue``;` `        ``// Check if factors of A[i]``        ``// haven't occurred previously``        ``if` `(findFactor(A[i], factors)) {``            ``all_coprime = ``false``;``            ``break``;``        ``}``    ``}``    ``return` `all_coprime;``}` `// Driver Code``let A = [ 3, 5, 11, 7, 19 ];``let arr_size = A.length``if` `(allCoprime(A, arr_size)) {``    ``console.log(``"Yes"``);``}``else` `{``    ``console.log(``"No"``);``}` `// This code is contributed by phasing17.`

Output:
`Yes`

Time Complexity: O(N3/2)
Auxiliary Space: O(N)

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next