Given an array arr[] of positive integers where 2 ? arr[i] ? 106 for all possible values of i. The task is to check whether there exists at least one element in the given array that forms co-prime pair with all other elements of the array. If no such element exists then print No else print Yes.
Examples:
Input: arr[] = {2, 8, 4, 10, 6, 7}
Output: Yes
7 is co-prime with all the other elements of the arrayInput: arr[] = {3, 6, 9, 12}
Output: No
Naive approach: A simple solution is to check whether the gcd of every element with all other elements is equal to 1. Time complexity of this solution is O(n2).
Efficient approach: An efficient solution is to generate all the prime factors of integers in the given array. Using hash, store the count of every element which is a prime factor of any of the number in the array. If the element does not contain any common prime factor with other elements, it always forms a co-prime pair with other elements.
For generating prime factors please go through the article Prime Factorization using Sieve in O(log n)
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MAXN 1000001// Stores smallest prime factor for every numberint spf[MAXN];// Hash to store prime factors countint hash1[MAXN] = { 0 };// Function to calculate SPF (Smallest Prime Factor)// for every number till MAXNvoid sieve(){ spf[1] = 1; for (int i = 2; i < MAXN; i++) // Marking smallest prime factor for every // number to be itself spf[i] = i; // Separately marking spf for every even // number as 2 for (int i = 4; i < MAXN; i += 2) spf[i] = 2; // Checking if i is prime for (int i = 3; i * i < MAXN; i++) { // Marking SPF for all numbers divisible by i if (spf[i] == i) { for (int j = i * i; j < MAXN; j += i) // Marking spf[j] if it is not // previously marked if (spf[j] == j) spf[j] = i; } }}// Function to store the prime factors after dividing// by the smallest prime factor at every stepvoid getFactorization(int x){ int temp; while (x != 1) { temp = spf[x]; if (x % temp == 0) { // Storing the count of // prime factors in hash hash1[spf[x]]++; x = x / spf[x]; } while (x % temp == 0) x = x / temp; }}// Function that returns true if there are// no common prime factors between x// and other numbers of the arraybool check(int x){ int temp; while (x != 1) { temp = spf[x]; // Checking whether it common // prime factor with other numbers if (x % temp == 0 && hash1[temp] > 1) return false; while (x % temp == 0) x = x / temp; } return true;}// Function that returns true if there is// an element in the array which is coprime// with all the other elements of the arraybool hasValidNum(int arr[], int n){ // Using sieve for generating prime factors sieve(); for (int i = 0; i < n; i++) getFactorization(arr[i]); // Checking the common prime factors // with other numbers for (int i = 0; i < n; i++) if (check(arr[i])) return true; return false;}// Driver codeint main(){ int arr[] = { 2, 8, 4, 10, 6, 7 }; int n = sizeof(arr) / sizeof(arr[0]); if (hasValidNum(arr, n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the approachclass GFG{ static int MAXN = 1000001;// Stores smallest prime factor for every numberstatic int[] spf = new int[MAXN];// Hash to store prime factors countstatic int[] hash1 = new int[MAXN];// Function to calculate SPF (Smallest Prime Factor)// for every number till MAXNstatic void sieve(){ spf[1] = 1; for (int i = 2; i < MAXN; i++) // Marking smallest prime factor for every // number to be itself spf[i] = i; // Separately marking spf for every even // number as 2 for (int i = 4; i < MAXN; i += 2) spf[i] = 2; // Checking if i is prime for (int i = 3; i * i < MAXN; i++) { // Marking SPF for all numbers divisible by i if (spf[i] == i) { for (int j = i * i; j < MAXN; j += i) // Marking spf[j] if it is not // previously marked if (spf[j] == j) spf[j] = i; } }}// Function to store the prime factors after dividing// by the smallest prime factor at every stepstatic void getFactorization(int x){ int temp; while (x != 1) { temp = spf[x]; if (x % temp == 0) { // Storing the count of // prime factors in hash hash1[spf[x]]++; x = x / spf[x]; } while (x % temp == 0) x = x / temp; }}// Function that returns true if there are// no common prime factors between x// and other numbers of the arraystatic boolean check(int x){ int temp; while (x != 1) { temp = spf[x]; // Checking whether it common // prime factor with other numbers if (x % temp == 0 && hash1[temp] > 1) return false; while (x % temp == 0) x = x / temp; } return true;}// Function that returns true if there is// an element in the array which is coprime// with all the other elements of the arraystatic boolean hasValidNum(int []arr, int n){ // Using sieve for generating prime factors sieve(); for (int i = 0; i < n; i++) getFactorization(arr[i]); // Checking the common prime factors // with other numbers for (int i = 0; i < n; i++) if (check(arr[i])) return true; return false;}// Driver codepublic static void main (String[] args) { int []arr = { 2, 8, 4, 10, 6, 7 }; int n = arr.length; if (hasValidNum(arr, n)) System.out.println("Yes"); else System.out.println("No");}}// This code is contributed by chandan_jnu |
Python3
# Python3 implementation of the approachMAXN = 1000001# Stores smallest prime factor for # every numberspf = [i for i in range(MAXN)]# Hash to store prime factors counthash1 = [0 for i in range(MAXN)]# Function to calculate SPF (Smallest # Prime Factor) for every number till MAXNdef sieve(): # Separately marking spf for # every even number as 2 for i in range(4, MAXN, 2): spf[i] = 2 # Checking if i is prime for i in range(3, MAXN): if i * i >= MAXN: break # Marking SPF for all numbers # divisible by i if (spf[i] == i): for j in range(i * i, MAXN, i): # Marking spf[j] if it is not # previously marked if (spf[j] == j): spf[j] = i# Function to store the prime factors # after dividing by the smallest prime # factor at every stepdef getFactorization(x): while (x != 1): temp = spf[x] if (x % temp == 0): # Storing the count of # prime factors in hash hash1[spf[x]] += 1 x = x // spf[x] while (x % temp == 0): x = x // temp# Function that returns true if there # are no common prime factors between x# and other numbers of the arraydef check(x): while (x != 1): temp = spf[x] # Checking whether it common # prime factor with other numbers if (x % temp == 0 and hash1[temp] > 1): return False while (x % temp == 0): x = x //temp return True# Function that returns true if there is# an element in the array which is coprime# with all the other elements of the arraydef hasValidNum(arr, n): # Using sieve for generating # prime factors sieve() for i in range(n): getFactorization(arr[i]) # Checking the common prime factors # with other numbers for i in range(n): if (check(arr[i])): return True return False# Driver codearr = [2, 8, 4, 10, 6, 7]n = len(arr)if (hasValidNum(arr, n)): print("Yes")else: print("No")# This code is contributed by mohit kumar |
C#
// C# implementation of the approachusing System;class GFG{ static int MAXN=1000001;// Stores smallest prime factor for every numberstatic int[] spf = new int[MAXN];// Hash to store prime factors countstatic int[] hash1 = new int[MAXN];// Function to calculate SPF (Smallest Prime Factor)// for every number till MAXNstatic void sieve(){ spf[1] = 1; for (int i = 2; i < MAXN; i++) // Marking smallest prime factor for every // number to be itself spf[i] = i; // Separately marking spf for every even // number as 2 for (int i = 4; i < MAXN; i += 2) spf[i] = 2; // Checking if i is prime for (int i = 3; i * i < MAXN; i++) { // Marking SPF for all numbers divisible by i if (spf[i] == i) { for (int j = i * i; j < MAXN; j += i) // Marking spf[j] if it is not // previously marked if (spf[j] == j) spf[j] = i; } }}// Function to store the prime factors after dividing// by the smallest prime factor at every stepstatic void getFactorization(int x){ int temp; while (x != 1) { temp = spf[x]; if (x % temp == 0) { // Storing the count of // prime factors in hash hash1[spf[x]]++; x = x / spf[x]; } while (x % temp == 0) x = x / temp; }}// Function that returns true if there are// no common prime factors between x// and other numbers of the arraystatic bool check(int x){ int temp; while (x != 1) { temp = spf[x]; // Checking whether it common // prime factor with other numbers if (x % temp == 0 && hash1[temp] > 1) return false; while (x % temp == 0) x = x / temp; } return true;}// Function that returns true if there is// an element in the array which is coprime// with all the other elements of the arraystatic bool hasValidNum(int []arr, int n){ // Using sieve for generating prime factors sieve(); for (int i = 0; i < n; i++) getFactorization(arr[i]); // Checking the common prime factors // with other numbers for (int i = 0; i < n; i++) if (check(arr[i])) return true; return false;}// Driver codestatic void Main(){ int []arr = { 2, 8, 4, 10, 6, 7 }; int n = arr.Length; if (hasValidNum(arr, n)) Console.WriteLine("Yes"); else Console.WriteLine("No");}}// This code is contributed by chandan_jnu |
PHP
<?php// PHP implementation of the approach$MAXN = 10001;// Stores smallest prime factor for every number$spf = array_fill(0, $MAXN, 0);// Hash to store prime factors count$hash1 = array_fill(0, $MAXN, 0);// Function to calculate SPF (Smallest Prime Factor)// for every number till MAXNfunction sieve(){ global $spf, $MAXN, $hash1; $spf[1] = 1; for ($i = 2; $i < $MAXN; $i++) // Marking smallest prime factor for every // number to be itself $spf[$i] = $i; // Separately marking spf for every even // number as 2 for ($i = 4; $i < $MAXN; $i += 2) $spf[$i] = 2; // Checking if i is prime for ($i = 3; $i * $i < $MAXN; $i++) { // Marking SPF for all numbers divisible by i if ($spf[$i] == $i) { for ($j = $i * $i; $j < $MAXN; $j += $i) // Marking spf[j] if it is not // previously marked if ($spf[$j] == $j) $spf[$j] = $i; } }}// Function to store the prime factors after dividing// by the smallest prime factor at every stepfunction getFactorization($x){ global $spf,$MAXN,$hash1; while ($x != 1) { $temp = $spf[$x]; if ($x % $temp == 0) { // Storing the count of // prime factors in hash $hash1[$spf[$x]]++; $x = (int)($x / $spf[$x]); } while ($x % $temp == 0) $x = (int)($x / $temp); }}// Function that returns true if there are// no common prime factors between x// and other numbers of the arrayfunction check($x){ global $spf,$MAXN,$hash1; while ($x != 1) { $temp = $spf[$x]; // Checking whether it common // prime factor with other numbers if ($x % $temp == 0 && $hash1[$temp] > 1) return false; while ($x % $temp == 0) $x = (int)($x / $temp); } return true;}// Function that returns true if there is// an element in the array which is coprime// with all the other elements of the arrayfunction hasValidNum($arr, $n){ global $spf,$MAXN,$hash1; // Using sieve for generating prime factors sieve(); for ($i = 0; $i < $n; $i++) getFactorization($arr[$i]); // Checking the common prime factors // with other numbers for ($i = 0; $i < $n; $i++) if (check($arr[$i])) return true; return false;}// Driver code $arr = array( 2, 8, 4, 10, 6, 7 ); $n = count($arr); if (hasValidNum($arr, $n)) echo "Yes"; else echo "No";// This code is contributed by chandan_jnu?> |
Javascript
<script>// Javascript implementation of the approachlet MAXN = 1000001;// Stores smallest prime factor for every numberlet spf = new Array(MAXN);// Hash to store prime factors countlet hash1 = new Array(MAXN);// Function to calculate SPF (Smallest Prime Factor)// for every number till MAXNfunction sieve(){ spf[1] = 1; for(let i = 2; i < MAXN; i++) // Marking smallest prime factor for // every number to be itself spf[i] = i; // Separately marking spf for every even // number as 2 for(let i = 4; i < MAXN; i += 2) spf[i] = 2; // Checking if i is prime for(let i = 3; i * i < MAXN; i++) { // Marking SPF for all numbers divisible by i if (spf[i] == i) { for(let j = i * i; j < MAXN; j += i) // Marking spf[j] if it is not // previously marked if (spf[j] == j) spf[j] = i; } }}// Function to store the prime factors // after dividing by the smallest prime// factor at every stepfunction getFactorization(x){ let temp; while (x != 1) { temp = spf[x]; if (x % temp == 0) { // Storing the count of // prime factors in hash hash1[spf[x]]++; x = x / spf[x]; } while (x % temp == 0) x = x / temp; }}// Function that returns true if there are// no common prime factors between x// and other numbers of the arrayfunction check(x){ let temp; while (x != 1) { temp = spf[x]; // Checking whether it common // prime factor with other numbers if (x % temp == 0 && hash1[temp] > 1) return false; while (x % temp == 0) x = x / temp; } return true;}// Function that returns true if there is// an element in the array which is coprime// with all the other elements of the arrayfunction hasValidNum(arr, n){ // Using sieve for generating prime factors sieve(); for(let i = 0; i < n; i++) getFactorization(arr[i]); // Checking the common prime factors // with other numbers for(let i = 0; i < n; i++) if (check(arr[i])) return true; return false;}// Driver codelet arr = [ 2, 8, 4, 10, 6, 7 ];let n = arr.length;if (hasValidNum(arr, n)) document.write("Yes");else document.write("No");// This code is contributed by unknown2108</script> |
Output:
Yes
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