Find set of size K such that any value of the set is co-prime with any Array element
Last Updated :
20 Dec, 2022
Given a set S of having numbers 1, 2, 3, . . ., N, and an integer K, the task is to form a set A by taking K values from S such that any pair formed by taking one value from S and another from A, is always coprime. (Two numbers are coprime if their GCD is 1).
Note: If multiple solutions exist, print any of them, and if no such set exists, print -1.
Examples:
Input: N = 4, K = 1
Output: 1
Explanation: We can choose [1] for set A and remaining numbers
[2, 3, 4] contained in set S. 1 is coprime with 2, 3 and 4 so the condition is satisfied.
Input: N = 4, K = 2
Output: 1 3
Approach: The problem can be solved based on the following observation:
Let’s focus on prime numbers since the gcd function works on primes independently of other primes.
- Consider all primes p such that p ∗ 2 ≤ N. Hence, 2 and p must be in same component for all primes ≤ N / 2. Let’s add this to a set called X.
- All non-prime integers shall also lie in the same set as their prime factors. So any integer > 1 which is not a prime shall also be added to this X.
We can claim that the elements present in X shall all be in either S or A.
- For example, for S = 13, the primes less than 6.5 are 2, 3, 5, so the set S formed is [2, 3, 4, 5, 6, 8, 9, 10, 12]. Elements not present in this set are 1 and all primes greater than N/2. Let’s say there are C such elements. For N = 13, we have [1, 7, 11, 13]. C = 4 here.
- So, if we can add elements from [1, 7, 11, 13] to make a set of size K or size N − K, then it is possible to find such S and A.
- It is only when we have either K ≤ C or K ≥ N−C that we can form set S and A.
The list of primes can be computed using the sieve of Eratosthenes.
Follow the steps mentioned below to implement the idea:
- Find the primes within the range provided above.
- Now check the value of C and N-C.
- Compare that with K as per the conditions mentioned.
- If the conditions are satisfied then form the sets, otherwise, no such set is possible.
Below is the implementation of the above approach:
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the set
void findSet(int n, int k)
{
int dp[n + 1];
for (int i = 1; i <= n; i++) {
dp[i] = -1;
}
// Can be in any set.
dp[1] = 0;
int together = 0;
int single = 1;
for (int i = 2; i <= n; i++) {
if (i % 2 == 0 && dp[i] == -1) {
dp[i] = 1;
together++;
}
else if (dp[i] == -1) {
int z = 2 * i;
int flag = -1;
while (z <= n) {
flag = 1;
if (dp[z] == -1) {
together++;
}
dp[z] = 1;
z += i;
}
if (flag == 1) {
dp[i] = 1;
together++;
}
else {
dp[i] = 0;
single++;
}
}
}
if (k <= single) {
for (int i = 1; i <= n; i++) {
if (k != 0 && dp[i] == 0) {
cout << i << " ";
k--;
}
}
}
else if (k >= together) {
k -= together;
for (int i = 1; i <= n; i++) {
if (k != 0 && dp[i] == 0) {
cout << i << " ";
k--;
}
else if (dp[i] == 1) {
cout << i << " ";
}
}
}
else {
cout << -1;
}
}
// Driver Code
int main()
{
int N = 4;
int K = 1;
// Function call
findSet(N, K);
return 0;
}
// This code is contributed by aarohirai2616.
// Java code to implement the approach
import java.io.*;
import java.util.*;
public class GFG {
// Function to find the set
public static void findSet(int n, int k)
{
int dp[] = new int[n + 1];
for (int i = 1; i <= n; i++) {
dp[i] = -1;
}
// Can be in any set.
dp[1] = 0;
int together = 0;
int single = 1;
for (int i = 2; i <= n; i++) {
if (i % 2 == 0 && dp[i] == -1) {
dp[i] = 1;
together++;
}
else if (dp[i] == -1) {
int z = 2 * i;
int flag = -1;
while (z <= n) {
flag = 1;
if (dp[z] == -1) {
together++;
}
dp[z] = 1;
z += i;
}
if (flag == 1) {
dp[i] = 1;
together++;
}
else {
dp[i] = 0;
single++;
}
}
}
if (k <= single) {
for (int i = 1; i <= n; i++) {
if (k != 0 && dp[i] == 0) {
System.out.print(i + " ");
k--;
}
}
}
else if (k >= together) {
k -= together;
for (int i = 1; i <= n; i++) {
if (k != 0 && dp[i] == 0) {
System.out.print(i + " ");
k--;
}
else if (dp[i] == 1) {
System.out.print(i + " ");
}
}
}
else {
System.out.println(-1);
}
}
// Driver code
public static void main(String[] args)
{
int N = 4;
int K = 1;
// Function call
findSet(N, K);
}
}
# Python code for the above approach
# Function to find the set
def findSet(n, k) :
dp = [0] * (n + 1)
for i in range(1, n+1):
dp[i] = -1
# Can be in any set.
dp[1] = 0
together = 0
single = 1
for i in range(2, n+1):
if (i % 2 == 0 and dp[i] == -1) :
dp[i] = 1
together += 1
elif (dp[i] == -1) :
z = 2 * i
flag = -1
while (z <= n) :
flag = 1
if (dp[z] == -1) :
together += 1
dp[z] = 1
z += i
if (flag == 1) :
dp[i] = 1
together += 1
else :
dp[i] = 0
single += 1
if (k <= single) :
for i in range(1, n+1):
if (k != 0 and dp[i] == 0) :
print(i, end = " ")
k -= 1
elif (k >= together) :
k -= together
for i in range(1, n+1):
if (k != 0 and dp[i] == 0) :
print(i, end = " ")
k -= 1
elif (dp[i] == 1) :
print(i, end = " ")
else :
print("-1")
# Driver Code
if __name__ == "__main__":
N = 4
K = 1
# Function call
findSet(N, K)
# This code is contributed by code_hunt.
// C# code to implement the approach
using System;
public class GFG {
// Function to find the set
public static void findSet(int n, int k)
{
int[] dp = new int[n + 1];
for (int i = 1; i <= n; i++) {
dp[i] = -1;
}
// Can be in any set.
dp[1] = 0;
int together = 0;
int single = 1;
for (int i = 2; i <= n; i++) {
if (i % 2 == 0 && dp[i] == -1) {
dp[i] = 1;
together++;
}
else if (dp[i] == -1) {
int z = 2 * i;
int flag = -1;
while (z <= n) {
flag = 1;
if (dp[z] == -1) {
together++;
}
dp[z] = 1;
z += i;
}
if (flag == 1) {
dp[i] = 1;
together++;
}
else {
dp[i] = 0;
single++;
}
}
}
if (k <= single) {
for (int i = 1; i <= n; i++) {
if (k != 0 && dp[i] == 0) {
Console.WriteLine(i + " ");
k--;
}
}
}
else if (k >= together) {
k -= together;
for (int i = 1; i <= n; i++) {
if (k != 0 && dp[i] == 0) {
Console.WriteLine(i + " ");
k--;
}
else if (dp[i] == 1) {
Console.WriteLine(i + " ");
}
}
}
else {
Console.WriteLine(-1);
}
}
// Driver Code
static public void Main()
{
int N = 4;
int K = 1;
// Function call
findSet(N, K);
}
}
// This code is contributed by Rohit Pradhan
<script>
// JavaScript code to implement the approach
// Function to find the set
function findSet(n, k)
{
let dp = new Array(n + 1);
for (let i = 1; i <= n; i++) {
dp[i] = -1;
}
// Can be in any set.
dp[1] = 0;
let together = 0;
let single = 1;
for (let i = 2; i <= n; i++) {
if (i % 2 == 0 && dp[i] == -1) {
dp[i] = 1;
together++;
}
else if (dp[i] == -1) {
let z = 2 * i;
let flag = -1;
while (z <= n) {
flag = 1;
if (dp[z] == -1) {
together++;
}
dp[z] = 1;
z += i;
}
if (flag == 1) {
dp[i] = 1;
together++;
}
else {
dp[i] = 0;
single++;
}
}
}
if (k <= single) {
for (let i = 1; i <= n; i++) {
if (k != 0 && dp[i] == 0) {
document.write(i + " ");
k--;
}
}
}
else if (k >= together) {
k -= together;
for (let i = 1; i <= n; i++) {
if (k != 0 && dp[i] == 0) {
document.write(i + " ");
k--;
}
else if (dp[i] == 1) {
document.write(i + " ");
}
}
}
else {
document.write(-1);
}
}
// Driver Code
let N = 4;
let K = 1;
// Function call
findSet(N, K);
// This code is contributed by sanjoy_62.
</script>
<?php
// Function to find the set
function findSet($n, $k)
{
$dp = array_fill(1, $n + 1, -1);
// Can be in any set.
$dp[1] = 0;
$together = 0;
$single = 1;
for ($i = 2; $i <= $n; $i++) {
if ($i % 2 == 0 && $dp[$i] == -1) {
$dp[$i] = 1;
$together++;
} else if ($dp[$i] == -1) {
$z = 2 * $i;
$flag = -1;
while ($z <= $n) {
$flag = 1;
if ($dp[$z] == -1) {
$together++;
}
$dp[$z] = 1;
$z += $i;
}
if ($flag == 1) {
$dp[$i] = 1;
$together++;
} else {
$dp[$i] = 0;
$single++;
}
}
}
if ($k <= $single) {
for ($i = 1; $i <= $n; $i++) {
if ($k != 0 && $dp[$i] == 0) {
echo $i . " ";
$k--;
}
}
} else if ($k >= $together) {
$k -= $together;
for ($i = 1; $i <= $n; $i++) {
if ($k != 0 && $dp[$i] == 0) {
echo $i . " ";
$k--;
} else if ($dp[$i] == 1) {
echo $i . " ";
}
}
} else {
echo -1;
}
}
// Driver Code
$N = 4;
$K = 1;
// Function call
findSet($N, $K);
?>
Time Complexity: O(N ∗ log(log(N)))
Auxiliary Space: O(N)
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