Given an array arr[] of N integers, the task is to find the largest divisor for each element in an array other than 1 and the number itself. If there is no such divisor, print -1.
Examples:
Input: arr[] = {5, 6, 7, 8, 9, 10}
Output: -1 3 -1 4 3 5
Divisors(5) = {1, 5}
-> Since there is no divisor other than 1 and the number itself, therefore largest divisor = -1
Divisors(6) = [1, 2, 3, 6]
-> largest divisor other than 1 and the number itself = 3
Divisors(7) = [1, 7]
-> Since there is no divisor other than 1 and the number itself, therefore largest divisor = -1
Divisors(8) = [1, 2, 4, 8]
-> largest divisor other than 1 and the number itself = 4
Divisors(9) = [1, 3, 9]
-> largest divisor other than 1 and the number itself = 3
Divisors(10) = [1, 2, 5, 10]
-> largest divisor other than 1 and the number itself = 5Input: arr[] = {15, 16, 17, 18, 19, 20, 21}
Output: 5 8 -1 9 -1 10 7
Naive approach: The idea is to iterate over all the array elements and find the largest divisor for each of the element using the approach discussed in this article.
Time Complexity: O(N * ?N)
Efficient approach: A better solution is to precompute the maximum divisor of the numbers from 2 to 105 and then just run a loop for array and print precomputed answer.
- Use Sieve of Eratosthenes to mark the prime numbers and store the smallest prime divisor of each number.
- Now largest divisor for any number will be number / smallest_prime_divisor.
- Find the Largest divisor for each number using the precomputed answer.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
#define int long long const int maxin = 100001;
// Divisors array to keep track // of the maximum divisor int divisors[maxin];
// Function to pre-compute the prime // numbers and largest divisors void Calc_Max_Div( int arr[], int n)
{ // Visited array to keep
// track of prime numbers
bool vis[maxin];
memset (vis, 1, maxin);
// 0 and 1 are not prime numbers
vis[0] = vis[1] = 0;
// Initialising divisors[i] = i
for ( int i = 1; i <= maxin; i++)
divisors[i] = i;
// For all the numbers divisible by 2
// the maximum divisor will be number / 2
for ( int i = 4; i <= maxin; i += 2) {
vis[i] = 0;
divisors[i] = i / 2;
}
for ( int i = 3; i <= maxin; i += 2) {
// If divisors[i] is not equal to i then
// this means that divisors[i] contains
// minimum prime divisor for the number
if (divisors[i] != i) {
// Update the answer to
// i / smallest_prime_divisor[i]
divisors[i] = i / divisors[i];
}
// Condition if i is a prime number
if (vis[i] == 1) {
for ( int j = i * i; j < maxin; j += i) {
vis[j] = 0;
// If divisors[j] is equal to j then
// this means that i is the first prime
// divisor for j so we update divi[j] = i
if (divisors[j] == j)
divisors[j] = i;
}
}
}
for ( int i = 0; i < n; i++) {
// If the current element is prime
// then it has no divisors
// other than 1 and itself
if (divisors[arr[i]] == arr[i])
cout << "-1 " ;
else
cout << divisors[arr[i]] << " " ;
}
} // Driver code int32_t main() { int arr[] = { 5, 6, 7, 8, 9, 10 };
int n = sizeof (arr) / sizeof ( int );
Calc_Max_Div(arr, n);
return 0;
} |
// Java implementation of the approach import java.io.*;
public class GFG
{ final static int maxin = 10001 ;
// Divisors array to keep track
// of the maximum divisor
static int divisors[] = new int [maxin + 1 ];
// Function to pre-compute the prime
// numbers and largest divisors
static void Calc_Max_Div( int arr[], int n)
{
// Visited array to keep
// track of prime numbers
int vis[] = new int [maxin + 1 ];
for ( int i = 0 ;i <maxin+ 1 ; i++)
vis[i] = 1 ;
// 0 and 1 are not prime numbers
vis[ 0 ] = vis[ 1 ] = 0 ;
// Initialising divisors[i] = i
for ( int i = 1 ; i <= maxin; i++)
divisors[i] = i;
// For all the numbers divisible by 2
// the maximum divisor will be number / 2
for ( int i = 4 ; i <= maxin; i += 2 )
{
vis[i] = 0 ;
divisors[i] = i / 2 ;
}
for ( int i = 3 ; i <= maxin; i += 2 )
{
// If divisors[i] is not equal to i then
// this means that divisors[i] contains
// minimum prime divisor for the number
if (divisors[i] != i)
{
// Update the answer to
// i / smallest_prime_divisor[i]
divisors[i] = i / divisors[i];
}
// Condition if i is a prime number
if (vis[i] == 1 )
{
for ( int j = i * i; j < maxin; j += i)
{
vis[j] = 0 ;
// If divisors[j] is equal to j then
// this means that i is the first prime
// divisor for j so we update divi[j] = i
if (divisors[j] == j)
divisors[j] = i;
}
}
}
for ( int i = 0 ; i < n; i++)
{
// If the current element is prime
// then it has no divisors
// other than 1 and itself
if (divisors[arr[i]] == arr[i])
System.out.print( "-1 " );
else
System.out.print(divisors[arr[i]] + " " );
}
}
// Driver code
public static void main (String[] args)
{
int []arr = { 5 , 6 , 7 , 8 , 9 , 10 };
int n = arr.length;
Calc_Max_Div(arr, n);
}
} // This code is contributed by AnkitRai01 |
# Python3 implementation of the approach maxin = 100001 ;
# Divisors array to keep track # of the maximum divisor divisors = [ 0 ] * (maxin + 1 );
# Function to pre-compute the prime # numbers and largest divisors def Calc_Max_Div(arr, n) :
# Visited array to keep
# track of prime numbers
vis = [ 1 ] * (maxin + 1 );
# 0 and 1 are not prime numbers
vis[ 0 ] = vis[ 1 ] = 0 ;
# Initialising divisors[i] = i
for i in range ( 1 , maxin + 1 ) :
divisors[i] = i;
# For all the numbers divisible by 2
# the maximum divisor will be number / 2
for i in range ( 4 , maxin + 1 , 2 ) :
vis[i] = 0 ;
divisors[i] = i / / 2 ;
for i in range ( 3 , maxin + 1 , 2 ) :
# If divisors[i] is not equal to i then
# this means that divisors[i] contains
# minimum prime divisor for the number
if (divisors[i] ! = i) :
# Update the answer to
# i / smallest_prime_divisor[i]
divisors[i] = i / / divisors[i];
# Condition if i is a prime number
if (vis[i] = = 1 ) :
for j in range ( i * i, maxin, i) :
vis[j] = 0 ;
# If divisors[j] is equal to j then
# this means that i is the first prime
# divisor for j so we update divi[j] = i
if (divisors[j] = = j) :
divisors[j] = i;
for i in range (n) :
# If the current element is prime
# then it has no divisors
# other than 1 and itself
if (divisors[arr[i]] = = arr[i]) :
print ( "-1 " , end = "");
else :
print (divisors[arr[i]], end = " " );
# Driver code if __name__ = = "__main__" :
arr = [ 5 , 6 , 7 , 8 , 9 , 10 ];
n = len (arr);
Calc_Max_Div(arr, n);
# This code is contributed by AnkitRai01 |
// C# implementation of the approach using System;
class GFG
{ static int maxin = 10001;
// Divisors array to keep track
// of the maximum divisor
static int []divisors = new int [maxin + 1];
// Function to pre-compute the prime
// numbers and largest divisors
static void Calc_Max_Div( int []arr, int n)
{
// Visited array to keep
// track of prime numbers
int []vis = new int [maxin + 1];
for ( int i = 0; i < maxin + 1 ; i++)
vis[i] = 1;
// 0 and 1 are not prime numbers
vis[0] = vis[1] = 0;
// Initialising divisors[i] = i
for ( int i = 1; i <= maxin; i++)
divisors[i] = i;
// For all the numbers divisible by 2
// the maximum divisor will be number / 2
for ( int i = 4; i <= maxin; i += 2)
{
vis[i] = 0;
divisors[i] = i / 2;
}
for ( int i = 3; i <= maxin; i += 2)
{
// If divisors[i] is not equal to i then
// this means that divisors[i] contains
// minimum prime divisor for the number
if (divisors[i] != i)
{
// Update the answer to
// i / smallest_prime_divisor[i]
divisors[i] = i / divisors[i];
}
// Condition if i is a prime number
if (vis[i] == 1)
{
for ( int j = i * i; j < maxin; j += i)
{
vis[j] = 0;
// If divisors[j] is equal to j then
// this means that i is the first prime
// divisor for j so we update divi[j] = i
if (divisors[j] == j)
divisors[j] = i;
}
}
}
for ( int i = 0; i < n; i++)
{
// If the current element is prime
// then it has no divisors
// other than 1 and itself
if (divisors[arr[i]] == arr[i])
Console.Write( "-1 " );
else
Console.Write(divisors[arr[i]] + " " );
}
}
// Driver code
public static void Main()
{
int []arr = { 5, 6, 7, 8, 9, 10 };
int n = arr.Length;
Calc_Max_Div(arr, n);
}
} // This code is contributed by AnkitRai01 |
<script> // Javascript implementation of the approach var maxin = 100001;
// Divisors array to keep track // of the maximum divisor var divisors = Array(maxin).fill(0);
// Function to pre-compute the prime // numbers and largest divisors function Calc_Max_Div(arr, n)
{ // Visited array to keep
// track of prime numbers
var vis = Array(maxin).fill(1);
// 0 and 1 are not prime numbers
vis[0] = vis[1] = 0;
var i,j;
// Initialising divisors[i] = i
for (i = 1; i <= maxin; i++)
divisors[i] = i;
// For all the numbers divisible by 2
// the maximum divisor will be number / 2
for (i = 4; i <= maxin; i += 2)
{
vis[i] = 0;
divisors[i] = i / 2;
}
for (i = 3; i <= maxin; i += 2)
{
// If divisors[i] is not equal to i then
// this means that divisors[i] contains
// minimum prime divisor for the number
if (divisors[i] != i)
{
// Update the answer to
// i / smallest_prime_divisor[i]
divisors[i] = i / divisors[i];
}
// Condition if i is a prime number
if (vis[i] == 1)
{
for (j = i * i; j < maxin; j += i)
{
vis[j] = 0;
// If divisors[j] is equal to j then
// this means that i is the first prime
// divisor for j so we update divi[j] = i
if (divisors[j] == j)
divisors[j] = i;
}
}
}
for (i = 0; i < n; i++)
{
// If the current element is prime
// then it has no divisors
// other than 1 and itself
if (divisors[arr[i]] == arr[i])
document.write( "-1 " );
else
document.write(divisors[arr[i]] + " " );
}
} // Driver code var arr = [ 5, 6, 7, 8, 9, 10 ];
var n = arr.length;
Calc_Max_Div(arr, n); // This code is contributed by bgangwar59 </script> |
-1 3 -1 4 3 5
Time Complexity: O(N)
Auxiliary Space: O(100001)