You are given the number N. Your task is to split this number into 3 positive integers x, y, and z, such that their sum is equal to ‘N’ and none of the 3 integers is a multiple of 3. Given that N >= 2.
Examples:
Input : N = 10
Output : x = 1, y = 2, z = 7
Note that x + y + z = N and x, y & z are not divisible by N.
Input : 18
Output :x = 1, y = 1, z = 16
Naive Approach: The idea is to iterate three nested loops from 1 to N-1 and choose three elements such that their sum is equal to N and they are not divisible by 3. Below is the implementation of the approach:
// CPP program to split a number into three // parts such than none of them is divisible by 3 #include <iostream> using namespace std;
void printThreeParts( int N)
{ // Traversing to choose first part
for ( int i = 1; i < N; i++) {
// Traversing to choose second part
for ( int j = 1; j < N; j++) {
// Traversing to choose third part
for ( int k = 1; k < N; k++) {
// if all three part's sum is N and
// they are not divisible by 3
// then print those
if ((i + j + k == N) && (i % 3 != 0)
&& (j % 3 != 0) && (k % 3 != 0)) {
cout << "x = " << i << ", "
<< "y = " << j << ", "
<< "z = " << k << endl;
return ;
}
}
}
}
} // Driver Code int main()
{ int N = 10;
printThreeParts(N);
return 0;
} |
// Java program to split a number into three // parts such than none of them is divisible by 3 import java.util.*;
class GFG {
public static void printThreeParts( int N) {
// Traversing to choose first part
for ( int i = 1 ; i < N; i++) {
// Traversing to choose second part
for ( int j = 1 ; j < N; j++) {
// Traversing to choose third part
for ( int k = 1 ; k < N; k++) {
// if all three parts' sum is N and
// they are not divisible by 3
// then print those
if ((i + j + k == N) && (i % 3 != 0 )
&& (j % 3 != 0 ) && (k % 3 != 0 )) {
System.out.println( "x = " + i + ", "
+ "y = " + j + ", "
+ "z = " + k);
return ;
}
}
}
}
}
// Driver Code
public static void main(String[] args) {
int N = 10 ;
printThreeParts(N);
}
} |
def print_three_parts(N):
# Traversing to choose first part
for i in range ( 1 , N):
# Traversing to choose second part
for j in range ( 1 , N):
# Traversing to choose third part
for k in range ( 1 , N):
# if all three part's sum is N and they are not divisible by 3
# then print those
if (i + j + k = = N) and (i % 3 ! = 0 ) and (j % 3 ! = 0 ) and (k % 3 ! = 0 ):
print (f "x = {i}, y = {j}, z = {k}" )
return
# Driver Code if __name__ = = "__main__" :
N = 10
print_three_parts(N)
|
using System;
namespace NumberSplit
{ class Program
{
static void PrintThreeParts( int N)
{
// Traversing to choose first part
for ( int i = 1; i < N; i++)
{
// Traversing to choose second part
for ( int j = 1; j < N; j++)
{
// Traversing to choose third part
for ( int k = 1; k < N; k++)
{
// if all three part's sum is N and
// they are not divisible by 3
// then print those
if ((i + j + k == N) && (i % 3 != 0)
&& (j % 3 != 0) && (k % 3 != 0))
{
Console.WriteLine($ "x = {i}, y = {j}, z = {k}" );
return ;
}
}
}
}
}
static void Main( string [] args)
{
int N = 10;
PrintThreeParts(N);
// Ensure the console window remains open until a key is pressed.
Console.ReadKey();
}
}
} |
// Function to split a number into three parts // such that none of them is divisible by 3 function printThreeParts(N)
{ // Traversing to choose first part
for (let i = 1; i < N; i++)
{
// Traversing to choose second part
for (let j = 1; j < N; j++)
{
// Traversing to choose third part
for (let k = 1; k < N; k++)
{
// if all three part's sum is N and
// they are not divisible by 3 then print those
if ((i + j + k == N) && (i % 3 != 0) &&
(j % 3 != 0) && (k % 3 != 0)) {
console.log( "x = " + i + ", y = " + j + ", z = " + k);
return ;
}
}
}
}
} // Driver Code let N = 10; printThreeParts(N); |
x = 1, y = 1, z = 8
Time Complexity: O(N3)
Auxiliary Space: O(1)
Efficient Approach: To split N into 3 numbers we split N as
- If N is divisible by 3, then the numbers x, y, and z can be 1, 1, and N-2, respectively. All x, y, and z are not divisible by 3. And (1)+(1)+(N-2)=N .
- If N is not divisible by 3 then N-3 will also not be divisible by 3. Therefore, we can have x=1, y=2, and z=N-3.Also, (1)+(2)+(N-3)=N.
Below is the implementation of the approach:
// CPP program to split a number into three parts such // than none of them is divisible by 3. #include <iostream> using namespace std;
void printThreeParts( int N)
{ // Print x = 1, y = 1 and z = N - 2
if (N % 3 == 0)
cout << " x = 1, y = 1, z = " << N - 2 << endl;
// Otherwise, print x = 1, y = 2 and z = N - 3
else
cout << " x = 1, y = 2, z = " << N - 3 << endl;
} // Driver code int main()
{ int N = 10;
printThreeParts(N);
return 0;
} |
// Java program to split a number into three parts such // than none of them is divisible by 3. import java.util.*;
class solution
{ static void printThreeParts( int N)
{ // Print x = 1, y = 1 and z = N - 2
if (N % 3 == 0 )
System.out.println( "x = 1, y = 1, z = " + (N- 2 ));
// Otherwise, print x = 1, y = 2 and z = N - 3
else
System.out.println( " x = 1, y = 2, z = " + (N- 3 ));
} // Driver code public static void main(String args[])
{ int N = 10 ;
printThreeParts(N);
} } |
# Python3 program to split a number into three parts such # than none of them is divisible by 3. def printThreeParts(N) :
# Print x = 1, y = 1 and z = N - 2
if (N % 3 = = 0 ) :
print ( " x = 1, y = 1, z = " ,N - 2 )
# Otherwise, print x = 1, y = 2 and z = N - 3
else :
print ( " x = 1, y = 2, z = " ,N - 3 )
# Driver code if __name__ = = "__main__" :
N = 10
printThreeParts(N)
# This code is contributed by Ryuga |
// C# program to split a number into three parts such // than none of them is divisible by 3. using System;
public class GFG{
static void printThreeParts( int N)
{ // Print x = 1, y = 1 and z = N - 2
if (N % 3 == 0)
Console.WriteLine( " x = 1, y = 1, z = " +(N - 2));
// Otherwise, print x = 1, y = 2 and z = N - 3
else
Console.WriteLine( " x = 1, y = 2, z = " +(N - 3));
} // Driver code static public void Main (){
int N = 10;
printThreeParts(N);
}
// This code is contributed by ajit. } |
<script> // javascript program to split a number into three parts such // than none of them is divisible by 3. function printThreeParts(N)
{ // Print x = 1, y = 1 and z = N - 2
if (N % 3 == 0)
document.write( "x = 1, y = 1, z = " + (N-2));
// Otherwise, print x = 1, y = 2 and z = N - 3
else
document.write( " x = 1, y = 2, z = " + (N-3));
} // Driver code var N = 10;
printThreeParts(N); // This code contributed by Princi Singh </script> |
<?php // PHP program to split a number into // three parts such than none of them // is divisible by 3. function printThreeParts( $N )
{ // Print x = 1, y = 1 and z = N - 2
if ( $N % 3 == 0)
echo " x = 1, y = 1, z = " .
( $N - 2) . "\n" ;
// Otherwise, print x = 1,
// y = 2 and z = N - 3
else
echo " x = 1, y = 2, z = " .
( $N - 3) . "\n" ;
} // Driver code $N = 10;
printThreeParts( $N );
// This code is contributed by ita_c ?> |
x = 1, y = 2, z = 7
Time Complexity: O(1)
Auxiliary Space: O(1)