Given a positive integer N, the task is to find a triple of three distinct positive integers (X, Y, Z) such that X + Y + Z = N and X = GCD (Y, Z).
Example:
Input: N = 12
Output:1 2 9
Explanation: The triplet (1, 2, 9) is set of distinct integers such that 1 + 2 + 9 = 12 and 1 = GCD(2, 9).
Input: N = 5675
Output:1 3 5671
Naive Approach: The basic idea is to iterate to find all possible triplets of (X, Y, Z) with sum N and for each triplet, check if GCD(Y, Z) = X.
Code-
C++
#include <bits/stdc++.h>
using namespace std;
void printTriplet(int N)
{
for(int i=1;i<N;i++){
for(int j=i+1;j<N;j++){
for(int k=j+1;k<N;k++){
if(i+j+k==N && __gcd(j,k)==i){
cout<<i<<" "<<j<<" "<<k<<endl;
return;
}
}
}
}
}
int main()
{
int N = 5875;
printTriplet(N);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static void printTriplet(int N)
{
for (int i = 1; i < N; i++) {
for (int j = i + 1; j < N; j++) {
for (int k = j + 1; k < N; k++) {
if (i + j + k == N && gcd(j, k) == i) {
System.out.println(i + " " + j + " "
+ k);
return;
}
}
}
}
}
static int gcd(int a, int b)
{
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
public static void main(String[] args)
{
int N = 5875;
printTriplet(N);
}
}
|
Python3
import math
def printTriplet(N):
for i in range(1, N):
for j in range(i+1, N):
for k in range(j+1, N):
if i+j+k == N and math.gcd(j, k) == i:
print(i, j, k)
return
if __name__ == "__main__":
N = 5875
printTriplet(N)
|
C#
using System;
class MainClass {
static void PrintTriplet(int N)
{
for (int i = 1; i < N; i++) {
for (int j = i + 1; j < N; j++) {
for (int k = j + 1; k < N; k++) {
if (i + j + k == N && gcd(j, k) == i) {
Console.WriteLine(i + " " + j + " "
+ k);
return;
}
}
}
}
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
public static void Main(string[] args)
{
int N = 5875;
PrintTriplet(N);
}
}
|
Javascript
function printTriplet(N)
{
for (let i = 1; i < N; i++)
{
for (let j = i + 1; j < N; j++)
{
for (let k = j + 1; k < N; k++)
{
if (i + j + k == N && gcd(j, k) == i)
{
console.log(i, j, k);
return;
}
}
}
}
}
function gcd(a, b) {
if (b == 0) return a;
return gcd(b, a % b);
}
let N = 5875;
printTriplet(N);
|
Time Complexity: O(N3*logN),logN for finding GCD, and N3 for three “for” loops
Auxiliary space: O(1)
Efficient Approach: The above approach can be further optimized using the observation that for any given N, there are the following three cases:
- Case 1: If N is even then, a valid triplet is (1, N/2, N/2 -1).
- Case 2: If N is odd and (N/2) is even then, a valid triplet is (1, N/2 + 1, N/2 -1).
- Case 3: If N is odd and (N/2) is also odd then, a valid triplet is (1, N/2 – 2, N/2 + 2).
Hence, for any given N, identify the case and print its respective triplet.
Below is the implementation of the approach:
C++
#include <bits/stdc++.h>
using namespace std;
int printTriplet(int N)
{
if (N % 2 == 0) {
cout << 1 << " " << (N / 2)
<< " " << (N / 2) - 1;
}
else {
if ((N / 2) % 2 == 0) {
cout << 1 << " "
<< (N / 2) - 1 << " "
<< (N / 2) + 1;
}
else {
cout << 1 << " "
<< (N / 2) - 2 << " "
<< (N / 2) + 2;
}
}
}
int main()
{
int N = 5875;
printTriplet(N);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static void printTriplet(int N)
{
if (N % 2 == 0) {
System.out.print(1 + " " + (N / 2) +
" " + ((N / 2) - 1));
} else {
if ((N / 2) % 2 == 0) {
System.out.print(1 + " " + ((N / 2) - 1) +
" " + ((N / 2) + 1));
}
else {
System.out.print(1 + " " + ((N / 2) - 2) +
" " + ((N / 2) + 2));
}
}
}
public static void main(String[] args) {
int N = 5875;
printTriplet(N);
}
}
|
Python3
def printTriplet(N):
if (N % 2 == 0):
print(f"{1} {(N / 2)} {(N / 2) - 1}")
else:
if ((N // 2) % 2 == 0):
print(f"{1} {(N // 2) - 1} {(N // 2) + 1}")
else:
print(f"{1} {(N // 2) - 2} {(N // 2) + 2}")
if __name__ == "__main__":
N = 5875
printTriplet(N)
|
C#
using System;
class GFG{
static void printTriplet(int N)
{
if (N % 2 == 0)
{
Console.Write(1 + " " + (N / 2) + " " +
((N / 2) - 1));
}
else
{
if ((N / 2) % 2 == 0)
{
Console.Write(1 + " " + ((N / 2) - 1) + " " +
((N / 2) + 1));
}
else
{
Console.Write(1 + " " + ((N / 2) - 2) + " " +
((N / 2) + 2));
}
}
}
public static void Main()
{
int N = 5875;
printTriplet(N);
}
}
|
Javascript
<script>
function printTriplet(N)
{
if (N % 2 == 0) {
document.write(1 + " " + Math.floor(N / 2)
+ " " + (Math.floor(N / 2) - 1));
}
else {
if ((N / 2) % 2 == 0) {
document.write(1 + " "
+ (Math.floor(N / 2) - 1) + " "
+ (Math.floor(N / 2) + 1));
}
else {
document.write(1 + " "
+ (Math.floor(N / 2) - 2) + " "
+ (Math.floor(N / 2) + 2));
}
}
}
let N = 5875;
printTriplet(N);
</script>
|
Time Complexity: O(1)
Auxiliary space: O(1)