Given an integer N, the task is to find out whether it can be written as a sum of 2 triangular numbers (which may or may not be distinct).
Examples:
Input: N = 24
Output: YES
24 can be represented as 3+21.
Input: N = 15
Output: NO
Approach: Consider all triangular numbers less than N i.e. sqrt(N) numbers. Let’s add them to a set, and for each triangular number X we check if N-X is present in the set. If it is true with any triangular number, then the answer is YES, otherwise the answer is NO.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool checkTriangularSumRepresentation( int n)
{
unordered_set< int > tri;
int i = 1;
while (1) {
int x = i * (i + 1) / 2;
if (x >= n)
break ;
tri.insert(x);
i++;
}
for ( auto tm : tri)
if (tri.find(n - tm ) != tri.end())
return true ;
return false ;
}
int main()
{
int n = 24;
checkTriangularSumRepresentation(n) ? cout << "Yes"
: cout << "No" ;
return 0;
}
|
Java
import java.util.*;
class GFG
{
static boolean checkTriangularSumRepresentation( int n)
{
HashSet<Integer> tri = new HashSet<>();
int i = 1 ;
while ( true )
{
int x = i * (i + 1 ) / 2 ;
if (x >= n)
{
break ;
}
tri.add(x);
i++;
}
for (Integer tm : tri)
{
if (tri.contains(n - tm) && (n - tm) !=
( int ) tri.toArray()[tri.size() - 1 ])
{
return true ;
}
}
return false ;
}
public static void main(String[] args)
{
int n = 24 ;
if (checkTriangularSumRepresentation(n))
{
System.out.println( "Yes" );
}
else
{
System.out.println( "No" );
}
}
}
|
Python 3
def checkTriangularSumRepresentation(n) :
tri = list ();
i = 1 ;
while ( 1 ) :
x = i * (i + 1 ) / / 2 ;
if (x > = n) :
break ;
tri.append(x);
i + = 1 ;
for tm in tri :
if n - tm in tri:
return True ;
return False ;
if __name__ = = "__main__" :
n = 24 ;
if checkTriangularSumRepresentation(n) :
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static bool checkTriangularSumRepresentation( int n)
{
HashSet< int > tri = new HashSet< int >();
int i = 1;
while ( true )
{
int x = i * (i + 1) / 2;
if (x >= n)
{
break ;
}
tri.Add(x);
i++;
}
foreach ( int tm in tri)
{
if (tri.Contains(n - tm))
{
return true ;
}
}
return false ;
}
public static void Main(String[] args)
{
int n = 24;
if (checkTriangularSumRepresentation(n))
{
Console.WriteLine( "Yes" );
}
else
{
Console.WriteLine( "No" );
}
}
}
|
PHP
<?php
function checkTriangularSumRepresentation( $n )
{
$tri = array ();
$i = 1;
while (true)
{
$x = $i * ( $i + 1);
if ( $x >= $n )
break ;
array_push ( $tri , $x );
$i += 1;
}
foreach ( $tri as $tm )
if (in_array( $n - $tm , $tri ))
return true;
return false;
}
$n = 24;
if (checkTriangularSumRepresentation( $n ))
print ( "Yes" );
else
print ( "No" );
?>
|
Javascript
<script>
function checkTriangularSumRepresentation(n)
{
var tri = new Set();
var i = 1;
while (1) {
var x = i * parseInt((i + 1) / 2);
if (x >= n)
break ;
tri.add(x);
i++;
}
var ans = false ;
tri.forEach(tm => {
if (tri.has(n - tm))
ans = true
});
return ans;
}
var n = 24;
checkTriangularSumRepresentation(n) ? document.write( "Yes" )
: document.write( "No" );
</script>
|
Complexity Analysis:
- Time Complexity: O(Sqrt(N))
- Auxiliary Space: O(sqrt(N))
Second approach: Its very clear that N is positive because triangular numbers are numbers that are representable as k*(k+1)/2 where k is some positive integer by this we also get that each term will be less than N. This means that we take and iterate over one of the other terms. If we iterate over the first term in increasing order, then we can use two pointers, which gives us a solution in O(Sqrt(N)) .
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
#define ll long long
int main()
{
ll n;
cin >> n;
ll dig = 2 * n;
bool flag= 0;
for (ll i = 1; i < sqrt (2 * n + 1); i++) {
if (i * i + i > dig)
break ;
ll rest = dig - i * i - i;
ll left = 1;
ll right = dig;
while (left <= right) {
ll mid = (left + right) / 2;
if (mid * mid + mid > rest)
right = mid - 1;
else if (mid * mid + mid == rest) {
flag = 1;
break ;
}
else
left = mid + 1;
}
if (flag)
break ;
}
if (flag)
cout<< "YES" <<endl;
else
cout<< "NO" <<endl;
return 0;
}
|
Java
import java.util.*;
class GFG {
public static void main(String[] args) {
int n = 24 ;
int dig = 2 * n;
boolean flag = false ;
for ( int i = 1 ; i < Math.sqrt( 2 * n + 1 ); i++) {
if (i * i + i > dig)
break ;
int rest = dig - i * i - i;
int left = 1 ;
int right = dig;
while (left <= right) {
int mid = (left + right) / 2 ;
if (mid * mid + mid > rest)
right = mid - 1 ;
else if (mid * mid + mid == rest) {
flag = true ;
break ;
} else
left = mid + 1 ;
}
if (flag)
break ;
}
if (flag)
System.out.print( "YES" );
else
System.out.print( "NO" );
}
}
|
Python3
import math
if __name__ = = '__main__' :
n = 24 ;
dig = 2 * n;
flag = False ;
for i in range ( 1 , int (math.sqrt( 2 * n + 1 ))):
if (i * i + i > dig):
break ;
rest = dig - i * i - i;
left = 1 ;
right = dig;
while (left < = right):
mid = (left + right) / / 2 ;
if (mid * mid + mid > rest):
right = mid - 1 ;
elif (mid * mid + mid = = rest):
flag = True ;
break ;
else :
left = mid + 1 ;
if (flag):
break ;
if (flag):
print ( "YES" );
else :
print ( "NO" );
|
C#
using System;
public class GFG
{
public static void Main(String[] args)
{
int n = 24;
int dig = 2 * n;
bool flag = false ;
for ( int i = 1; i < Math.Sqrt(2 * n + 1); i++)
{
if (i * i + i > dig)
break ;
int rest = dig - i * i - i;
int left = 1;
int right = dig;
while (left <= right) {
int mid = (left + right) / 2;
if (mid * mid + mid > rest)
right = mid - 1;
else if (mid * mid + mid == rest) {
flag = true ;
break ;
} else
left = mid + 1;
}
if (flag)
break ;
}
if (flag)
Console.Write( "YES" );
else
Console.Write( "NO" );
}
}
|
Javascript
<script>
var n = 24;
var dig = 2 * n;
var flag = false ;
for (i = 1; i < Math.sqrt(2 * n + 1); i++) {
if (i * i + i > dig)
break ;
var rest = dig - i * i - i;
var left = 1;
var right = dig;
while (left <= right) {
var mid = parseInt((left + right) / 2);
if (mid * mid + mid > rest)
right = mid - 1;
else if (mid * mid + mid == rest) {
flag = true ;
break ;
} else
left = mid + 1;
}
if (flag)
break ;
}
if (flag)
document.write( "YES" );
else
document.write( "NO" );
</script>
|
Time Complexity: O(Sqrt(N))
Auxiliary space: O(1)
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Last Updated :
20 Sep, 2022
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