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# Check if a number can be represented as a sum of 2 triangular numbers

• Last Updated : 03 Jun, 2021

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 implemnetation of the above approach:

## C++

 `// C++ implementation of the above approach``#include ``using` `namespace` `std;` `// Function to check if it is possible or not``bool` `checkTriangularSumRepresentation(``int` `n)``{``    ``unordered_set<``int``> tri;``    ``int` `i = 1;` `    ``// Store all triangular numbers up to N in a Set``    ``while` `(1) {``        ``int` `x = i * (i + 1) / 2;``        ``if` `(x >= n)``            ``break``;``        ``tri.insert(x);``        ``i++;``    ``}`` ` `    ``// Check if a pair exists``    ``for` `(``auto` `tm` `: tri)``        ``if` `(tri.find(n - ``tm``) != tri.end())``            ``return` `true``;``    ``return` `false``;``}` `// Driver Code``int` `main()``{``    ``int` `n = 24;``    ``checkTriangularSumRepresentation(n) ? cout << ``"Yes"``                                        ``: cout << ``"No"``;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `    ``// Function to check if it is possible or not``    ``static` `boolean` `checkTriangularSumRepresentation(``int` `n)``    ``{``        ``HashSet tri = ``new` `HashSet<>();``        ``int` `i = ``1``;` `        ``// Store all triangular numbers up to N in a Set``        ``while` `(``true``)``        ``{``            ``int` `x = i * (i + ``1``) / ``2``;``            ``if` `(x >= n)``            ``{``                ``break``;``            ``}``            ``tri.add(x);``            ``i++;``        ``}` `        ``// Check if a pair exists``        ``for` `(Integer tm : tri)``        ``{``            ``if` `(tri.contains(n - tm) && (n - tm) !=``                ``(``int``) tri.toArray()[tri.size() - ``1``])``            ``{``                ``return` `true``;``            ``}``        ``}``        ``return` `false``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``24``;``        ``if` `(checkTriangularSumRepresentation(n))``        ``{``            ``System.out.println(``"Yes"``);``        ``}``        ``else``        ``{``            ``System.out.println(``"No"``);``        ``}``    ``}``}` `// This code has been contributed by 29AjayKumar`

## Python 3

 `# Python3 implementation of the above approach` `# Function to check if it is possible or not``def` `checkTriangularSumRepresentation(n) :``    ` `    ``tri ``=` `list``();``    ``i ``=` `1``;` `    ``# Store all triangular numbers``    ``# up to N in a Set``    ``while` `(``1``) :``        ``x ``=` `i ``*` `(i ``+` `1``) ``/``/` `2``;``        ``if` `(x >``=` `n) :``            ``break``;``            ` `        ``tri.append(x);``        ``i ``+``=` `1``;` `    ``# Check if a pair exists``    ``for` `tm ``in` `tri :``        ``if` `n ``-` `tm ``in` `tri:``            ``return` `True``;``            ` `    ``return` `False``;` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:``    ``n ``=` `24``;``    ` `    ``if` `checkTriangularSumRepresentation(n) :``        ``print``(``"Yes"``)``    ``else` `:``        ``print``(``"No"``)` `# This code is contributed by Ryuga       `

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `    ``// Function to check if it is possible or not``    ``static` `bool` `checkTriangularSumRepresentation(``int` `n)``    ``{``        ``HashSet<``int``> tri = ``new` `HashSet<``int``>();``        ``int` `i = 1;` `        ``// Store all triangular numbers up to N in a Set``        ``while` `(``true``)``        ``{``            ``int` `x = i * (i + 1) / 2;``            ``if` `(x >= n)``            ``{``                ``break``;``            ``}``            ``tri.Add(x);``            ``i++;``        ``}` `        ``// Check if a pair exists``        ``foreach` `(``int` `tm ``in` `tri)``        ``{``            ``if` `(tri.Contains(n - tm))``            ``{``                ``return` `true``;``            ``}``        ``}``        ``return` `false``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int` `n = 24;``        ``if` `(checkTriangularSumRepresentation(n))``        ``{``            ``Console.WriteLine(``"Yes"``);``        ``}``        ``else``        ``{``            ``Console.WriteLine(``"No"``);``        ``}``    ``}``}` `// This code contributed by Rajput-Ji`

## PHP

 `= ``\$n``)``            ``break``;``            ` `        ``array_push``(``\$tri``, ``\$x``);``        ``\$i` `+= 1;``    ``}``    ` `    ``// Check if a pair exists``    ``foreach``(``\$tri` `as` `\$tm``)``        ``if` `(in_array(``\$n` `- ``\$tm``, ``\$tri``))``            ``return` `true;``            ` `    ``return` `false;``}` `// Driver Code``\$n` `= 24;` `if` `(checkTriangularSumRepresentation(``\$n``))``    ``print``(``"Yes"``);``else``    ``print``(``"No"``);` `// This code is contributed by mits``?>`

## Javascript

 ``
Output:
`Yes`

Time Complexity: 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)) .

## C++

 `// C++ implementation of the above approach``#include ``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"``<

Output:

`YES`

Time Complexity: O(Sqrt(N))

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