Check if a number can be represented as a sum of 2 triangular numbers
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 <bits/stdc++.h> 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; } |
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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<Integer> 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 |
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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 |
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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 |
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PHP
<?php // PHP implementation of the above approach // Function to check if it is possible or not function checkTriangularSumRepresentation($n) { $tri = array(); $i = 1; // Store all triangular numbers // up to N in a Set while (true) { $x = $i * ($i + 1); if ($x >= $n) break; array_push($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 ?> |
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Output:
Yes
Time Complexity: O(Sqrt(N))
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