Check whether a number can be represented as sum of K distinct positive integers
Given two integers N and K, the task is to check whether N can be represented as sum of K distinct positive integers.
Examples:
Input: N = 12, K = 4
Output: Yes
N = 1 + 2 + 4 + 5 = 12 (12 as sum of 4 distinct integers)Input: N = 8, K = 4
Output: No
Approach: Consider the series 1 + 2 + 3 + … + K which has exactly K distinct integers with minimum possible sum i.e. Sum = (K * (K – 1)) / 2. Now, if N < Sum then it is not possible to represent N as the sum of K distinct positive integers but if N ≥ Sum then any integer say X ≥ 0 can be added to Sum to generate the sum equal to N i.e. 1 + 2 + 3 + … + (K – 1) + (K + X) ensuring that there are exactly K distinct positive integers.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <iostream> using namespace std; // Function that returns true if n // can be represented as the sum of // exactly k distinct positive integers bool solve(int n, int k) { // If n can be represented as // 1 + 2 + 3 + ... + (k - 1) + (k + x) if (n >= (k * (k + 1)) / 2) { return true; } return false; } // Driver code int main() { int n = 12, k = 4; if (solve(n, k)) cout << "Yes"; else cout << "No"; return 0; } |
Java
// Java implementation of the approach import java.io.*; class GFG { // Function that returns true if n // can be represented as the sum of // exactly k distinct positive integers static boolean solve(int n, int k) { // If n can be represented as // 1 + 2 + 3 + ... + (k - 1) + (k + x) if (n >= (k * (k + 1)) / 2) { return true; } return false; } // Driver code public static void main(String[] args) { int n = 12, k = 4; if (solve(n, k)) System.out.println("Yes"); else System.out.println("No"); } } // This code is contributed by anuj_67.. |
Python3
# Python 3 implementation of the approach # Function that returns true if n # can be represented as the sum of # exactly k distinct positive integers def solve(n,k): # If n can be represented as # 1 + 2 + 3 + ... + (k - 1) + (k + x) if (n >= (k * (k + 1)) // 2): return True return False # Driver code if __name__ == '__main__': n = 12 k = 4 if (solve(n, k)): print("Yes") else: print("No") # This code is contributed by # Surendra_Gangwar |
C#
// C# implementation of the approach using System; class GFG { // Function that returns true if n // can be represented as the sum of // exactly k distinct positive integers static bool solve(int n, int k) { // If n can be represented as // 1 + 2 + 3 + ... + (k - 1) + (k + x) if (n >= (k * (k + 1)) / 2) { return true; } return false; } // Driver code static public void Main () { int n = 12, k = 4; if (solve(n, k)) Console.WriteLine("Yes"); else Console.WriteLine("No"); } } // This code is contributed by ajit. |
PHP
<?php // PHP implementation of the approach // Function that returns true if n // can be represented as the sum of // exactly k distinct positive integers function solve($n, $k) { // If n can be represented as // 1 + 2 + 3 + ... + (k - 1) + (k + x) if ($n >= ($k * ($k + 1)) / 2) { return true; } return false; } // Driver code $n = 12; $k = 4; if (solve($n, $k)) echo "Yes"; else echo "No"; // This code is contributed by ihritik ?> |
Output:
Yes
