Given an integer N, the task is to check whether the elements from the range [1, N] can be divided into three non-empty equal sum subsets. If possible then print Yes else print No.
Examples:
Input: N = 5
Output: Yes
The possible subsets are {1, 4}, {2, 3} and {5}.
(1 + 4) = (2 + 3) = (5)Input: N = 3
Output: No
Approach: There are two cases:
- If N ? 3: In this case, it is not possible to divide the elements in the subsets that satisfy the given condition. So, print No.
- If N > 3: In this case, it is only possible when the sum of all the elements of the range [1, N] is divisible by 3 which can be easily calculated as sum = (N * (N + 1)) / 2. Now, if sum % 3 = 0 then print Yes else print No.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <iostream>using namespace std;// Function that returns true// if the subsets are possiblebool possible(int n){ // If n <= 3 then it is not possible // to divide the elements in three subsets // satisfying the given conditions if (n > 3) { // Sum of all the elements // in the range [1, n] int sum = (n * (n + 1)) / 2; // If the sum is divisible by 3 // then it is possible if (sum % 3 == 0) { return true; } } return false;}// Driver codeint main(){ int n = 5; if (possible(n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the approach import java.math.*;class GFG{ // Function that returns true // if the subsets are possible public static boolean possible(int n) { // If n <= 3 then it is not possible // to divide the elements in three subsets // satisfying the given conditions if (n > 3) { // Sum of all the elements // in the range [1, n] int sum = (n * (n + 1)) / 2; // If the sum is divisible by 3 // then it is possible if (sum % 3 == 0) { return true; } } return false; } // Driver code public static void main(String[] args) { int n = 5; if (possible(n)) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by Naman_Garg |
Python3
# Python3 implementation of the approach # Function that returns true # if the subsets are possible def possible(n) : # If n <= 3 then it is not possible # to divide the elements in three subsets # satisfying the given conditions if (n > 3) : # Sum of all the elements # in the range [1, n] sum = (n * (n + 1)) // 2; # If the sum is divisible by 3 # then it is possible if (sum % 3 == 0) : return True; return False; # Driver code if __name__ == "__main__" : n = 5; if (possible(n)) : print("Yes"); else : print("No"); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approach using System;class GFG{ // Function that returns true // if the subsets are possible public static bool possible(int n) { // If n <= 3 then it is not possible // to divide the elements in three subsets // satisfying the given conditions if (n > 3) { // Sum of all the elements // in the range [1, n] int sum = (n * (n + 1)) / 2; // If the sum is divisible by 3 // then it is possible if (sum % 3 == 0) { return true; } } return false; } // Driver code static public void Main (){ int n = 5; if (possible(n)) Console.Write("Yes"); else Console.Write("No");}}// This code is contributed by ajit |
Javascript
<script>// Javascript implementation of the approach// Function that returns true// if the subsets are possiblefunction possible(n){ // If n <= 3 then it is not possible // to divide the elements in three subsets // satisfying the given conditions if (n > 3) { // Sum of all the elements // in the range [1, n] let sum = parseInt((n * (n + 1)) / 2); // If the sum is divisible by 3 // then it is possible if (sum % 3 == 0) { return true; } } return false;}// Driver codelet n = 5;if (possible(n)) document.write("Yes");else document.write("No"); // This code is contributed by rishavmahato348</script> |
Output:
Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
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