Given an integer N, our task is to print N distinct numbers such that their sum is 0.
Examples:
Input: N = 3
Output: 1, -1, 0
Explanation:
On adding the numbers that is 1 + (-1) + 0 the sum is 0.Input: N = 4
Output: 1, -1, 2, -2
Explanation:
On adding the numbers that is 1 + (-1) + 2 + (-2) the sum is 0.
Approach: To solve the problem mentioned above the main idea is to print Symmetric Pairs like (+x, -x) so that the sum will always be 0. The edge case to the problem is to observe that if integer N is odd, then print one 0 along with the numbers so that sum is not affected.
Below is the implementation of the above approach:
C++
// C++ implementation to Print N distinct// numbers such that their sum is 0 #include <bits/stdc++.h>using namespace std; // Function to print distinct n// numbers such that their sum is 0void findNumbers(int N){ for (int i = 1; i <= N / 2; i++) { // Print 2 symmetric numbers cout << i << ", " << -i << ", "; } // print a extra 0 if N is odd if (N % 2 == 1) cout << 0;} // Driver codeint main(){ int N = 10; findNumbers(N);} |
Java
// Java implementation to Print N distinct// numbers such that their sum is 0 class GFG{ // Function to print distinct n// numbers such that their sum is 0static void findNumbers(int N){ for (int i = 1; i <= N / 2; i++) { // Print 2 symmetric numbers System.out.print(i + ", " + -i + ", "); } // Print a extra 0 if N is odd if (N % 2 == 1) System.out.print(0);} // Driver codepublic static void main(String[] args){ int N = 10; findNumbers(N);}} // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation to print N distinct# numbers such that their sum is 0 # Function to print distinct n# numbers such that their sum is 0def findNumbers(N): for i in range(1, N // 2 + 1): # Print 2 symmetric numbers print(i, end = ', ') print(-i, end = ', ') # Print a extra 0 if N is odd if N % 2 == 1: print(0, end = '') # Driver code if __name__=='__main__': N = 10 findNumbers(N) # This code is contributed by rutvik_56 |
C#
// C# implementation to print N distinct // numbers such that their sum is 0 using System; class GFG { // Function to print distinct n // numbers such that their sum is 0 static void findNumbers(int N) { for(int i = 1; i <= (N / 2); i++) { // Print 2 symmetric numbers Console.Write(i + ", " + -i + ", "); } // Print a extra 0 if N is odd if (N % 2 == 1) Console.Write(0);} // Driver codestatic void Main() { int N = 10; findNumbers(N); }} // This code is contributed by divyeshrabadiya07 |
1, -1, 2, -2, 3, -3, 4, -4, 5, -5,
Time Complexity: O(log N)
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