Print first N natural numbers using an iterative approach i.e. using for loop. For loop has three parameters initialization, testing condition, and increment/decrement.
Input: N = 10
Output: First 10 Numbers = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Sum of first 10 Natural Number = 55
Input: N = 5
Output: First 5 Numbers = 1, 2, 3, 4, 5
Sum of first 5 Natural Number = 15
Approach
- Start for loop initialization with i = 1.
- Write testing condition as i <= N.
- Add increment statement as i++ or i+=1.
- Initialize a variable sum with 0.
- Start adding i with the sum at each iteration of for loop and print i.
- Print sum at the end for loop.
Below is the implementation of above approach
Java
import java.io.*;
class GFG {
public static void main(String[] args)
{
int N = 10 ;
int sum = 0 ;
System.out.print( "First " + N + " Numbers = " );
for ( int i = 1 ; i <= N; i++) {
System.out.print(i + " " );
sum += i;
}
System.out.println();
System.out.println( "Sum of first " + N
+ " Natural Number = " + sum);
}
}
|
Output
First 10 Numbers = 1 2 3 4 5 6 7 8 9 10
Sum of first 10 Natural Number = 55
Time Complexity: O(n)
Auxiliary Space: O(1) because constant space for variables is being used
Alternate Approach
- Start for loop initialization with i = 1.
- Write testing condition as i <= N.
- Add increment statement as i++ or i+=1.
- Start Printing i for each iteration.
- Print sum using first N natural number formula at the end of for loop.
Below is the implementation of the above approach
Java
import java.io.*;
class GFG {
public static void main(String[] args)
{
int N = 5 ;
System.out.print( "First " + N + " Numbers = " );
for ( int i = 1 ; i <= N; i++) {
System.out.print(i + " " );
}
System.out.println();
System.out.println( "Sum of first " + N
+ " Natural Number = " + (N*(N+ 1 ))/ 2 );
}
}
|
Output
First 5 Numbers = 1 2 3 4 5
Sum of first 5 Natural Number = 15
Time Complexity: O(n)
Auxiliary Space: O(1) as it is using constant space for variables
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Last Updated :
05 Jan, 2023
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