711654. Pattern 3
Pattern 3 Slug: triangle-number Difficulty: Easy Id: 711654 Topic Tags: pattern-printing Company Tags: None
Summary
This problem involves printing a triangular number pattern. The pattern starts with 1, then each subsequent row is one more than the previous row's sum. For example, the first three rows are 1, 2+1=3, and 4+3+1=8. We need to write an algorithm that prints this pattern up to a given number of rows.
Detailed Explanation
The problem can be solved by using a simple loop to generate the triangular numbers. The key concept involved is understanding how triangular numbers are calculated. A triangular number or triangle number counts objects arranged in an equilateral triangle.
Here's a step-by-step breakdown of the solution:
- Initialize a variable
rowto 1, which will be the first row. - Start a loop that runs up to the given number of rows.
- Inside the loop, calculate the sum of all numbers from 1 to
row. This is the formula for triangular numbers:n*(n+1)/2, wherenis the current row. - Print the calculated sum as the current row.
- Increment
rowby 1 and repeat steps 3-4 until the loop ends.
Here's a diagram of how the pattern looks like:
1
2 1
3 2 1
4 3 2 1
The time complexity of this algorithm is O(n), where n is the number of rows. The space complexity is O(1), as we only use a constant amount of space to store the variables.
Optimized Solutions
Java
public class Main {
public static void main(String[] args) {
int rows = 5; // Number of rows
for (int i = 1; i <= rows; i++) {
int sum = 0;
for (int j = 1; j <= i; j++) {
sum += j;
}
System.out.println(sum);
}
}
}