⚡ TL;DR
Print the right pascal star pattern in Dart. Complete working code, expected output, algorithm explanation, and practice variations for beginners learning loops.
The Right Pascal Star pattern is a classic loop exercise. Here’s the complete implementation in Dart with working code and output.
Pattern to Print
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*Implementation
void main() {
int n = 5;
// Top half
for (int i = 1; i <= n; i++) {
print(' ' * (n - i) + '*' * i);
}
// Bottom half
for (int i = n - 1; i >= 1; i--) {
print(' ' * (n - i) + '*' * i);
}
}Output
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*How It Works
Two loops: the first builds rows from 1 to n stars (right-aligned), the second mirrors it back down from n-1 to 1. Result is a sideways diamond.
Practice Variations
- Try printing the pattern with a different character (e.g.,
#or+) - Modify the code to accept the pattern size as user input
- Combine this pattern with its inverse to create a more complex shape
Complexity
- Time: O(n²) — the grid or line count grows quadratically
- Space: O(1) — only loop counters are needed
Related Patterns
- Right Pascal Star in Python
- Right Pascal Star in Go
- Right Pascal Star in Swift
- Hollow Square Star in Dart
- Solid Square Star in Dart
- Cross/Plus Sign Star in Dart
FAQ
How do you print the Right Pascal Star pattern in Dart?
Use nested loops: the outer loop walks through the rows while the inner loop prints the characters or values for each row. The complete Dart implementation with expected output is shown in the sections above.
What is the time complexity of the Right Pascal Star pattern in Dart?
The time complexity is O(n²) and the space complexity is O(1), since the pattern is built with a fixed number of loop counters and printed row by row.
Happy coding!
