Pattern:
1
AA
123
AAAA
Solution
Note: Don't jump directly to the solution, try it out yourself first.
Approach:
1)Require minimum 2 nested loop
2)Outer loop is for row
3)Inner loop is for column
4)Print in inner loop
Pattern Breakdown
Row 1 → 1
Row 2 → AA
Row 3 → 123
Row 4 → AAAA
Observations:
1)Odd rows contain numbers (1, 123).
2)Even rows contain uppercase As (AA, AAAA).
3)The count of elements in each row corresponds to the row number.
Video Solution
| Java |
public class Main { public static void main(String[] args) { int rows = 4; // Number of rows // Outer loop for rows for (int i = 1; i <= rows; i++) { // Check if row is odd if (i % 2 != 0) { // Print numbers from 1 to row number for (int j = 1; j <= i; j++) { System.out.print(j); } } else { // Print 'A' row number of times for (int j = 1; j <= i; j++) { System.out.print("A"); } } System.out.println(); // Move to the next line } } } |
| C++ |
#include <iostream> using namespace std; int main() { int rows = 4; // Number of rows // Outer loop for rows for (int i = 1; i <= rows; i++) { // Check if row is odd if (i % 2 != 0) { // Print numbers from 1 to row number for (int j = 1; j <= i; j++) { cout << j; } } else { // Print 'A' row number of times for (int j = 1; j <= i; j++) { cout << "A"; } } cout << endl; // Move to the next line } return 0; } |
| Python |
rows = 4 # Number of rows # Outer loop for rows for i in range(1, rows + 1): if i % 2 != 0: # Print numbers from 1 to row number for j in range(1, i + 1): print(j, end="") else: # Print 'A' row number of times print("A" * i, end="") # Repeat 'A' i times print() # Move to the next line |
