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Pattern Printing – Till N & Reverse

Print a symmetric pattern of numbers increasing from 1 to N and then decreasing back to 1

Understand the Problem

Problem Statement

Problem Statement

Write a program that accepts an integer N and prints a total of 2N lines, following the pattern shown in the examples below.

The pattern consists of:

  • Lines 1 to N: Each line i contains the digit i repeated i times
  • Lines N+1 to 2N: Each line contains the digit (2N - line_number) repeated the appropriate number of times

Constraints

  • 2 ≤ N ≤ 100
  • N must be a positive integer
  • Output must contain exactly 2N lines

Examples

Example 1
Input
4
Output
1
22
333
4444
4444
333
22
1
Explanation

For N=4, we print 8 lines total. First 4 lines show digits 1, 22, 333, 4444. Next 4 lines show the same pattern in reverse: 4444, 333, 22, 1.

Example 2
Input
7
Output
1
22
333
4444
55555
666666
7777777
7777777
666666
55555
4444
333
22
1
Explanation

For N=7, we print 14 lines total. First 7 lines show digits 1, 22, 333, 4444, 55555, 666666, 7777777. Next 7 lines show the same pattern in reverse order.

Solution

#include <stdio.h>

int main() {
    int n;
    scanf("%d", &n);
    
    // Print ascending pattern
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= i; j++) {
            printf("%d", i);
        }
        printf("\n");
    }
    
    // Print descending pattern
    for (int i = n; i >= 1; i--) {
        for (int j = 1; j <= i; j++) {
            printf("%d", i);
        }
        printf("\n");
    }
    
    return 0;
}
Time:O(N²) - We iterate through N lines, and for each line i, we print i characters
Space:O(1) - We only use a constant amount of extra space for variables
Approach:

The C solution uses nested loops:

  1. First loop (i = 1 to N): For each line i, prints the digit i repeated i times using an inner loop
  2. Second loop (i = N down to 1): Prints the same pattern in reverse order

Each line ends with a newline character to move to the next line.

Visual Explanation

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