C Palindrome Number Triangle Pattern (Outer Peak)

What Is This Pattern?

A palindrome number triangle prints a mirrored sequence on each row: descend from i to 2, then ascend from 1 to i. Every row reads the same both ways.

Rule: for i from 1 to rows
      print j from i down to 2
      print j from 1 up to i

1
212
32123
4321234
543212345     ← rows = 5

Two inner loops per row build the symmetry. Follows the decreasing triangle in Program 36; next is Program 38.

How to Solve It

Start with fixed rows = 5, then generalize with scanf — optionally demo a smaller count for tracing.

Pseudocode

for i from 1 to rows:
    for j from i down to 2:
        print j (no newline)
    for j from 1 to i:
        print j (no newline)
    print newline

Cheat sheet

Printing Numbers vs Starting a New Line

Print all digits without a newline, then end the row once.

Live Preview

Change the row count and the palindrome triangle updates instantly — capped at 9 so every value stays a single character.

1
212
32123
4321234
543212345

Worked Walkthrough — rows = 4

Trace each outer-loop value of i and list both halves of the palindrome.

Total digits for n rows = 1 + 3 + … + (2n - 1) = n² (here 16). That square sum is why time is O(n²).

C Programs

Three complete programs: fixed rows, scanf input, and a small demo. Use View Output to reveal sample results.

#include <stdio.h>

int main(void)
{
    int rows = 5;
    int i, j;

    for (i = 1; i <= rows; i++)
    {
        for (j = i; j > 1; j--)
            printf("%d", j);

        for (j = 1; j <= i; j++)
            printf("%d", j);

        printf("\n");
    }

    return 0;
}

1
212
32123
4321234
543212345

#include <stdio.h>

int main(void)
{
    int rows;
    int i, j;

    printf("Enter rows: ");
    if (scanf("%d", &rows) != 1 || rows < 1)
        return 0;

    for (i = 1; i <= rows; i++)
    {
        for (j = i; j > 1; j--)
            printf("%d", j);

        for (j = 1; j <= i; j++)
            printf("%d", j);

        printf("\n");
    }

    return 0;
}

Enter rows: 3
1
212
32123

if (scanf("%d", &rows) != 1 || rows < 1)
{
    printf("Enter a positive whole number of rows.\n");
    return 1;
}

#include <stdio.h>

int main(void)
{
    int rows = 3;
    int i, j;

    for (i = 1; i <= rows; i++)
    {
        for (j = i; j > 1; j--)
            printf("%d", j);

        for (j = 1; j <= i; j++)
            printf("%d", j);

        printf("\n");
    }

    return 0;
}

1
212
32123

Time and Space Complexity

Total digits = n² because odd numbers sum to a square. For rows = 5 that is 25 printed digits.

Key Takeaways