C++ Palindrome Number Pattern (Increasing-Decreasing)

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6 min read
Updated: Sep 2026
3 programs
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Increasing-decreasing number pyramid pattern in C++ programming (1, 232, 34543...)
Definition

What Is This Pattern?

An increasing-decreasing number pyramid prints each row as a palindrome: count up from i to the peak 2i - 1, then back down to i. With rows = 5 you get 1, 232, 34543, 4567654, 567898765.

Remember
Rule: m = i; print up with m++; then m -= 2; print down with m--

1
232
34543
4567654
567898765     ← rows = 5

Unlike Program 51 (one continuous counter across rows), this shape resets m = i on every row and builds a self-contained palindrome.

Approach

How to Solve It

Set m = i. Print the ascending half, step back with m = m - 2, then print the descending half.

MethodIdeaBest for
Two half-loopsAscend with m++, skip peak, descend with m--Learning, interviews, demos
Compact traceSame logic with rows = 3Quick dry-runs on paper

Pseudocode

Pseudocode
for i from 1 to rows:
    m = i
    for j from 1 to i:
        print m
        m = m + 1
    m = m - 2
    for k from 1 to (i - 1):
        print m
        m = m - 1
    print newline

Cheat sheet

GoalPattern
Start of rowm = i;
Ascending halffor (j = 1; j <= i; j++) cout << m++;
Skip peakm = m - 2;
Descending halffor (k = 1; k < i; k++) cout << m--;
Digits on row i2 * i - 1
End the rowcout << "\n";

Printing Numbers vs Starting a New Line

APIEffectUse for
cout << m++ / cout << m--Stays on the same lineEach digit
cout << "\n"Ends the current lineAfter both inner loops

Print digits without a newline, then end the row once. Putting "\n" inside either half-loop turns the pyramid into a vertical list.

Try it

Live Preview

Change the row count and the palindromic pyramid updates instantly.

Use rows from 1 to 9 (single digits stay tidy). Tap a chip or type a value — the preview redraws as you go.

Live result 5 rows · 25 digits
1
232
34543
4567654
567898765
Trace

Worked Walkthrough

Trace row i = 3 — ascend to the peak, skip it, then descend.

StepmActionLine so far
Start3m = i
Ascend3 → 4 → 5print m++ three times345
Skip peak6 → 4m = m - 2345
Descend4 → 3print m-- twice34543

Without m = m - 2, the peak 5 would print twice. The descending loop runs i - 1 times so the peak stays unique.

Code

C++ Programs

Three complete programs: fixed rows = 5, cin input, and a compact rows = 3 demo. Use View Output to reveal sample results.

Example 1 — Fixed rows = 5

Hard-coded height — ascend with m++, skip the peak, descend with m--.

C++
#include <iostream>
using namespace std;

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

    for (i = 1; i <= rows; i++)
    {
        m = i;

        for (j = 1; j <= i; j++)
            cout << m++;

        m = m - 2;

        for (k = 1; k < i; k++)
            cout << m--;

        cout << "\n";
    }

    return 0;
}

How It Works

1. Seed each row at i. m = i starts the ascending half at the row index.

2. Print up to the peak. The first loop runs i times with m++, reaching 2i - 1.

3. Skip and mirror. m = m - 2 avoids a double peak; the second loop prints i - 1 digits downward, then "\n" ends the row.

Example 2 — User Input Rows

Read the row count at runtime — both halves scale automatically.

C++
#include <iostream>
using namespace std;

int main()
{
    int rows, i, j, k, m;

    cout << "Enter the number of rows: ";
    cin >> rows;

    for (i = 1; i <= rows; i++)
    {
        m = i;

        for (j = 1; j <= i; j++)
            cout << m++;

        m = m - 2;

        for (k = 1; k < i; k++)
            cout << m--;

        cout << "\n";
    }

    return 0;
}

How It Works

1. Same two-loop core. Only the source of rows changes from a literal to cin.

2. Entering 4 stops early. You get four palindromic rows ending at 4567654.

3. Validate in real apps. Prefer checking cin failure and requiring rows >= 1 (tip below).

Safer input tip
if (!(cin >> rows) || rows < 1)
{
    cout << "Please enter a positive integer.\n";
    return 1;
}

Example 3 — Compact rows = 3

Smaller height for quick tracing of m++, m - 2, and m--.

C++
#include <iostream>
using namespace std;

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

    for (i = 1; i <= rows; i++)
    {
        m = i;

        for (j = 1; j <= i; j++)
            cout << m++;

        m = m - 2;

        for (k = 1; k < i; k++)
            cout << m--;

        cout << "\n";
    }

    return 0;
}

How It Works

1. Only three rows. Easy to dry-run every m change on paper.

2. Same formula. Nothing changes except rows — proving the pattern scales.

3. Row 1 has no descent. When i = 1, the decreasing loop never runs — output is just 1.

Edge Cases & Pitfalls

Check these before calling the solution done.

m - 2

Do not skip the peak step-back

Without m = m - 2, the peak digit prints twice and the palindrome breaks.

Descend bound

Use k < i, not k <= i

The descending half must run i - 1 times so the peak stays unique.

Newline

Do not put "\n" inside either half-loop

Call cout << "\n" only after both halves finish.

Bad cin

Validate input

Check cin >> rows and require rows >= 1 before the loops.

Analysis

Time and Space Complexity

ProgramTimeExtra space
Nested loops (Examples 1–3)O(n²)O(1)

Row i prints 2i - 1 digits, and 1 + 3 + … + (2n - 1) = n² — quadratic in n. Only a few integers of extra memory.

Remember

Key Takeaways

  • Rule: set m = i, print up with m++, then m -= 2, then print down with m--.
  • Skip the peak: m = m - 2 prevents a duplicated middle digit.
  • Break the row: call cout << "\n" only after both half-loops.
  • Complexity: O(n²) time; O(1) extra space.

One line: for each row i, print i..(2i-1) then back down to i, skipping a second peak, then end the line.

Frequently Asked Questions

Row 3 starts at 3, prints up to 5 (345), then prints back down to 3 (43) after m = m - 2 — producing 34543.
Each row counts up from i to the peak 2i-1, then counts back down to i. The sequence reads the same left-to-right on each line.
After the increasing loop, m is one past the peak. Subtracting 2 moves it to the value just before the peak so the decreasing loop does not repeat the peak digit.
Program 51 uses a continuous counter with alternating direction across rows. Program 52 resets m = i each row and builds a palindromic line per row.
Printing a digit stays on the same line. Printing a newline ends the current row. Digits use cout without a newline; the row break uses cout << "\n" after both inner loops.
O(n²) for n rows because row i prints 2i-1 digits and 1+3+5+...+(2n-1) = n² total prints.
Step back with m = m - 2 before the decreasing loop. The decreasing loop then runs i-1 times, skipping the peak.
After cin >> rows, check failure and require rows >= 1 before the loops.

Did you know?

Each row is palindromic: print i..(2i-1) ascending, then back down with m = m - 2 to skip the peak. Total digits = 1+3+5+…+(2n-1) = n² — time O(n²).

Next: Mirror Diagonal Pattern

Continue with the next pattern in the C++ number-pattern series.

Program 53 tutorial →

About the author

Mari Selvan M P
Mari Selvan M P 🔗

Developer, cloud engineer, and technical writer

  • Experience 12 years building web and cloud systems
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I write practical tutorials so students and working developers can learn by doing—from databases and APIs to deployment on AWS.

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