C++ Number Triangle Pattern (Alternating Direction)
Beginner
6 min read
Updated: Sep 2026
3 programs
Live preview
Definition
What Is This Pattern?
An alternating number triangle prints continuous integers across rows: odd rows go left-to-right, even rows go right-to-left. With rows = 5 you get 1, 3 2, 4 5 6, 10 9 8 7, 11 12 13 14 15.
Unlike Program 50 (fixed digit halves), this shape uses a running counter and flips direction with i % 2.
Approach
How to Solve It
Keep a counter next. On each row compute end, then print ascending or descending based on odd/even i.
Method
Idea
Best for
Counter + direction
next++ always; print next or end-- by parity
Learning, interviews, demos
Compact trace
Same logic with rows = 3
Quick dry-runs on paper
Pseudocode
Pseudocode
next = 1
for i from 1 to rows:
end = next + i - 1
for j from 1 to i:
if i is odd:
print next
else:
print end
end = end - 1
next = next + 1
print newline
Cheat sheet
Goal
Pattern
Start counter
int next = 1;
Row end value
int end = next + i - 1;
Odd row
cout << next << " ";
Even row
cout << end-- << " ";
Advance always
next++; after each print
End the row
cout << "\n";
Printing Numbers vs Starting a New Line
API
Effect
Use for
cout << n << " "
Stays on the same line
Each value
cout << "\n"
Ends the current line
After the inner loop
Print values without a newline, then end the row once. Putting "\n" inside the inner loop turns the triangle into a vertical list.
Try it
Live Preview
Change the row count and the alternating triangle updates instantly.
Use rows from 1 to 9. Tap a chip or type a value — the preview redraws as you go.
Live result5 rows · 15 values
1
3 2
4 5 6
10 9 8 7
11 12 13 14 15
Trace
Worked Walkthrough
Trace the first four rows — watch next, end, and direction.
Row i
Parity
next start
end
Printed
1
odd / asc
1
1
1
2
even / desc
2
3
3 2
3
odd / asc
4
6
4 5 6
4
even / desc
7
10
10 9 8 7
Even on even rows, next still increments once per cell — so numbering stays continuous for the next row.
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 — odd rows print next ascending, even rows print end-- descending.
C++
#include <iostream>
using namespace std;
int main()
{
int rows = 5;
int i, j, next = 1;
for (i = 1; i <= rows; i++)
{
int end = next + i - 1;
for (j = 1; j <= i; j++)
{
if (i % 2 == 1)
cout << next << " ";
else
cout << end-- << " ";
next++;
}
cout << "\n";
}
return 0;
}
Output
1
3 2
4 5 6
10 9 8 7
11 12 13 14 15
How It Works
1. Compute the row’s last value.end = next + i - 1 is the reverse start for even rows.
2. Branch on parity. Odd rows print next; even rows print end--.
3. Always advance next. Increment after every cell so the next row continues the sequence, then end the line with "\n".
Example 2 — User Input Rows
Read the row count at runtime — counter and direction logic stay the same.
C++
#include <iostream>
using namespace std;
int main()
{
int rows, i, j, next = 1;
cout << "Enter the number of rows: ";
cin >> rows;
for (i = 1; i <= rows; i++)
{
int end = next + i - 1;
for (j = 1; j <= i; j++)
{
if (i % 2 == 1)
cout << next << " ";
else
cout << end-- << " ";
next++;
}
cout << "\n";
}
return 0;
}
Output (when user enters 4)
Enter the number of rows: 4
1
3 2
4 5 6
10 9 8 7
How It Works
1. Same counter core. Only the source of rows changes from a literal to cin.
2. Entering 4 stops early. You get four rows ending at 10 9 8 7.
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 next, end, and odd/even branches.
C++
#include <iostream>
using namespace std;
int main()
{
int rows = 3;
int i, j, next = 1;
for (i = 1; i <= rows; i++)
{
int end = next + i - 1;
for (j = 1; j <= i; j++)
{
if (i % 2 == 1)
cout << next << " ";
else
cout << end-- << " ";
next++;
}
cout << "\n";
}
return 0;
}
Output
1
3 2
4 5 6
How It Works
1. Only three rows. Easy to dry-run every increment and both parity branches on paper.
2. Same formula. Nothing changes except rows — proving the pattern scales.
3. One even row in the middle. Row 2 is the only reverse print — the zig-zag is obvious.
Edge Cases & Pitfalls
Check these before calling the solution done.
Forget end
Compute end before the inner loop
Even rows need end = next + i - 1 as the reverse start. Skipping it breaks the zig-zag.
next++
Always increment next
Even when printing end--, still call next++ so the next row continues correctly.
Newline
Do not put "\n" inside the inner loop
Call cout << "\n" only after all values on the row are printed.
Bad cin
Validate input
Check cin >> rows and require rows >= 1 before the loops.
Analysis
Time and Space Complexity
Program
Time
Extra space
Nested loops (Examples 1–3)
O(n²)
O(1)
Total prints are 1 + 2 + … + n = n(n + 1) / 2 — quadratic in n. Only a few integers of extra memory.
Remember
Key Takeaways
Rule: continuous next; odd rows ascending, even rows end--.
Set end:end = next + i - 1 before printing an even row.
Break the row: call cout << "\n" only after the inner loop.
Complexity:O(n²) time; O(1) extra space.
One line: for each row, print i continuous numbers — left-to-right if odd, right-to-left if even — then end the line.
Frequently Asked Questions
Row 2 is even, so it prints in reverse. The row contains numbers 2 and 3, but they are printed as 3 2 using end--.
Even rows print right-to-left for the zig-zag effect. We compute end = next + i - 1 and decrement while printing.
A running counter next increments once per printed value and is never reset between rows.
Program 50 concatenates fixed digit sequences per row. Program 51 uses a continuous counter and alternates print direction on odd/even rows.
Printing a number (and space) stays on the same line. Printing a newline ends the current row. Values use cout without a newline; the row break uses cout << "\n" after the inner loop.
O(n²) for n rows because total printed numbers are 1+2+...+n = n(n+1)/2.
Before printing row i, the last number is next + i - 1 — the reverse start when the row is even.
After cin >> rows, check failure and require rows >= 1 before the loops.
🤔
Did you know?
Numbers stay continuous via a running counter next. Odd rows print ascending; even rows print descending using end = next + i - 1. Total prints = n(n+1)/2 — time O(n²).
