A centered continuous number pyramid grows with odd row widths (1, 3, 5…), leading spaces for centering, and a running counter that never resets.
Remember
Rule: k = 1 before loops
for i from 1 to max step 2
for j from max down to 1
if j > i: print space
else: print k, then k++
1
2 3 4
5 6 7 8 9 ← max = 5 (spaces shown)
Follows the number & asterisk mirror in Program 23; next is the bidirectional number triangle in Program 25.
Approach
How to Solve It
Outer loop walks odd widths with i += 2. Reverse inner loop prints a space when j > i, else k++.
Method
Idea
Best for
Spaces + k++
Reverse scan: space if j > i, else print and bump k
Learning, interviews, exams
Separate space loop
Print (max - i) / 2 spaces, then i numbers
Clearer centering when you prefer two loops
Pseudocode
Pseudocode
k = 1
for i from 1 to max step 2:
for j from max down to 1:
if j > i:
print space
else:
print k and a space
k = k + 1
print newline
Cheat sheet
Goal
Pattern
Odd row widths
for (i = 1; i <= max; i += 2)
Init counter once
int k = 1; before the outer loop
Reverse columns
for (j = max; j >= 1; j--)
Space vs number
if (j > i) cout << " "; else cout << k++ << " ";
Force odd max
if (max % 2 == 0) max -= 1;
Printing Numbers vs Starting a New Line
API
Effect
Use for
cout << " " / k++ << " "
Stays on the same line
Spaces and numbers on the row
cout << "\n"
Ends the current line
After the inner loop finishes
Print characters without a newline, then end the row once.
Try it
Live Preview
Change the max odd width and the pyramid updates instantly — even values snap down to odd; capped at 9 for readable demos.
Whole numbers from 1 to 9. Even values are adjusted down by 1. Tap a chip or type — the preview redraws as you go.
Live resultmax = 5 · 9 numbers
1
2 3 4
5 6 7 8 9
Trace
Worked Walkthrough — max = 5
Trace how odd i, leading spaces, and persistent k build each row.
i
Spaces
Numbers
k after
Prints
1
4
1
2
1
3
2
2 3 4
5
2 3 4
5
0
5 6 7 8 9
10
5 6 7 8 9
Spaces when j > i = max - i. Total numbers = sum of odd widths → O(n²) for max width n.
Code
C++ Programs
Three complete programs: fixed max = 5, cin input, and a compact max = 3 demo. Use View Output to reveal sample results.
Example 1 — Fixed max = 5
Hard-coded width — odd outer loop, reverse inner loop with space vs k++.
C++
#include <iostream>
using namespace std;
int main()
{
int max = 5;
int i, j;
int k = 1;
for (i = 1; i <= max; i += 2)
{
for (j = max; j >= 1; j--)
{
if (j > i)
cout << " ";
else
cout << k++ << " ";
}
cout << "\n";
}
return 0;
}
Output
1
2 3 4
5 6 7 8 9
How It Works
1. Odd widths.i becomes 1, 3, 5 — each value is how many numbers that row prints.
2. Space or number. When j > i print a space; otherwise print k++.
3. Keep k alive.k sits outside the outer loop so numbers continue across rows.
Example 2 — User Input Max
Read max with cin, validate, and force odd so every row width stays odd.
C++
#include <iostream>
using namespace std;
int main()
{
int max;
int i, j;
int k = 1;
cout << "Enter the maximum odd width: ";
if (!(cin >> max) || max <= 0)
{
cout << "Please enter a positive integer.\n";
return 1;
}
if (max % 2 == 0)
max -= 1;
for (i = 1; i <= max; i += 2)
{
for (j = max; j >= 1; j--)
{
if (j > i)
cout << " ";
else
cout << k++ << " ";
}
cout << "\n";
}
return 0;
}
Output (when user enters 5)
Enter the maximum odd width: 5
1
2 3 4
5 6 7 8 9
How It Works
1. Prompt and validate. Reject failed reads and non-positive values before printing.
2. Force odd.if (max % 2 == 0) max -= 1; keeps row widths odd.
3. Safer input tip. Cap demos for readable output:
Safer input
if (!(cin >> max) || max < 1 || max > 9)
{
cout << "Enter a whole number from 1 to 9.\n";
return 1;
}
Example 3 — Compact max = 3
Same structure with only two rows — easy to confirm spaces and the running counter.
C++
#include <iostream>
using namespace std;
int main()
{
int max = 3;
int i, j;
int k = 1;
for (i = 1; i <= max; i += 2)
{
for (j = max; j >= 1; j--)
{
if (j > i)
cout << " ";
else
cout << k++ << " ";
}
cout << "\n";
}
return 0;
}
Output
1
2 3 4
How It Works
1. Two rows.i = 1 → two spaces then 1; i = 3 → 2 3 4.
2. Trace on paper. If you reset k = 1 inside the outer loop, every row restarts — that is not this pattern.
3. Scale up next. Once the small demo is clear, use Examples 1–2 for max = 5 or user input.
Edge Cases & Pitfalls
Check these before calling the solution done.
reset k
Counter restarts each row
Declaring k = 1 inside the outer loop restarts the sequence. Keep k outside both loops.
no spaces
Left-aligned pyramid
Skipping if (j > i) leaves numbers flush left. Print a space when j > i.
i++
Wrong step
Using i++ prints every width. For odd lengths only, keep i += 2.
even max
Even maximum
If max is even, subtract 1 so the last row stays odd-width — see Example 2.
\n inside
Broken rows
If cout << "\n" sits inside the inner loop, each column lands on its own line. Call it only after the row finishes.
cin
Check the stream
Validate cin >> max before looping — a failed read leaves max unset or stale.
Analysis
Time and Space Complexity
Program
Time
Extra space
Fixed / input (Examples 1–2)
O(n²)
O(1)
Compact max = 3 (Example 3)
O(n²)
O(1)
About n/2 rows each scan n columns → O(n²) time. Only a few loop variables are needed.
Remember
Key Takeaways
Odd widths:i += 2 gives row lengths 1, 3, 5…
Center with spaces: print a space when j > i, else print k++.
Keep k alive: init once before the outer loop so numbers continue across rows.
Complexity:O(n²) time from scanning n columns per odd row; O(1) extra space.
One line: for each odd i, scan columns reverse — space if j > i, else k++, then newline.
Frequently Asked Questions
A centered continuous pyramid: for max=5 you get a centered 1, then 2 3 4, then 5 6 7 8 9 — odd row widths with a running counter that never resets.
The program prints a space when j > i, which shifts the numbers right and centers each row within max columns.
Because the counter k is declared once before the outer loop and increments with k++ every print — it never resets per row.
Row widths are odd (1, 3, 5, …) so each row adds two more numbers than the previous row.
cout << " " stays on the same line for spaces and numbers. cout << "\n" ends the current row after the inner loop finishes.
The reverse inner loop prints leading spaces first (when j > i) and numbers afterward — a common centering trick.
Subtract 1 to force an odd width so every row length stays odd.
O(n²) for max width n because each of about n/2 rows iterates across n columns.
🤔
Did you know?
This centered pyramid prints numbers continuously using a counter k. An if inside a reverse loop prints leading spaces when j > i, then prints k++ once the column reaches the row boundary.
