A spaced mirror number pattern prints a fixed-width left half 1..i (spaces after), then a fixed-width right half i..1 (spaces before) — until the last row joins with no gap.
Remember
Rule: for i from 1 to rows
left: j=1..rows → j if j<=i else space
right: k=rows..1 → space if k>i else k
1 1
12 21
123 321
1234 4321
1234554321 ← rows = 5
Follows the 0-centered descending mirror in Program 28; next is the right-aligned descending triangle in Program 30.
Approach
How to Solve It
Outer loop grows i. Both inner loops always run rows times — print a digit or a space so columns stay aligned.
Method
Idea
Best for
Fixed-width if/else
Left j<=i ? j : " "; right k>i ? " " : k
Learning, interviews, exams
Ternary form
Same logic in one expression per position
Shorter demos once if/else is clear
Pseudocode
Pseudocode
for i from 1 to rows:
for j from 1 to rows:
if j <= i: print j
else: print space
for k from rows down to 1:
if k > i: print space
else: print k
print newline
Cheat sheet
Goal
Pattern
Grow each row
for (i = 1; i <= rows; i++)
Left half
if (j <= i) cout << j; else cout << " ";
Right half
if (k > i) cout << " "; else cout << k;
Fixed width
Both inners always loop 1..rows / rows..1
End of row
cout << "\n";
Printing Numbers vs Starting a New Line
API
Effect
Use for
cout << j / k / " "
Stays on the same line
Each digit or space on the row
cout << "\n"
Ends the current line
After both fixed-width loops finish
Print characters without a newline, then end the row once.
Try it
Live Preview
Change the row count and the spaced mirror updates instantly — capped at 9 for readable demos.
Whole numbers from 1 to 9. Tap a chip or type a value — the preview redraws as you go.
Live resultrows = 5 · 50 chars
1 1
12 21
123 321
1234 4321
1234554321
Trace
Worked Walkthrough — rows = 5
Trace how left digits grow, gap spaces shrink, and the right mirror closes in.
i
Left
Gap
Right
Prints
1
1
8 spaces
1
1 1
2
12
6 spaces
21
12 21
3
123
4 spaces
321
123 321
4
1234
2 spaces
4321
1234 4321
5
12345
0 spaces
54321
1234554321
Gap spaces = 2 × (rows − i). Each row prints 2 × rows characters → O(n²) total.
Code
C++ Programs
Three complete programs: fixed rows = 5, cin input with ternary form, and a compact rows = 3 demo. Use View Output to reveal sample results.
Example 1 — Fixed rows = 5
Hard-coded height — fixed-width left and right halves with if/else.
C++
#include <iostream>
using namespace std;
int main()
{
int i, j, k;
for (i = 1; i <= 5; i++)
{
for (j = 1; j <= 5; j++)
{
if (j <= i)
cout << j;
else
cout << " ";
}
for (k = 5; k >= 1; k--)
{
if (k > i)
cout << " ";
else
cout << k;
}
cout << "\n";
}
return 0;
}
Output
1 1
12 21
123 321
1234 4321
1234554321
How It Works
1. Outer grows the filled count.i is how many digits appear on each half.
2. Fixed width. Both inners always run 5 times — unused slots become spaces.
3. Last row joins. When i = 5, no spaces remain — output 1234554321.
Example 2 — User Input Rows
Read rows with cin, validate, then use ternary conditions for each position.
C++
#include <iostream>
using namespace std;
int main()
{
int rows;
int i, j, k;
cout << "Enter rows: ";
if (!(cin >> rows) || rows <= 0)
{
cout << "Please enter a positive integer.\n";
return 1;
}
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= rows; j++)
(j <= i) ? cout << j : cout << " ";
for (k = rows; k >= 1; k--)
(k > i) ? cout << " " : cout << k;
cout << "\n";
}
return 0;
}
Output (when user enters 4)
Enter rows: 4
1 1
12 21
123 321
12344321
How It Works
1. Prompt and validate. Reject failed reads and non-positive values before printing.
2. Same core. Ternary form matches Example 1’s if/else — only rows comes from the user.
3. Safer input tip. Cap demos for readable output:
Safer input
if (!(cin >> rows) || rows < 1 || rows > 9)
{
cout << "Enter a whole number from 1 to 9.\n";
return 1;
}
Example 3 — Compact rows = 3
Same structure with only three rows — easy to confirm gap spaces shrink by 2 each time.
C++
#include <iostream>
using namespace std;
int main()
{
int rows = 3;
int i, j, k;
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= rows; j++)
{
if (j <= i)
cout << j;
else
cout << " ";
}
for (k = rows; k >= 1; k--)
{
if (k > i)
cout << " ";
else
cout << k;
}
cout << "\n";
}
return 0;
}
Output
1 1
12 21
123321
How It Works
1. Three rows.i = 1 → 1 1; i = 2 → 12 21; i = 3 → 123321.
2. Trace on paper. If either loop stops at i instead of rows, columns misalign.
3. Scale up next. Once the small demo is clear, use Examples 1–2 for five rows or user input.
Edge Cases & Pitfalls
Check these before calling the solution done.
no spaces
Missing else branches
Skipping cout << " " collapses the halves — you get a tight palindrome, not a spaced mirror.
width ≠ rows
Mismatched loop bounds
Both inners must run exactly rows times. Looping only to i breaks column alignment.
\n inside
Broken rows
If cout << "\n" sits inside either inner loop, the mirror splits across lines. Call it only after both finish.
rows = 1
Joined single row
Output is 11 — no gap spaces. A good sanity check for input validation.
rows > 9
Multi-digit glue
Digits run together (…91011…). Cap demos or add extra spacing carefully.
cin
Check the stream
Validate cin >> rows before looping — a failed read leaves rows unset or stale.
Analysis
Time and Space Complexity
Program
Time
Extra space
Fixed / input (Examples 1–2)
O(n²)
O(1)
Compact rows = 3 (Example 3)
O(n²)
O(1)
Each of n rows prints 2n characters → O(n²) time. Only a few loop variables are needed.
Remember
Key Takeaways
Two fixed halves: left fills 1..i; right fills i..1; unused slots are spaces.
Always width rows: both inners loop rows times — that is what keeps columns aligned.
Break the row: digits and spaces in both inners; cout << "\n" after both finish.
Complexity:O(n²) time from 2n² characters; O(1) extra space.
One line: for each row i, print fixed-width 1..i then spaces, then spaces then i..1, then newline.
Frequently Asked Questions
A spaced mirror: for rows=5 you get 1 1 / 12 21 / 123 321 / 1234 4321 / 1234554321 — left 1..i and right i..1 with alignment spaces.
Spaces keep the left and right halves aligned so the pattern looks symmetric. Without them, the right half shifts left each row.
The right loop runs k from rows down to 1. When k > i it prints a space; otherwise it prints k — building i..1 on the right.
Both inner loops always run rows times. Extra positions are filled with spaces so columns stay aligned.
Program 27 prints a tight palindrome with no alignment spaces. Program 29 uses fixed-width loops and spaces for a symmetric spaced shape.
When i equals rows, both halves fill all columns — 12345 on the left and 54321 on the right meet with no space between.
After prompting, use if (!(cin >> rows) || rows <= 0) to reject bad input before printing.
O(n²) for n rows because each row runs two inner loops of width n.
🤔
Did you know?
This pattern prints an increasing left half (1..i), then a mirrored right half (i..1). Spaces in the fixed-width loops keep both halves aligned until the final row joins without a gap.
