Unlike Program 5 (1..i ascending) or Program 6 (i..rows left-growing), this pattern counts the outer loop up and prints each row descending.
Approach
How to Solve It
Outer loop from 1 to rows. Inner loop from current i down to 1. Print each digit, then end the line.
Method
Idea
Best for
Nested loops
Outer up, inner i..1
Learning, interviews, demos
Compact trace
Same logic with rows = 3
Quick dry-runs on paper
Pseudocode
Pseudocode
for i from 1 to rows:
for j from i down to 1:
print j
print newline
Cheat sheet
Goal
Pattern
Outer (row grows)
for (i = 1; i <= rows; i++)
Inner (countdown)
for (j = i; j >= 1; j--)
Print digit
cout << j;
End the row
cout << "\n";
Digits on row i
i digits
Printing Numbers vs Starting a New Line
API
Effect
Use for
cout << j
Stays on the same line
Each digit
cout << "\n"
Ends the line
After the inner loop
Print digits without a newline, then end the row once.
Try it
Live Preview
Change the row count and the reverse row 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 digits
1
21
321
4321
54321
Trace
Worked Walkthrough
Trace three outer values when rows = 5 — watch each row count down to 1.
Outer i
Inner j
Prints
1
1..1
1
3
3..1
321
5
5..1
54321
Each new row adds one digit on the left while still ending at 1.
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 — outer up from 1, inner down from i to 1.
C++
#include <iostream>
using namespace std;
int main()
{
int rows = 5;
int i, j;
for (i = 1; i <= rows; i++)
{
for (j = i; j >= 1; j--)
cout << j;
cout << "\n";
}
return 0;
}
Output
1
21
321
4321
54321
How It Works
1. Outer grows the row.i runs from 1 to rows — that is both the row number and the first digit.
2. Inner counts down.j runs from i to 1, printing the reverse sequence for that row.
3. End the row. Call cout << "\n" only after the inner loop finishes.
Example 2 — User Input Rows
Read the row count at runtime with a simple validation tip.
C++
#include <iostream>
using namespace std;
int main()
{
int rows, i, j;
cout << "Enter the number of rows: ";
cin >> rows;
for (i = 1; i <= rows; i++)
{
for (j = i; j >= 1; j--)
cout << j;
cout << "\n";
}
return 0;
}
Output (when user enters 4)
Enter the number of rows: 4
1
21
321
4321
How It Works
1. Same loop core. Only the source of rows changes from a literal to cin.
2. Entering 4 stops early. You get four rows ending at 4321.
3. Validate in real apps. Prefer checking cin failure and requiring a positive height (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 a quick paper trace of both loop bounds.
C++
#include <iostream>
using namespace std;
int main()
{
int rows = 3;
int i, j;
for (i = 1; i <= rows; i++)
{
for (j = i; j >= 1; j--)
cout << j;
cout << "\n";
}
return 0;
}
Output
1
21
321
How It Works
1. Only three rows. Easy to dry-run every i and j value on paper.
2. Same formula. Nothing changes except rows — proving the pattern scales.
3. Countdown still holds. Row 2 prints 21 — start at 2, end at 1.
Edge Cases & Pitfalls
Check these before calling the solution done.
Inner direction
Count down with j--
Using j++ from 1 to i reprints Program 5 (1, 12, 123) instead of this pattern.
Newlines
Do not put "\n" inside the inner loop
That prints one digit per line and destroys the triangle.
Start value
Start the inner loop at i, not at rows
Starting at rows every time prints the same full countdown on every line.
Bad cin
Validate input
Check cin >> rows and require rows >= 1 before the loops.
Analysis
Time and Space Complexity
Program
Time
Extra space
Triangle (Examples 1–3)
O(n²)
O(1)
Digits printed are 1 + 2 + … + n = n(n + 1) / 2 — quadratic in n. Extra memory is only the loop variables.
Remember
Key Takeaways
Rule: outer i = 1..rows, inner j = i..1.
Growth: each row adds one digit on the left and still ends at 1.
Break the row: call cout << "\n" only after the inner loop finishes.
Complexity:O(n²) time, O(1) extra space.
One line: grow the row index, then count that index down to 1.
Frequently Asked Questions
A reverse row number triangle: row 1 shows 1, row 2 shows 21, row 3 shows 321, and so on until the last row shows digits from rows down to 1.
The inner loop starts at j = i and decrements to 1. That prints i, i-1, ..., 1 on each row.
When i = rows (5), the inner loop prints j from 5 down to 1 — giving 54321 on the last line.
Program 6 outer counts down and prints i..rows (5, 45, 345). Program 7 outer counts up and prints i down to 1 (1, 21, 321).
Program 5 prints 1..i ascending per row. Program 7 prints i..1 descending 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 the inner loop.
O(n²) for n rows because total prints are 1 + 2 + ... + n = n(n+1)/2.
One row prints a single digit 1.
🤔
Did you know?
Each row prints digits in reverse order — outer i runs 1..rows, inner j counts down from i to 1 — producing 1, 21, 321, and so on. Total prints grow as O(n²).
