An ascending number triangle grows one digit per row: row i prints 1 through i with no spaces between digits. With rows = 5 you get 1, 12, 123, 1234, 12345.
Remember
Rule: on row i, print j for j = 1..i
1
12
123
1234
12345 ← rows = 5
Unlike Program 4 (descending rows..i shrink), this shape counts the outer loop up so each row is longer than the last.
Approach
How to Solve It
Outer loop picks row i from 1 to rows. Inner loop prints digits 1..i.
Method
Idea
Best for
Tight digits
cout << j with no spaces
Classic textbook demos
Spaced digits
cout << j << " "
Easier reading when digits get wider
Pseudocode
Pseudocode
for i from 1 to rows:
for j from 1 to i:
print j (no newline)
print newline
Cheat sheet
Goal
Pattern
Outer rows
for (i = 1; i <= rows; i++)
Inner digits
for (j = 1; j <= i; j++)
Print digit
cout << j;
Spaced form
cout << j << " ";
End the row
cout << "\n";
Printing Numbers vs Starting a New Line
API
Effect
Use for
cout << j
Stays on the same line
Each digit
cout << "\n"
Ends the current line
After the inner loop
Print digits 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 ascending triangle 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 result5 rows · 15 digits
1
12
123
1234
12345
Trace
Worked Walkthrough
Trace row i = 4 — four digits glued together as 1234.
j
Print
Line so far
1
1
1
2
2
12
3
3
123
4
4
1234
After j finishes, cout << "\n" starts the next row. Row 5 will print one more digit — 12345.
Code
C++ Programs
Three complete programs: fixed rows = 5, cin input, and spaced digits. Use View Output to reveal sample results.
Example 1 — Fixed rows = 5
Hard-coded height — outer loop grows the row, inner loop prints 1..i.
C++
#include <iostream>
using namespace std;
int main()
{
int rows = 5;
int i, j;
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= i; j++)
cout << j;
cout << "\n";
}
return 0;
}
Output
1
12
123
1234
12345
How It Works
1. Outer loop grows the row.i runs from 1 to 5 — one more digit each line.
2. Inner loop prints 1..i.j runs from 1 to i; each digit uses cout << j with no space.
3. Newline after the row.cout << "\n" runs only after the inner loop finishes.
Example 2 — User Input Rows
Read the row count at runtime so the triangle grows or shrinks with input.
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 = 1; j <= i; j++)
cout << j;
cout << "\n";
}
return 0;
}
Output (when user enters 4)
Enter the number of rows: 4
1
12
123
1234
How It Works
1. Same nested-loop core. Only the source of rows changes from a literal to cin.
2. Entering 4 stops early. You get four rows ending at 1234.
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 — Spaced Digits
Same loops — add a space after each digit for easier reading.
C++
#include <iostream>
using namespace std;
int main()
{
int rows = 5;
int i, j;
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= i; j++)
cout << j << " ";
cout << "\n";
}
return 0;
}
Output
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
How It Works
1. Only the print changes.cout << j << " " instead of cout << j.
2. Loop bounds stay the same. Outer and inner limits match Example 1 exactly.
3. Useful for wider values. Spaces keep columns readable when digits become two characters.
Edge Cases & Pitfalls
Check these before calling the solution done.
Newline
Do not put "\n" inside the inner loop
That prints one digit per line. Call cout << "\n" only after the inner loop.
Outer direction
Count up, not down
Outer i = rows..1 with the same inner loop is Program 1 (shrinking). This pattern uses i = 1..rows.
Missing newline
Omitting the row break
Without cout << "\n", every digit glues onto one endless line.
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: on row i, print j for j = 1..i.
Grow up: outer i = 1..rows makes each row one digit longer.
Break the row: call cout << "\n" only after the inner loop.
Complexity:O(n²) time; O(1) extra space.
One line: for each row i from 1 to rows, print digits 1..i, then end the line.
Frequently Asked Questions
The outer loop runs i from 1 to rows. For each row i, the inner loop runs j from 1 to i and prints j. Row 1 prints 1, row 2 prints 12, and so on until row rows prints 1..rows.
Because the outer loop counts up and the inner bound equals i. When i increases, each row prints one more digit than the previous row.
Program 4 prints rows..i in descending order (54321, 5432, ...). Program 5 prints 1..i in ascending order — a growing triangle instead of a shrinking one.
Program 1 counts the outer loop down and prints 12345, 1234, ... Program 5 counts up and prints 1, 12, 123, ... — same inner loop, opposite outer direction.
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.
Use cout << j << " " instead of cout << j — see Example 3.
After cin >> rows, check failure and require rows >= 1 before the loops.
🤔
Did you know?
Row i prints digits 1 through i. The outer loop counts up from 1 to rows, so each row grows by one digit — total prints = n(n+1)/2, time O(n²).
