An increasing number triangle from 1 prints i values on row i, each computed as i + j - 1 — so each row starts with its own row number.
Remember
Rule: for i from 1 to rows
for j from 1 to i
print (i + j - 1) and a space
1
2 3
3 4 5
4 5 6 7
5 6 7 8 9 ← rows = 5
Follows the triangle from 11 in Program 32; next is the variant from 0 in Program 34.
Approach
How to Solve It
Outer loop grows i. Inner loop prints i values with the formula i + j - 1, separated by spaces.
Method
Idea
Best for
Formula i + j - 1
When j = 1, value equals i — row starts at its index
Learning, interviews, exams
Compare offsets
Same shape as Program 32, but no base-9 shift
Seeing how formulas change the start
Pseudocode
Pseudocode
for i from 1 to rows:
for j from 1 to i:
print (i + j - 1) and a space
print newline
Cheat sheet
Goal
Pattern
Grow each row
for (i = 1; i <= rows; i++)
Print i values
for (j = 1; j <= i; j++)
Start at 1
cout << i + j - 1 << " ";
Row starts at i
When j = 1, i + j - 1 = i
End of row
cout << "\n";
Printing Numbers vs Starting a New Line
API
Effect
Use for
cout << value << " "
Stays on the same line
Each number on the row
cout << "\n"
Ends the current line
After the inner loop finishes
Print values without a newline, then end the row once.
Try it
Live Preview
Change the row count and the increasing triangle 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 · 15 numbers
1
2 3
3 4 5
4 5 6 7
5 6 7 8 9
Trace
Worked Walkthrough — rows = 5
Trace how each (i, j) pair maps to i + j - 1.
i
Values
Count
Prints
1
1
1
1
2
2, 3
2
2 3
3
3, 4, 5
3
3 4 5
4
4..7
4
4 5 6 7
5
5..9
5
5 6 7 8 9
Total numbers = 1 + 2 + … + n = n(n+1)/2 → O(n²).
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 — print i + j - 1 with a trailing space on each cell.
C++
#include <iostream>
using namespace std;
int main()
{
int i, j;
for (i = 1; i <= 5; i++)
{
for (j = 1; j <= i; j++)
cout << i + j - 1 << " ";
cout << "\n";
}
return 0;
}
Output
1
2 3
3 4 5
4 5 6 7
5 6 7 8 9
How It Works
1. Outer grows the width. Row i prints exactly i numbers.
2. Formula fills the cells.i + j - 1 makes row i start at i and count upward.
3. Newline once. Call cout << "\n" only after the inner loop finishes.
Example 2 — User Input Rows
Read rows with cin, validate, then use the same formula core.
C++
#include <iostream>
using namespace std;
int main()
{
int rows;
int i, j;
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 <= i; j++)
cout << i + j - 1 << " ";
cout << "\n";
}
return 0;
}
Output (when user enters 4)
Enter rows: 4
1
2 3
3 4 5
4 5 6 7
How It Works
1. Prompt and validate. Reject failed reads and non-positive values before printing.
2. Same core.i + j - 1 matches Example 1 — 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 the formula on paper.
C++
#include <iostream>
using namespace std;
int main()
{
int rows = 3;
int i, j;
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= i; j++)
cout << i + j - 1 << " ";
cout << "\n";
}
return 0;
}
Output
1
2 3
3 4 5
How It Works
1. Three rows.i = 1 → 1; i = 2 → 2 3; i = 3 → 3 4 5.
2. Trace on paper. If you use i + j instead of i + j - 1, every value shifts up by one.
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.
i + j
Missing −1
Using i + j or 9 + i + j shifts every value — rows no longer start at the row number. Keep i + j - 1.
j <= rows
Rectangle instead of triangle
Inner bound j <= rows makes every row the same width. Use j <= i.
no space
Glued numbers
Without << " ", values run together (23). Always print a trailing space.
\n inside
Broken rows
If cout << "\n" sits inside the inner loop, you get one number per line. Call it only after the loop.
rows = 1
Single value
Output is just 1. A good sanity check for input validation.
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)
Row i prints i numbers. Total = 1 + 2 + … + n = n(n+1)/2 → O(n²) time. Only a few loop variables are needed.
Remember
Key Takeaways
Formula: each cell is i + j - 1 — row i starts at i.
Growing width: inner loop runs j = 1..i so row i has i numbers.
Break the row: values with spaces in the inner loop; cout << "\n" after it finishes.
Complexity:O(n²) time from n(n+1)/2 prints; O(1) extra space.
One line: for each row i, print i + j - 1 for j = 1..i with spaces, then newline.
Frequently Asked Questions
An increasing triangle from 1: for rows=5 you get 1 / 2 3 / 3 4 5 / 4 5 6 7 / 5 6 7 8 9 — each value is i + j - 1.
Because when j = 1, the expression i + j - 1 becomes i. Row 4 therefore starts with 4.
j increases by 1, so i + j - 1 increases by 1 as well — producing consecutive numbers on each row.
Program 32 uses 9 + i + j (starts at 11). Program 33 uses i + j - 1 (starts at 1).
Program 33 uses i + j - 1 with i starting at 1. Program 34 uses i + j with i starting at 0.
cout << i + j - 1 << " " keeps values separated on the same row. cout << "\n" ends the row.
After prompting, use if (!(cin >> rows) || rows <= 0) to reject bad input before printing.
O(n²) for n rows because total prints are 1 + 2 + … + n = n(n+1)/2.
🤔
Did you know?
Each printed value is computed as i + j - 1. Row i = 1 prints 1; row i = 4 prints 4, 5, 6, 7 — a left-shifted increasing triangle starting at 1.
