A continuous decreasing-row triangle prints numbers from 1 onward without resetting, while each row is one number shorter. Row 1 has n values; the last row has one.
Remember
Rule: k = 1
for i from 1 to rows
for j from rows down to i
print k with %3d; k++
print newline
1 2 3 4 5
6 7 8 9
10 11 12
13 14
15 ← rows = 5
Same continuous k idea as Program 35, but left-aligned and shrinking. Follows the palindrome triangle in Program 37; next is Program 39.
Approach
How to Solve It
Start with fixed rows = 5, then generalize with scanf — optionally demo a smaller count for tracing.
Method
Idea
Best for
Counter + shrink
k++ inside j = rows..i
Learning, interviews, exams
User input rows
Same loops; read rows with scanf
Reusable demos and labs
Pseudocode
Pseudocode
k = 1
for i from 1 to rows:
for j from rows down to i:
print k in width 3; k++
print newline
Cheat sheet
Goal
Pattern
Walk each row
for (i = 1; i <= rows; i++)
Shrink the row
for (j = rows; j >= i; j--) printf("%3d", k++);
End the row
printf("\n");
Count on row i
rows - i + 1
Total numbers
n(n + 1) / 2
Printing Numbers vs Starting a New Line
API
Effect
Use for
printf("%3d", k++)
Stays on the same line
Each number in the row
printf("\n")
Ends the current line
After the inner loop
Print all numbers without a newline, then end the row once.
Try it
Live Preview
Change the row count and the decreasing-row continuous triangle updates instantly — each number uses a width-3 field like %3d.
Whole numbers from 1 to 9. Tap a chip or type a value — the preview redraws as you go.
Live result5 rows · 15 numbers
1 2 3 4 5
6 7 8 9
10 11 12
13 14
15
Trace
Worked Walkthrough — rows = 4
Trace each outer-loop value of i, how many times the inner loop runs, and which values k takes.
i
Inner count
Numbers from k
Printed row
1
4
1, 2, 3, 4
1 2 3 4
2
3
5, 6, 7
5 6 7
3
2
8, 9
8 9
4
1
10
10
Total numbers for n rows = n(n + 1)/2 (here 10). That triangular sum is why time is O(n²).
Code
C Programs
Three complete programs: fixed rows, scanf input, and a small demo. Use View Output to reveal sample results.
Example 1 — Fixed rows = 5
Hard-coded row count — continuous counter with shrinking inner loop and %3d.
C
#include <stdio.h>
int main(void)
{
int rows = 5;
int i, j, k = 1;
for (i = 1; i <= rows; i++)
{
for (j = rows; j >= i; j--)
printf("%3d", k++);
printf("\n");
}
return 0;
}
Output
1 2 3 4 5
6 7 8 9
10 11 12
13 14
15
How It Works
1. Start the counter.k = 1 is declared once before the loops so the sequence never resets.
2. Shrink each row. Inner loop j = rows..i prints rows - i + 1 numbers with %3d.
3. End the line.printf("\n") runs only after the inner loop finishes.
When i = 2, four numbers start at 6; when i = 5, only 15 prints.
Example 2 — User Input Version
Read rows with scanf and reject invalid input. Same counter and shrink core.
C
#include <stdio.h>
int main(void)
{
int rows;
int i, j, k = 1;
printf("Enter rows: ");
if (scanf("%d", &rows) != 1 || rows < 1)
return 0;
for (i = 1; i <= rows; i++)
{
for (j = rows; j >= i; j--)
printf("%3d", k++);
printf("\n");
}
return 0;
}
Output (when user enters 3)
Enter rows: 3
1 2 3
4 5
6
How It Works
1. Prompt and validate. Exit early if scanf fails or rows < 1.
2. Same core. Only the hard-coded 5 becomes the user value.
3. Safer input tip. Prefer an explicit message instead of a silent exit:
Safer input
if (scanf("%d", &rows) != 1 || rows < 1)
{
printf("Enter a positive whole number of rows.\n");
return 1;
}
Example 3 — Compact rows = 3
Same counter and shrinking inner loop with a smaller row count for quick paper tracing.
C
#include <stdio.h>
int main(void)
{
int rows = 3;
int i, j, k = 1;
for (i = 1; i <= rows; i++)
{
for (j = rows; j >= i; j--)
printf("%3d", k++);
printf("\n");
}
return 0;
}
Output
1 2 3
4 5
6
How It Works
1. Same structure. Only rows changes from 5 to 3 — k and the inner loop stay identical.
2. Trace on paper. For i = 2: inner runs twice → 4 then 5.
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.
reset k
Broken sequence
If you set k = 1 inside the outer loop, every row restarts at 1. Declare k once before both loops.
j = 1..i
Growing instead of shrinking
An ascending inner bound prints a widening triangle. Keep j = rows; j >= i; j--.
\n inside
Broken rows
If printf("\n") sits inside the inner loop, each number lands on its own line. Call the newline only after the row finishes.
no %3d
Misaligned columns
Plain printf("%d", k++) makes values like 10 crowd earlier columns. Keep %3d.
rows = 1
Smallest triangle
Output is a single width-3 1. A good sanity check for the counter and newline.
scanf
Check the return value
If scanf fails, rows may be uninitialized — always test scanf(...) == 1.
Analysis
Time and Space Complexity
Program
Time
Extra space
Fixed / input (Examples 1–2)
O(rows²)
O(1)
Small demo (Example 3)
O(rows²)
O(1)
Total numbers = n(n + 1)/2 (triangular). For rows = 5 that is 15 printed values.
Remember
Key Takeaways
Rule:k = 1 once; inner j = rows..i prints %3d with k++.
Row length: row i prints rows - i + 1 numbers — always one fewer than the row above.
Break the row: call printf("\n") only after the inner loop.
Complexity:O(n²) time from the triangular sum; O(1) extra space.
One line: for each i, run j = rows..i printing %3d with k++, then printf("\n").
Frequently Asked Questions
Continuous numbers starting from 1, but each row has one fewer number: 5 on row 1, then 4, 3, 2, and 1 — totaling 15 numbers for 5 rows.
Counter k starts at 1 before the loops and increments with k++ each time a number prints — it is never reset inside the outer loop.
Row i prints rows - i + 1 numbers — the inner loop runs from j = rows down to i.
The format specifier reserves 3 columns per number, keeping columns aligned when values become two digits.
Program 35 is right-aligned with leading spaces. Program 38 is left-aligned with decreasing row width and the same continuous counter.
Replace 5 with rows in the outer loop bound — see Example 2.
O(n²) for n rows because total prints are 1 + 2 + ... + n = n(n+1)/2.
Check scanf's return value and reject non-positive row counts so the outer loop has a valid range.
🤔
Did you know?
A counter k starts at 1 and increments every time a number prints. Row i prints rows - i + 1 numbers with %3d — total prints = n(n+1)/2.
