An alphabet right-angled triangle prints a left-aligned staircase of letters: row i runs from A through the i-th letter (A, AB, ABC, …).
Remember
Rule: on row i, print A through the i-th letter
A
AB
ABC
ABCD
ABCDE ← 5 rows
In C++ you solve it with two nested for loops over char: the outer loop picks the end letter, the inner loop restarts at A and prints through that end letter, then cout << "\n" moves to the next line. Once this clicks, inverted letter triangles and other alphabet shapes become much easier.
Approach
How to Solve It
Two ways to emit the same shape — start with nested char loops and cout, then optionally polish with substr.
Method
Idea
Best for
Nested loops + cout
Outer = end letter; inner = A..end via cout << j
Learning, interviews, exams
substr(0, i)
Slice an A–Z string for each row length
Short demos once loops click
Pseudocode
Pseudocode
for end from 'A' to lastLetter: // lastLetter depends on rows
for ch from 'A' to end:
print ch (no newline)
print newline
Print characters without a newline, then end the row once. (cout << endl also ends the line (and flushes); "\n" is enough for these demos.
Try it
Live Preview
Change the row count and the alphabet triangle updates instantly — capped at 26 letters (A–Z).
Whole numbers from 1 to 26. Tap a chip or type a value — the preview redraws as you go.
Live result5 rows · 15 letters
A
AB
ABC
ABCD
ABCDE
Trace
Worked Walkthrough — rows = 4
Trace each outer-loop end letter and count how many times the inner loop prints.
End i
Inner j
Printed row
Letters
'A'
A..A
A
1
'B'
A..B
AB
2
'C'
A..C
ABC
3
'D'
A..D
ABCD
4
Total letter prints: 1 + 2 + 3 + 4 = 10 = 4×5/2. That triangular sum is why time is O(n²).
Code
C++ Programs
Three complete programs: fixed end letter, cin input, and a substr shortcut. Use View Output to reveal sample results.
Example 1 — Fixed through 'E'
Hard-coded end letter — ideal for first demos and screenshots (5 rows).
C++
#include <iostream>
using namespace std;
int main() {
char i, j;
for (i = 'A'; i <= 'E'; i++) {
for (j = 'A'; j <= i; j++) {
cout << j;
}
cout << "\n";
}
return 0;
}
Output
A
AB
ABC
ABCD
ABCDE
How It Works
1. Outer loop picks the end letter.i runs from 'A' to 'E' — one row per end letter.
2. Inner loop restarts at A. For each i, j runs from 'A' to i, so the row is A..i.
3. Print letters, then break the line.cout << j stays on the row; cout << "\n" after the inner loop starts the next row.
When i = 'A' you get A; when i = 'B' you get AB; up through ABCDE.
Example 2 — User Input Version
Read the row count at runtime with cin. Always check failure in real apps (shown in the tip below).
C++
#include <iostream>
using namespace std;
int main() {
int rows;
char i, j, last;
cout << "Enter the number of rows: ";
cin >> rows;
last = (char)('A' + rows - 1);
for (i = 'A'; i <= last; i++) {
for (j = 'A'; j <= i; j++) {
cout << j;
}
cout << "\n";
}
return 0;
}
Output (when user enters 4)
Enter the number of rows: 4
A
AB
ABC
ABCD
How It Works
1. Prompt and read. Ask for a row count, then store it with cin >> rows.
2. Map rows to an end letter.last = (char)('A' + rows - 1) — for rows = 4, last is 'D'.
3. Same nested-loop core. Only the source of last changes — the print logic matches Example 1.
4. Safer input tip. Unchecked cin leaves rows unset on bad input. Prefer:
Safer input
if (!(cin >> rows) || rows < 1 || rows > 26) {
cout << "Enter a whole number from 1 to 26.\n";
return 1;
}
Example 3 — substr(0, i)
Slice an A–Z string for each row length — same shape without an explicit inner letter loop.
C++
#include <iostream>
#include <string>
using namespace std;
int main() {
int rows = 5;
string letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
for (int i = 1; i <= rows; i++) {
cout << letters.substr(0, i) << "\n";
}
return 0;
}
Output
A
AB
ABC
ABCD
ABCDE
How It Works
1. One alphabet string.letters holds A through Z once.
2. Slice by row length.letters.substr(0, i) returns the first i characters — exactly row i.
3. Print the whole row. One cout per row replaces the inner letter loop.
Learn the nested-loop version first (Examples 1–2) so you can explain both bounds in an interview; treat substr as a polish shortcut afterward.
Edge Cases & Pitfalls
Check these before calling the solution done.
\n inside
Column of letters
If cout << "\n" is inside the inner loop, each letter lands on its own line. Print letters without a newline; end the row only after the inner loop.
No restart
Letters keep advancing
The inner loop must start at 'A' every row. A single advancing counter across rows produces a different pattern.
No newline
One endless line
Omitting the row break glues every letter onto a single line.
rows > 26
Past Z
'A' + rows - 1 leaves A–Z when rows > 26. Clamp or reject in interactive programs.
rows = 1
Single A
Output is just A on one line — a good sanity check.
cin fail
Check the stream
If cin fails, rows may be unset — always test if (!(cin >> rows)) and prefer 1–26.
Analysis
Time and Space Complexity
Program
Time
Extra space
Nested loops + cout (Examples 1–2)
O(rows²)
O(1)
substr (Example 3)
O(rows²)
O(1) for the fixed alphabet string*
Total letters printed = 1 + 2 + … + n = n(n+1)/2, which is still quadratic in n. *Each substr builds a temporary of length i; overall work stays O(n²).
Remember
Key Takeaways
Rule: row with end letter i prints A through i.
Two loops: outer = end letter, inner = A..end with cout << j.
Break the row: call cout << "\n" only after the inner loop.
Complexity:O(n²) time from the triangular letter count; O(1) extra space for the loop form.
One line: for each end letter i, print A through i with cout, then cout << "\n".
Frequently Asked Questions
The outer loop picks the last letter on each row. The inner loop always restarts at A and prints up through that last letter, so row 1 is A, row 2 is AB, row 3 is ABC, and so on.
Each row is a fresh sequence from A to the current end letter. If you kept advancing a single char across rows, you would get a different pattern (not this right-angled alphabet triangle).
cout << j stays on the same line. cout << "\n" ends the current line. Letters use cout << j; the row break uses cout << "\n" after the inner loop.
Shrink the end letter each row — for example walk the outer loop from the last letter down toward A. See Alphabet Pattern Program 2.
O(n²) where n is the number of rows. Total letter prints equal 1+2+…+n = n(n+1)/2.
Yes. Keep a string of A–Z and print letters.substr(0, i) for row length i. Nested char loops are better for learning; substr is a handy shortcut later.
After cin >> rows, check failure: if (!(cin >> rows)) handle bad input. Clamp rows between 1 and 26 so you stay within A–Z.
Char math can produce characters beyond Z. Clamp to 26 for A–Z demos, or define a clear wrap/error policy.
🤔
Did you know?
Row i prints letters from A through the i-th letter (A, AB, ABC, …). Total letters for n rows is the triangular number n(n+1)/2 — the same count that makes this pattern O(n²).
