Top 5 Pattern Problems in C++: Crack Your Next Coding Interview

Problem #5: Number Pyramid

🔹 1. Problem Description

The Number Pyramid is a common C++ pattern problem where numbers are arranged in a pyramid-like structure. Unlike star patterns, this one focuses on number progression and formatting. It tests your understanding of:

• Nested loops
• Output alignment
• Arithmetic progression
• Conditional placement

✅ Objective:

Given an input n, print a pyramid of n rows using increasing numbers row-wise.

Example (n = 5):

    1

   1 2

  1 2 3

 1 2 3 4

1 2 3 4 5

Each row starts from 1 and ends at the row number.

🔹 2. Flowchart

Start

|

|-- Input n

|

|-- For i = 1 to n:

|     Print (n - i) spaces

|     For j = 1 to i:

|         Print j

|     Print newline

|

End

✅ Focus on two main tasks:

• Leading spaces for centering
• Incremental numbers per row

🔹 3. Algorithm

🔽 Step-by-Step:

1. Start
2. Read input n (number of rows)
3. Loop i from 1 to n:
• Print n - i spaces
• Print numbers from 1 to i
4. Print newline after each row
5. End

✅ Pseudocode:

Input: n

for i = 1 to n:

    print (n - i) spaces

    for j = 1 to i:

        print j

    print newline

🔹 4. C++ Code — Number Pyramid

#include <iostream>

using namespace std;

int main() {

    int n;

    cout << "Enter number of rows: ";

    cin >> n;

    for (int i = 1; i <= n; i++) {

        // Print leading spaces

        for (int space = 1; space <= n - i; space++)

            cout << " ";

        // Print numbers

        for (int j = 1; j <= i; j++)

            cout << j << " ";

        cout << endl;

    }

    return 0;

}

🔹 5. Output for n = 5

    1

   1 2

  1 2 3

 1 2 3 4

1 2 3 4 5

✅ Numbers increase per row, aligned in a centered pyramid.

🔹 6. Output Breakdown Table

🔹 7. Alternate Variations

🔁 Reverse Numbers:

    1

   2 1

  3 2 1

 4 3 2 1

5 4 3 2 1

Replace cout << j with cout << i - j + 1.

🔁 Continuous Number Pyramid:

    1

   2 3

  4 5 6

 7 8 9 10

11 12 13 14 15

Add a counter:

int num = 1;

for (int i = 1; i <= n; i++) {

    for (int space = 1; space <= n - i; space++)

        cout << " ";

    for (int j = 1; j <= i; j++)

        cout << num++ << " ";

    cout << endl;

}

🔹 8. Complexity & Considerations

✅ This pattern builds logic around triangular formatting and number progression.

🔹 9. Common Pitfalls