Pattern 3 - Number Triangle Pattern

Difficulty: Easy

Problem Statement

You are given an integer n. You need to recreate the pattern shown below for any value of N.

The pattern should be a right triangle of numbers where each row i contains exactly i numbers, starting from 1 and going up to i in sequence.

Examples

Example 1:
Input:  n = 4
Output:
1
12
123
1234

Example 2:
Input:  n = 5
Output:
1
12
123
1234
12345

Example 3:
Input:  n = 3
Output:
1
12
123

Example 4:
Input:  n = 1
Output:
1

Constraints

• 1 ≤ n ≤ 20
• Print the pattern in the function given to you.

1. Brute Force Approach

Algorithm / Intuition

Solution1: Nested Loop Approach

Intuition:

To create a number triangle pattern, we need to print n rows, where row i contains numbers from 1 to i. This forms a right triangle where each row builds upon the previous one by adding one more number. This can be achieved using nested loops - the outer loop handles the rows and the inner loop handles the numbers in each row.

Approach:

• Use an outer loop to iterate through rows (from 1 to n).
• Use an inner loop to iterate through numbers (from 1 to current row number i).
• In the inner loop, print the current number j.
• After completing each row (inner loop), print a newline to move to the next row.

DryRun:

Input: n = 4

Row 1: Print numbers 1 to 1: 1
Row 2: Print numbers 1 to 2: 12
Row 3: Print numbers 1 to 3: 123
Row 4: Print numbers 1 to 4: 1234

Final Output:
1
12
123
1234

Code.

Java

class Solution {
  public void pattern3(int n) {
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= i; j++) {
        System.out.print(j);
      }
      System.out.println();
    }
  }
}

JavaScript

class Solution {
  pattern3(n) {
    for (let i = 1; i <= n; i++) {
      for (let j = 1; j <= i; j++) {
        process.stdout.write(j.toString());
      }
      console.log();
    }
  }
}

Python

class Solution:
    def pattern3(self, n):
        for i in range(1,n+1):
            for j in range(1,i+1):
                print(j, end="")
            print();

Complexity Analysis

Time Complexity: O(n²)

We have nested loops where the outer loop runs n times and the inner loop runs 1, 2, 3, ..., n times respectively. Total iterations = 1 + 2 + 3 + ... + n = n(n+1)/2 = O(n²).

Space Complexity: O(1)

We only use a constant amount of extra space for loop variables. The output space is not counted in auxiliary space complexity.

---

Alternative Approaches

Using String Building

Java

class Solution {
    public void pattern3(int n) {
        for (int i = 1; i <= n; i++) {
            StringBuilder row = new StringBuilder();
            for (int j = 1; j <= i; j++) {
                row.append(j);
            }
            System.out.println(row.toString());
        }
    }
}

JavaScript

class Solution {
  pattern3(n) {
    for (let i = 1; i <= n; i++) {
      let row = "";
      for (let j = 1; j <= i; j++) {
        row += j;
      }
      console.log(row);
    }
  }
}

Python

class Solution:
    def pattern3(self, n):
        for i in range(1, n+1):
            row = ""
            for j in range(1, i+1):
                row += str(j)
            print(row)

---

Edge Cases to Consider

1. n = 1: Should print a single "1"
2. Small Values (n = 2, 3): Verify correct triangle formation
3. Larger Values: Ensure pattern maintains right triangle shape
4. Maximum Constraint Value: n = 20 should work efficiently

Pattern Analysis

Pattern Characteristics:

• Shape: Right Triangle
• Dimensions: n rows, with row i having i elements
• Fill: Sequential numbers (1, 2, 3, ...)
• Growth: Each row adds one more number than the previous row

Key Observations:

• Row i has exactly i numbers
• Each row starts from 1 and goes up to the row number
• Total numbers printed = 1 + 2 + 3 + ... + n = n(n+1)/2
• No spaces between numbers in each row

Mathematical Pattern

For any row i:

• Start: 1
• End: i
• Count: i numbers
• Sequence: 1, 2, 3, ..., i

Follow-up Questions

1. Reverse Number Triangle: How would you print numbers in descending order?
2. Different Starting Point: How to start from a number other than 1?
3. Left-Aligned Triangle: How to create a left-aligned version with spaces?
4. Alphabetical Version: How to replace numbers with letters (A, B, C...)?

Related Patterns

This pattern is fundamental for understanding triangular patterns:

• Pattern 1: Square of Stars
• Pattern 4: Repeated Number Triangle (each row prints the row number)
• Pattern 5: Reverse Number Triangle
• Pattern 6: Inverted Number Triangle

Summary

Approach	Time Complexity	Space Complexity	Pros	Cons
Nested Loops	O(n²)	O(1)	Simple, direct approach, space optimal	Two loops needed
String Builder	O(n²)	O(n)	More readable for complex formatting	Uses extra space for row data

Recommended Solution: Use the nested loop approach for optimal space complexity and clear understanding of the pattern logic. This approach directly builds the pattern character by character.

Tips for Number Pattern Problems

1. Identify the Sequence: Determine what numbers appear in each row
2. Find the Range: Understand the start and end values for each row
3. Loop Boundaries: Pay attention to whether loops start from 0 or 1
4. Number Formatting: Consider how numbers should be displayed (spacing, alignment)
5. Practice Variations: Try with different number sequences or ranges

Debugging Tips

1. Check Loop Bounds: Ensure i and j start and end at correct values
2. Verify Row Length: Each row i should have exactly i numbers
3. Number Sequence: Confirm each row prints 1, 2, 3, ..., i
4. Newline Placement: Ensure println() is called after each row completion

Pattern Variations to Practice

1. Pattern 3a: Same triangle but with spaces between numbers
2. Pattern 3b: Right-aligned number triangle
3. Pattern 3c: Number triangle with leading zeros (01, 012, 0123...)
4. Pattern 3d: Reverse number triangle (n, n-1, ..., 1 in first row)