Problem Description

You have an equilateral triangle with side length n that is divided into n² smaller unit equilateral triangles. The large triangle has n rows (indexed from 1), where row i contains 2i - 1 unit triangles. Each unit triangle is identified by coordinates (i, j) where i is the row number and j ranges from 1 to 2i - 1.

Two unit triangles are considered neighbors if they share a common side. For example:

• Triangles (1,1) and (2,2) are neighbors
• Triangles (3,2) and (3,3) are neighbors
• Triangles (2,2) and (3,3) are NOT neighbors (they don't share a side)

Initially, all unit triangles are white. You need to select k triangles and color them red, then run this algorithm:

1. Find a white triangle that has at least 2 red neighbors
   • If no such triangle exists, stop
2. Color that triangle red
3. Return to step 1

The goal is to find the minimum value of k and determine which k triangles to initially color red such that when the algorithm completes, all unit triangles become red.

Return a 2D list containing the coordinates [i, j] of the triangles to initially color red. If multiple valid solutions exist with the minimum k, return any one of them.

Intuition

To understand the pattern, let's think about how the coloring propagates. Once we color some triangles red initially, the algorithm will automatically color white triangles that have at least 2 red neighbors. This creates a chain reaction where red triangles spread throughout the entire triangle.

The key insight is that we need to strategically place the initial red triangles so that the propagation can reach every unit triangle. If we observe the structure, we notice that:

1. The top triangle (1,1) is isolated - it only has neighbors in row 2, so it must be colored initially since it can never have 2 red neighbors from other triangles.

2. For the remaining rows (from row n down to row 2), we need to ensure the red color can propagate both horizontally within each row and vertically between rows.

By experimenting with different patterns, we can discover that there's a repeating 4-row pattern that works efficiently:

• In every 4th row (counting from the bottom), we color alternating triangles: positions 1, 3, 5, ...
• In the next row up, we color just one triangle at position 2
• In the next row up, we color alternating triangles starting from position 3: positions 3, 5, 7, ...
• In the next row up, we color just one triangle at position 1

This pattern ensures that:

1. Each row gets enough "seed" triangles to trigger propagation
2. The colored triangles in adjacent rows are positioned to help propagate the color both within their own row and to neighboring rows
3. The pattern repeats every 4 rows, making it a systematic approach

The algorithm works by processing rows from bottom to top (row n to row 2), applying this 4-row coloring pattern cyclically. The variable k in the solution tracks which step of the 4-row cycle we're in (0, 1, 2, or 3).

Learn more about Math patterns.

Solution Approach

Based on the pattern discovered, we implement the solution by coloring triangles in a specific order:

1. Always color the top triangle: Start by adding [1, 1] to our answer list since this triangle must be colored initially (it can never get 2 red neighbors from other triangles).

2. Process rows from bottom to top: We iterate from row n down to row 2, applying a 4-row repeating pattern. We use a variable k to track which step of the pattern we're in (cycles through 0, 1, 2, 3).

3. Apply the 4-row coloring pattern:

   • When k = 0: Color all odd-positioned triangles in the row. For row i, color triangles at positions (i, 1), (i, 3), (i, 5), ..., up to (i, 2i-1). This is implemented with for j in range(1, i << 1, 2).

   • When k = 1: Color only one triangle at position 2. Add [i, 2] to the answer.

   • When k = 2: Color odd-positioned triangles starting from position 3. For row i, color triangles at positions (i, 3), (i, 5), ..., up to (i, 2i-1). This is implemented with for j in range(3, i << 1, 2).

   • When k = 3: Color only one triangle at position 1. Add [i, 1] to the answer.

4. Update the pattern counter: After processing each row, increment k and take modulo 4 to cycle through the pattern: k = (k + 1) % 4.

The expression i << 1 is a bit shift operation equivalent to i * 2, giving us 2i which is one more than the maximum column index 2i - 1 for row i.

This systematic approach ensures that:

• Every unit triangle either gets colored initially or can be reached by the propagation algorithm
• We use the minimum number of initially colored triangles
• The red color spreads efficiently through the triangle structure

The final answer is the list of coordinates [row, column] of all triangles that need to be colored red initially.

Example Walkthrough

Let's walk through a small example with n = 5 to illustrate the solution approach.

Step 1: Visualize the triangle structure

Row 1:        (1,1)
Row 2:     (2,1) (2,2) (2,3)
Row 3:   (3,1) (3,2) (3,3) (3,4) (3,5)
Row 4: (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (4,7)
Row 5: (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (5,7) (5,8) (5,9)

Step 2: Apply the coloring pattern

We start with k = 0 and process from row 5 down to row 2:

• Always color (1,1): Add [1, 1] to our answer

• Row 5 (k = 0): Color odd positions 1, 3, 5, 7, 9

  • Add [5, 1], [5, 3], [5, 5], [5, 7], [5, 9]
  • Update: k = (0 + 1) % 4 = 1

• Row 4 (k = 1): Color only position 2

  • Add [4, 2]
  • Update: k = (1 + 1) % 4 = 2

• Row 3 (k = 2): Color odd positions starting from 3 → positions 3, 5

  • Add [3, 3], [3, 5]
  • Update: k = (2 + 1) % 4 = 3

• Row 2 (k = 3): Color only position 1

  • Add [2, 1]
  • Update: k = (3 + 1) % 4 = 0

Step 3: Initial red triangles

Row 1:        [R]           (1,1) is red
Row 2:     [R] [ ] [ ]       (2,1) is red
Row 3:   [ ] [ ] [R] [ ] [R] (3,3) and (3,5) are red
Row 4: [ ] [R] [ ] [ ] [ ] [ ] [ ]  (4,2) is red
Row 5: [R] [ ] [R] [ ] [R] [ ] [R] [ ] [R]  (5,1), (5,3), (5,5), (5,7), (5,9) are red

Step 4: Verify propagation works

Now let's verify a few propagation steps:

• (5,2) has red neighbors (5,1) and (5,3) → becomes red
• (5,4) has red neighbors (5,3) and (5,5) → becomes red
• (4,1) has red neighbors (5,1) and (5,2) → becomes red
• (3,1) has red neighbors (4,1) and (4,2) → becomes red
• (2,2) has red neighbors (1,1) and (3,3) → becomes red

This chain reaction continues until all triangles are colored red.

Final answer: [[1,1], [5,1], [5,3], [5,5], [5,7], [5,9], [4,2], [3,3], [3,5], [2,1]]

Total initially colored triangles: k = 10

Solution Implementation

Time and Space Complexity

The time complexity is O(n²), where n is the parameter given in the problem.

The algorithm iterates through values from n down to 2 (which is n-1 iterations). In each iteration:

• When k = 0: The inner loop runs from 1 to 2*i with step 2, resulting in i iterations (since 2*i/2 = i)
• When k = 1: Only one append operation
• When k = 2: The inner loop runs from 3 to 2*i with step 2, resulting in approximately i-1 iterations
• When k = 3: Only one append operation

Since the pattern repeats every 4 iterations, roughly half of the outer loop iterations will have inner loops that execute O(i) times. The total number of operations is approximately:

• For k = 0 cases: n + (n-4) + (n-8) + ... ≈ n²/8
• For k = 2 cases: (n-2) + (n-6) + (n-10) + ... ≈ n²/8
• For k = 1 and k = 3 cases: O(n)

The sum gives us O(n²) time complexity.

The space complexity is O(1) when ignoring the space consumption of the answer array ans. The only extra variables used are k, i, and j, which consume constant space regardless of the input size n.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Understanding of Triangle Adjacency

Many developers mistakenly assume that triangles sharing a vertex are neighbors. In this problem, only triangles sharing a complete edge are neighbors. This misunderstanding leads to incorrect propagation logic and may result in some triangles never turning red.

Solution: Carefully verify neighbor relationships by checking if triangles share a full side, not just a vertex. Draw out small examples (n=2, n=3) to visualize the actual adjacency relationships.

2. Off-by-One Errors in Row/Column Indexing

The problem uses 1-based indexing where row i contains columns from 1 to 2i-1. Common mistakes include:

• Using 0-based indexing accidentally
• Iterating up to 2i instead of 2i-1
• Confusing the range boundaries when generating odd-positioned triangles

Solution: Be consistent with 1-based indexing throughout. When iterating odd positions, use range(1, 2*i, 2) which correctly generates 1, 3, 5, ..., up to but not including 2*i.

3. Incorrect Pattern Cycle Management

The 4-row repeating pattern must be applied correctly from bottom to top. Common errors:

• Starting the cycle at the wrong position (not starting with cycle_index = 0)
• Applying the pattern top-to-bottom instead of bottom-to-top
• Forgetting to update the cycle counter or updating it incorrectly

Solution: Initialize cycle_index = 0 before the loop, process rows from n down to 2, and update with cycle_index = (cycle_index + 1) % 4 after each row.

4. Missing the Top Triangle

The triangle at position [1, 1] must always be colored initially because it has no triangles above it to provide the required 2 red neighbors. Forgetting to include this leads to an incomplete solution.

Solution: Always start the result list with [[1, 1]] before processing other rows.

5. Using Bit Shift Without Understanding

The code uses row << 1 as an optimization for row * 2. If modified incorrectly (like row << 2 thinking it means row * 2), it produces wrong bounds.

Solution: Either stick with the clear multiplication row * 2 for readability, or understand that << 1 means shift left by 1 bit (multiply by 2). Don't modify the shift amount unless you understand binary operations.