Problem Description

You are given two integers, red and blue, representing the count of red and blue colored balls. These balls need to be arranged to form a triangle where:

• The 1st row has 1 ball, the 2nd row has 2 balls, the 3rd row has 3 balls, and so on.
• All balls in a particular row should be the same color.
• Adjacent rows should have different colors.

The goal is to determine the maximum height of such a triangle that can be constructed using the given balls.

Intuition

To solve this problem, we need to find the maximum number of rows that can be formed while adhering to the color constraints. The construction begins by selecting the color for the first row and then alternating between the colors for subsequent rows due to the adjacency constraint.

Here’s the step-by-step intuition behind the approach used in the solution:

1. Enumerate Starting Colors: We try both possibilities of starting the triangle with a red ball or a blue ball. This is because the result might differ based on the initial choice due to different counts of available balls.

2. Iterative Construction:

   • We initialize the row counter i starting at 1, indicating the number of balls required for the current row.
   • Iterate until we can no longer form a complete row with the given number of balls of the selected color.
   • Deduct the used balls for each row from the respective color count.

3. Maximize Height:

   • Switch rows to the opposite color after placing each row to ensure adjacent rows have different colors.
   • Keep track of the maximum height achieved during each attempt by using a variable ans to store the maximum of current and previous heights.

Thus, by simulating the triangle construction considering both possible starting colors, we can find the maximum possible height of the triangle.

Solution Approach

To achieve the desired outcome of finding the maximum height of the triangle, we use a simulation-based approach, as outlined in the solution:

1. Initialize Variables:

   • Start by setting ans to 0. This variable will store the highest triangle height found during the simulation process.

2. Simulate Two Starting Conditions:

   • Conduct two separate simulations, one starting with a red ball in the first row and the other starting with a blue ball. This helps determine which starting condition allows for a taller triangle.

3. Loop Through Potential Rows:

   • Use a loop that continues forming rows as long as possible, starting with i = 1 (for the first row).
   • Track the color of the next row using j, which alternates between 0 and 1, indicating which color is used for the current row.

4. Deduct Balls from the Count:

   • For each potential row, check if there are enough balls of the current color (c[j]) to form the row. If so, decrease the count of that color by the number of balls in the row (i).

5. Switch Colors:

   • Flip j using j ^= 1 to alternate between red and blue for the next row, ensuring that adjacent rows use different colors.

6. Update Maximum Height:

   • After forming each row, update ans with the current row number to record the maximum height achieved so far.

7. Return Maximum Height:

   • Once both simulations are complete, return ans, which holds the maximum height of the triangle that can be constructed with the given constraints.

Here is the reference solution code:

class Solution:
    def maxHeightOfTriangle(self, red: int, blue: int) -> int:
        ans = 0
        for k in range(2):
            c = [red, blue]
            i, j = 1, k
            while i <= c[j]:
                c[j] -= i
                j ^= 1
                ans = max(ans, i)
                i += 1
        return ans

This code effectively simulates the building process of the triangle for both starting conditions, ensuring that the solution accounts for all potential maximum heights based on different initial configurations.

Example Walkthrough

Let's walk through an example using the problem-solving approach to understand the maximum height of a triangle made of red and blue balls.

Example: Suppose you have 5 red balls and 4 blue balls.

1. Simulate Starting with Red Balls:

   • Row 1: Place 1 red ball (1 row with 1 ball requires 1 ball). Remaining: 4 red, 4 blue.
   • Row 2: Place 2 blue balls (2 row with 2 balls requires 2 balls). Remaining: 4 red, 2 blue.

   At this point, for row 3, we need 3 balls. Since there are only 2 blue balls left (and it is the red ball's turn), we cannot form the next row.

   Maximum height starting with red: 2

2. Simulate Starting with Blue Balls:

   • Row 1: Place 1 blue ball (1 row with 1 ball requires 1 ball). Remaining: 5 red, 3 blue.
   • Row 2: Place 2 red balls (2 row with 2 balls requires 2 balls). Remaining: 3 red, 3 blue.
   • Row 3: Place 3 blue balls (3 row with 3 balls requires 3 balls). Remaining: 3 red, 0 blue.

   Here, we reach row 3. For row 4, we would need 4 balls; since no blue balls are left, further rows cannot be formed.

   Maximum height starting with blue: 3

3. Conclusion:

   Comparing both starting scenarios, the maximum achievable height for this setup is 3 rows when starting with a blue ball. Therefore, the maximum height of the triangle is 3.

This walkthrough illustrates how switching starting conditions and alternating the use of colored balls in adjacent rows ensures adherence to constraints while maximizing the triangle's height.

Solution Implementation

Time and Space Complexity

The code attempts to find the maximum height of a triangle that can be constructed using red and blue balls. The loop iterates incrementally by i until the total number of balls of one color is exhausted. This involves adjusting the number of balls used at each level.

In each iteration, the number of balls added at a certain height i increases. This means that the number of iterations needed before failing to place more balls corresponds to the smallest triangular number that exceeds the sum of red and blue. Triangular numbers grow as O(\sqrt{n}), where n is the sum of red and blue. Hence, the time complexity is O(\sqrt{n}).

The algorithm only uses a constant amount of extra space for variables like ans, c, i, and j. Thus, the space complexity is O(1).

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