LeetCode #3200 — EASY

Maximum Height of a Triangle

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given two integers red and blue representing the count of red and blue colored balls. You have to arrange these balls to form a triangle such that the 1^st row will have 1 ball, the 2^nd row will have 2 balls, the 3^rd row will have 3 balls, and so on.

All the balls in a particular row should be the same color, and adjacent rows should have different colors.

Return the maximum height of the triangle that can be achieved.

Example 1:

Input: red = 2, blue = 4

Output: 3

Explanation:

The only possible arrangement is shown above.

Example 2:

Input: red = 2, blue = 1

Output: 2

Explanation:

The only possible arrangement is shown above.

Example 3:

Input: red = 1, blue = 1

Output: 1

Example 4:

Input: red = 10, blue = 1

Output: 2

Explanation:

The only possible arrangement is shown above.

Constraints:

1 <= red, blue <= 100

Step 01

Brute Force Baseline

Problem summary: You are given two integers red and blue representing the count of red and blue colored balls. You have to arrange these balls to form a triangle such that the 1st row will have 1 ball, the 2nd row will have 2 balls, the 3rd row will have 3 balls, and so on. All the balls in a particular row should be the same color, and adjacent rows should have different colors. Return the maximum height of the triangle that can be achieved.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

2
4

Example 2

2
1

Example 3

1
1

Step 02

Core Insight

What unlocks the optimal approach

Count the max height using both possibilities. That is, red ball as top and blue ball as top.
For counting the max height, use a simple for loop and remove the number of balls required at this level.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3200: Maximum Height of a Triangle
class Solution {
    public int maxHeightOfTriangle(int red, int blue) {
        int ans = 0;
        for (int k = 0; k < 2; ++k) {
            int[] c = {red, blue};
            for (int i = 1, j = k; i <= c[j]; j ^= 1, ++i) {
                c[j] -= i;
                ans = Math.max(ans, i);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3200: Maximum Height of a Triangle
func maxHeightOfTriangle(red int, blue int) (ans int) {
	for k := 0; k < 2; k++ {
		c := [2]int{red, blue}
		for i, j := 1, k; i <= c[j]; i, j = i+1, j^1 {
			c[j] -= i
			ans = max(ans, i)
		}
	}
	return
}

# Accepted solution for LeetCode #3200: Maximum Height of a Triangle
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

// Accepted solution for LeetCode #3200: Maximum Height of a Triangle
/**
 * [3200] Maximum Height of a Triangle
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn max_height_of_triangle(red: i32, blue: i32) -> i32 {
        let mut result = 0;
        let mut row = 1;
        let mut switch = true;
        let (mut first_count, mut second_count) = (0, 0);

        loop {
            if switch {
                first_count += row;
            } else {
                second_count += row;
            }

            if first_count > red || second_count > blue {
                if second_count > red || first_count > blue {
                    break;
                }
            }

            switch = !switch;
            row += 1;
            result += 1;
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3200() {
        assert_eq!(3, Solution::max_height_of_triangle(2, 4));
        assert_eq!(2, Solution::max_height_of_triangle(2, 1));
        assert_eq!(1, Solution::max_height_of_triangle(1, 1));
        assert_eq!(2, Solution::max_height_of_triangle(10, 1));
        assert_eq!(5, Solution::max_height_of_triangle(10, 10));
    }
}

// Accepted solution for LeetCode #3200: Maximum Height of a Triangle
function maxHeightOfTriangle(red: number, blue: number): number {
    let ans = 0;
    for (let k = 0; k < 2; ++k) {
        const c: [number, number] = [red, blue];
        for (let i = 1, j = k; i <= c[j]; ++i, j ^= 1) {
            c[j] -= i;
            ans = Math.max(ans, i);
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.