Problem Description

You are given n coins and need to build a staircase with them. The staircase follows a specific pattern where:

• Row 1 contains exactly 1 coin
• Row 2 contains exactly 2 coins
• Row 3 contains exactly 3 coins
• And so on...

Each row i must contain exactly i coins. You can only count a row as complete if it has the exact required number of coins. The last row might not have enough coins to be complete, in which case it doesn't count toward your answer.

Your task is to determine how many complete rows you can build with the given n coins.

For example:

• If n = 5, you can build 2 complete rows (row 1 uses 1 coin, row 2 uses 2 coins, leaving 2 coins which isn't enough to complete row 3 that needs 3 coins)
• If n = 8, you can build 3 complete rows (row 1 uses 1 coin, row 2 uses 2 coins, row 3 uses 3 coins, totaling 6 coins, leaving 2 coins which isn't enough for row 4)

The solution uses a mathematical formula derived from the fact that the sum of the first k natural numbers is k * (k + 1) / 2. By solving the inequality k * (k + 1) / 2 ≤ n for k, we get the formula k = int(sqrt(2) * sqrt(n + 0.125) - 0.5), which directly calculates the number of complete rows.

Intuition

The key insight is recognizing that this is a problem about finding the largest triangular number that doesn't exceed n. When we build the staircase, we're essentially using 1 + 2 + 3 + ... + k coins for k complete rows.

The sum of the first k natural numbers follows the formula k * (k + 1) / 2. So we need to find the largest k where k * (k + 1) / 2 ≤ n.

To solve this, we can rearrange the inequality:

• k * (k + 1) / 2 ≤ n
• k² + k ≤ 2n
• k² + k - 2n ≤ 0

This is a quadratic inequality. Using the quadratic formula where a = 1, b = 1, and c = -2n, we get:

• k = (-1 ± sqrt(1 + 8n)) / 2

Since k must be positive (we can't have negative rows), we take the positive root:

• k = (-1 + sqrt(1 + 8n)) / 2

This can be rewritten as:

• k = sqrt(2n + 0.25) - 0.5
• k = sqrt(2) * sqrt(n + 0.125) - 0.5

Since we need the number of complete rows, we take the floor of this value using int(). This mathematical formula directly gives us the answer in constant time without needing to iterate or use binary search.

The beauty of this approach is that it transforms what seems like a counting problem into a mathematical equation that can be solved instantly, regardless of how large n is.

Solution Approach

The implementation is remarkably concise, using a direct mathematical formula to solve the problem in constant time.

def arrangeCoins(self, n: int) -> int:
    return int(math.sqrt(2) * math.sqrt(n + 0.125) - 0.5)

Let's break down the implementation step by step:

1. Mathematical Formula Application: The formula sqrt(2) * sqrt(n + 0.125) - 0.5 is derived from solving the quadratic inequality k * (k + 1) / 2 ≤ n for k.

2. Breaking Down the Computation:

   • First, we add 0.125 (which is 1/8) to n. This comes from rewriting the discriminant 1 + 8n as 8(n + 1/8).
   • We take the square root of (n + 0.125).
   • We multiply by sqrt(2) to account for the factoring of the discriminant.
   • We subtract 0.5 to complete the quadratic formula transformation.

3. Floor Operation: The int() function effectively performs a floor operation on the result, giving us the largest integer k that satisfies our constraint. This ensures we only count complete rows.

4. Time and Space Complexity:

   • Time Complexity: O(1) - We perform a constant number of mathematical operations regardless of input size.
   • Space Complexity: O(1) - We only use a fixed amount of extra space for the calculation.

The elegance of this solution lies in avoiding iterative approaches or binary search. Instead of checking row by row or searching for the answer, we directly compute it using the mathematical relationship between the number of coins and the number of complete rows that can be formed.

Example Walkthrough

Let's walk through the solution with n = 11 coins.

Step 1: Understanding what we're looking for We need to find how many complete staircase rows we can build. Each row i needs exactly i coins:

• Row 1: 1 coin
• Row 2: 2 coins
• Row 3: 3 coins
• Row 4: 4 coins
• And so on...

Step 2: Apply the mathematical formula Using our formula: k = int(sqrt(2) * sqrt(n + 0.125) - 0.5)

Let's substitute n = 11:

• First, calculate n + 0.125 = 11 + 0.125 = 11.125
• Take the square root: sqrt(11.125) ≈ 3.335
• Multiply by sqrt(2): sqrt(2) * 3.335 ≈ 1.414 * 3.335 ≈ 4.716
• Subtract 0.5: 4.716 - 0.5 = 4.216
• Apply int() to get the floor: int(4.216) = 4

Step 3: Verify the result With 4 complete rows, we use:

• Row 1: 1 coin
• Row 2: 2 coins
• Row 3: 3 coins
• Row 4: 4 coins
• Total: 1 + 2 + 3 + 4 = 10 coins

We've used 10 coins out of 11, leaving 1 coin. Row 5 would need 5 coins, but we only have 1 remaining, so we can't complete it.

Therefore, we can build 4 complete rows with 11 coins.

Why the formula works: The formula solves the inequality k * (k + 1) / 2 ≤ n. For our example:

• 4 * 5 / 2 = 10 ≤ 11 ✓ (4 rows work)
• 5 * 6 / 2 = 15 > 11 ✗ (5 rows would need too many coins)

The mathematical formula directly finds this breaking point without any iteration.

Solution Implementation

Time and Space Complexity

Time Complexity: O(1)

The solution uses a mathematical formula to directly calculate the number of complete rows that can be formed with n coins. The formula int(math.sqrt(2) * math.sqrt(n + 0.125) - 0.5) is derived from solving the quadratic equation for the sum of an arithmetic sequence.

Given that 1 + 2 + 3 + ... + k = k(k+1)/2 ≤ n, we need to find the largest k such that k(k+1)/2 ≤ n. Solving the quadratic equation k² + k - 2n = 0 using the quadratic formula gives us k = (-1 + sqrt(1 + 8n)) / 2. The code implements this formula in a slightly rearranged form.

All operations involved (math.sqrt(), multiplication, addition, subtraction, and integer conversion) are constant time operations, making the overall time complexity O(1).

Space Complexity: O(1)

The solution only uses a constant amount of extra space to store intermediate calculation results and the final return value. No additional data structures or recursive calls are used, resulting in constant space complexity.

Common Pitfalls

1. Floating-Point Precision Issues

The mathematical formula involves square root operations and decimal arithmetic, which can lead to precision errors for very large values of n. When n approaches the maximum integer value, the intermediate calculations might lose precision, potentially causing the formula to return an incorrect result that's off by 1.

Example of the issue:

# For very large n, floating-point errors might accumulate
n = 2**31 - 1  # Maximum 32-bit integer
result = int(math.sqrt(2) * math.sqrt(n + 0.125) - 0.5)
# The result might be off by 1 due to precision loss

Solution: Add a verification step to ensure correctness:

def arrangeCoins(self, n: int) -> int:
    k = int(math.sqrt(2) * math.sqrt(n + 0.125) - 0.5)
  
    # Verify the result and adjust if necessary
    # Check if k is valid (sum of 1 to k should be <= n)
    if k * (k + 1) // 2 > n:
        return k - 1
    # Check if we can actually fit one more row
    elif (k + 1) * (k + 2) // 2 <= n:
        return k + 1
  
    return k

2. Integer Overflow in Alternative Implementations

If someone tries to verify or implement this using the direct sum formula k * (k + 1) / 2, they might encounter integer overflow when calculating k * (k + 1) for large values of k.

Example of the issue:

# This verification might overflow for large k
def verify_result(k, n):
    total = k * (k + 1) // 2  # This multiplication can overflow
    return total <= n

Solution: Use a comparison that avoids overflow:

def verify_result(k, n):
    # Rearrange to avoid overflow: k*(k+1)/2 <= n becomes k*(k+1) <= 2*n
    # But even better, subtract as we go
    remaining = n
    for i in range(1, k + 1):
        remaining -= i
        if remaining < 0:
            return False
    return True

3. Direct Formula Application Without Understanding

Blindly applying the formula without understanding its derivation can lead to mistakes when modifying the problem or debugging edge cases.

Solution: Always keep a simple iterative solution as a fallback or for testing:

def arrangeCoins_iterative(self, n: int) -> int:
    # Simple O(sqrt(n)) solution for verification
    k = 0
    while n >= k + 1:
        k += 1
        n -= k
    return k

This can be used to verify the mathematical solution for test cases and helps understand the problem intuitively.