LeetCode 441: Arranging Coins Solution in Python – A Step-by-Step Guide

What Is LeetCode 441: Arranging Coins?

In LeetCode 441: Arranging Coins, you’re given an integer n representing the number of coins. Your task is to determine how many complete rows you can form in a staircase pattern, where the k-th row has exactly k coins. The total coins used for k rows is the sum of the first k natural numbers (1 + 2 + … + k), and you want the largest k where this sum is ≤ n. For example, with n = 5, you can form 2 rows (1 + 2 = 3), leaving 2 coins unused. It’s like building a pyramid, one row at a time, until you can’t fit another full layer.

Problem Statement

• Input: n (int)—number of coins.
• Output: int—number of complete rows.
• Rules:
• Row i has i coins.
• Total coins used for k rows = k * (k + 1) / 2.
• Find max k where sum ≤ n.

Constraints

• 0 <= n <= 2^31 - 1.

Examples to Get Us Started

• Input: n = 5
• Output: 2 (1 + 2 = 3 ≤ 5, 1 + 2 + 3 = 6 > 5).
• Input: n = 8
• Output: 3 (1 + 2 + 3 = 6 ≤ 8, 1 + 2 + 3 + 4 = 10 > 8).
• Input: n = 0
• Output: 0 (No coins, no rows).

Understanding the Problem: Building the Staircase

To solve LeetCode 441: Arranging Coins in Python, we need to find the largest integer k where the sum 1 + 2 + … + k (or k * (k + 1) / 2) is less than or equal to n. A naive approach—adding coins row by row—works but could be slow for large n. The sum is a triangular number, so we can use math or optimization. We’ll explore:

• Best Solution (Mathematical Formula with Binary Search): O(1) or O(log n) time, O(1) space—fast and smart.
• Alternative Solution (Iterative Subtraction): O(sqrt(n)) time, O(1) space—simple but less efficient.

Let’s dive into the best solution—it’s the golden key to this coin puzzle.

Best Solution: Mathematical Formula with Binary Search Fallback

Why This Is the Best Solution

The mathematical formula approach is the top choice because it can solve this in O(1) time using the quadratic equation for triangular numbers, with a binary search fallback in O(log n) for precision and overflow safety. It leverages the fact that k * (k + 1) / 2 = n can be solved directly, then adjusted. It’s like using a calculator to instantly figure out how many rows fit, with a double-check for big numbers—elegant and robust!

How It Works

Here’s the plan:

• Step 1: Use the formula for triangular numbers:
• k * (k + 1) / 2 ≈ n → k² + k ≈ 2n.
• Solve quadratic: k = (-1 ± √(1 + 8n)) / 2, take positive root.
• Floor the result to get the largest k where sum ≤ n.
• Step 2: For safety (overflow, precision):
• Use binary search to find k where k * (k + 1) / 2 ≤ n.
• Step 3: Return k.
• Why It Works:
• Math gives the exact boundary.
• Binary search ensures correctness for large n.

Step-by-Step Example

Example: n = 8

• Formula:
• 2n = 16, 1 + 8n = 65.
• k = (-1 + √65) / 2 ≈ 3.54, floor to 3.
• Check: 3 * 4 / 2 = 6 ≤ 8, 4 * 5 / 2 = 10 > 8.
• Result: 3.

Example: n = 5

• Formula:
• 2n = 10, 1 + 8n = 41.
• k = (-1 + √41) / 2 ≈ 2.70, floor to 2.
• Check: 2 * 3 / 2 = 3 ≤ 5, 3 * 4 / 2 = 6 > 5.
• Result: 2.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, blending formula and binary search for clarity:

Copy codeclass Solution:
    def arrangeCoins(self, n: int) -> int:
        # Mathematical formula (O(1))
        # k * (k + 1) / 2 <= n
        # k^2 + k <= 2n, solve quadratic
        k = int((-1 + (1 + 8 * n) ** 0.5) // 2)
        if k * (k + 1) // 2 <= n and (k + 1) * (k + 2) // 2 > n:
            return k

        # Binary search fallback (O(log n))
        left, right = 0, n
        while left <= right:
            mid = (left + right) // 2
            coins = mid * (mid + 1) // 2
            if coins == n:
                return mid
            elif coins < n:
                left = mid + 1
            else:
                right = mid - 1
        return right  # Largest k where coins <= n

# No TreeNode needed for this problem

• Line 5-7: Formula:
• Compute k, check if exact.
• Integer division ensures floor.
• Line 10-19: Binary Search:
• Search k where k * (k + 1) / 2 ≤ n.
• Adjust bounds based on sum.
• Return right as max valid k.
• Time Complexity: O(1) (formula) or O(log n) (binary search).
• Space Complexity: O(1)—just variables.

It’s like a coin-counting wizard with a safety net!

Alternative Solution: Iterative Subtraction

Why an Alternative Approach?

The iterative subtraction method builds rows one by one, subtracting coins until none remain—O(sqrt(n)) time, O(1) space. It’s intuitive, like stacking coins by hand, but slower for large n. Great for understanding the basics!

How It Works

• Step 1: Start with row = 1, total coins used = 0.
• Step 2: While coins remain:
• Add row’s coins, increment row count.
• Subtract from n.
• Step 3: Return rows - 1 (last row incomplete).

Step-by-Step Example

Example: n = 8

• Iterate:
• Row 1: 1 coin, n = 8 - 1 = 7, rows = 1.
• Row 2: 2 coins, n = 7 - 2 = 5, rows = 2.
• Row 3: 3 coins, n = 5 - 3 = 2, rows = 3.
• Row 4: 4 coins > 2, stop.
• Result: 3 rows.

Code for Iterative Approach

Copy codeclass Solution:
    def arrangeCoins(self, n: int) -> int:
        if n == 0:
            return 0

        rows = 0
        remaining = n

        while remaining >= rows + 1:
            rows += 1
            remaining -= rows

        return rows

• Time Complexity: O(sqrt(n))—runs k times, k ≈ √n.
• Space Complexity: O(1)—few variables.

It’s a hands-on coin stacker!

Comparing the Two Solutions

• Formula/Binary Search (Best):
• Pros: O(1) or O(log n), fast, scalable.
• Cons: Math or binary search logic.
• Iterative (Alternative):
• Pros: O(sqrt(n)), simple.
• Cons: Slower for large n.

Formula wins for speed.

Edge Cases and More Examples

• Input: n=0 → 0.
• Input: n=1 → 1.
• Input: n=3 → 2.

Both handle these perfectly.

Complexity Recap

• Formula: Time O(1), Space O(1).
• Binary Search: Time O(log n), Space O(1).
• Iterative: Time O(sqrt(n)), Space O(1).

Formula’s the champ.

Key Takeaways

• Formula: Math magic.
• Binary Search: Precision backup.
• Iterative: Step-by-step.
• Python Tip: Math rocks—see [Python Basics](/python/basics).

Final Thoughts: Stack Those Coins

LeetCode 441: Arranging Coins in Python is a fun math-meets-coding challenge. The formula with binary search is your express ticket to the answer, while iteration builds intuition. Want more mathy fun? Try LeetCode 69: Sqrt(x) or LeetCode 367: Valid Perfect Square. Ready to arrange some coins? Head to Solve LeetCode 441 on LeetCode and build that staircase today—your coding skills are stacking up!