Arranging Coins Leetcode Problem 441 [Python Solution]

Today, we’re going to tackle the LeetCode problem Arranging Coins Don’t be fooled by the “Easy” label.

This problem offers a great opportunity to dive into a variety of solutions, some of which are quite challenging.

We’ll be exploring an efficient binary search solution that operates in O(log N) time complexity.

So, let’s get started.

Problem Overview

The problem goes like this: You have n coins, and your task is to arrange them in a staircase pattern.

Each row of the staircase contains an increasing number of coins, starting with 1 in the first row, 2 in the second row, and so on.

The last row may not be complete, but you want to know how many complete rows you can build with the given number of coins.

Here’s a visual example:

Example 1:

• Input: n = 5
• Output: 2
• Explanation: In this case, we can build two complete rows, with the third row left incomplete.

Example 2:

• Input: n = 8
• Output: 3
• Explanation: With 8 coins, we can build three complete rows, and the fourth row is incomplete.

The problem statement also provides a constraint: 1 <= n <= 2^31 - 1.

Now, let’s dive deeper into the problem.

Understanding the Constraints

Before we jump into the solutions, it’s essential to understand the constraints.

In this problem, the critical constraint is the range of n, which is between 1 and 2^31 – 1. This means we might be dealing with a large number of coins, and we need to find a solution that can handle such large inputs efficiently.

Arranging Coins LeetCode Problem Solution

1. Bruteforce Approach

Let’s begin by discussing a straightforward approach, which is essentially a brute force method.

This approach involves simulating the process of arranging coins row by row until we run out of coins.

Here’s how it works:

def arrangeCoins(n: int) -> int:
    rows = 0
    while n >= rows + 1:
        rows += 1
        n -= rows
    return rows

This solution keeps adding rows until we don’t have enough coins to complete the next row.

It’s intuitive but not the most efficient method, especially for larger values of n.

Its time complexity is O(n).

2. Efficient Binary Search Approach

Now, let’s explore a much more efficient solution using binary search.

We’ll use a binary search approach to find the maximum number of complete rows we can build within the given number of coins.

Here’s the Python code for this approach:

def arrangeCoins(n: int) -> int:
    left, right = 1, n
    result = 0

    while left <= right:
        mid = left + (right - left) // 2
        coins_needed = (mid * (mid + 1)) // 2

        if coins_needed == n:
            return mid
        elif coins_needed > n:
            right = mid - 1
        else:
            left = mid + 1
            result = max(mid, result)

    return result

This efficient binary search approach operates in O(log N) time complexity.

It finds the maximum number of complete rows by searching for the midpoint and adjusting the search boundaries based on whether we need more or fewer coins.

Time and Space Complexity

• The brute force approach has a time complexity of O(n), as it simulates arranging coins row by row.
• The efficient binary search approach has a time complexity of O(log N), making it significantly faster, especially for larger values of n.

Both solutions have a space complexity of O(1) as they use a constant amount of extra space.

Reasoning Behind Our Approach

To understand the binary search solution’s logic, we need to remember Gauss’s formula, which helps us calculate the sum of a series of consecutive numbers efficiently.

Gauss’s formula states that the sum of consecutive numbers from 1 to n is (n * (n + 1)) / 2.

In our problem, we’re trying to find the maximum number of rows (r) we can complete with a given number of coins.

We can use Gauss’s formula to calculate the number of coins needed to complete r rows, which is r * (r + 1) / 2.

Now, using binary search, we can determine the maximum number of rows within our given number of coins.

We set up a search space between 1 and n (the upper bound of rows), and we iteratively calculate the number of coins needed for the midpoint row.

If we have exactly enough coins, we return mid.

If we need more coins, we update the search range to the left half.

If we have more coins than needed, we update the result and continue the search in the right half.

This process continues until we’ve found the maximum number of rows.

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Conclusion

In this blog post, we’ve explored the LeetCode problem Arranging Coins We discussed two different solutions: a brute force approach that simulates the coin arrangement and an efficient binary search approach that leverages Gauss’s formula to calculate the maximum number of rows.

The binary search approach is highly efficient, with a time complexity of O(log N), making it suitable for large inputs.

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Happy coding!

Link to the problem on LeetCode

Now, it’s your turn to practice!

Try solving the Arranging Coins problem on LeetCode using the binary search approach we discussed in this post.

And as always, keep coding and learning.