Leetcode – Arranging Coins Solution in Python

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In this article, we are going to deep-dive into a popular problem on Leetcode named “Arranging Coins”. We will dissect different approaches to solving this problem using Python, and analyze the time and space complexity for each approach.

Table of Contents

  1. Problem Statement
  2. Approach 1: Using Binary Search
  3. Approach 2: Mathematical Solution
  4. Approach 3: Simple Iteration
  5. Time and Space Complexity Analysis
  6. Comparing the Approaches
  7. Conclusion

1. Problem Statement

You have a total of n coins that you want to form in a staircase shape, where every k-th row must have exactly k coins. Given n, find the total number of full rows of coins that can be formed.

Example:

Input: n = 5

Output: 2

Understanding the Problem

We need to form rows where the first row contains 1 coin, the second row contains 2 coins, and so on. We need to find out how many full rows can be formed with the given n coins.

2. Approach 1: Using Binary Search

Given that the total number of coins required to form k rows is k * (k + 1) / 2, we can use binary search to find k.

def arrangeCoins(n):
    left, right = 1, n
    while left <= right:
        mid = (left + right) // 2
        total = mid * (mid + 1) // 2
        
        if total == n:
            return mid
        elif total < n:
            left = mid + 1
        else:
            right = mid - 1
            
    return right

3. Approach 2: Mathematical Solution

Using the formula for the sum of the first k natural numbers, k * (k + 1) / 2, we can solve for k.

import math

def arrangeCoins(n):
    return int((math.sqrt(8 * n + 1) - 1) // 2)

Here, we solve the quadratic equation k * (k + 1) / 2 = n for k and take the integer part of the result.

4. Approach 3: Simple Iteration

Another approach is to keep subtracting the number of coins required at each level until we don’t have enough coins for the next level.

def arrangeCoins(n):
    k = 0
    while n > 0:
        k += 1
        n -= k
        
    return k if n == 0 else k - 1

5. Time and Space Complexity Analysis

  • Approach 1: Binary search has a time complexity of O(log n) and a space complexity of O(1) since it doesn’t use any additional data structures.
  • Approach 2: The mathematical solution has a constant time complexity of O(1) as it involves simple arithmetic operations to find the value of k.
  • Approach 3: The simple iteration approach has a time complexity of O(sqrt(n)) as the number of iterations is equal to the result, and a space complexity of O(1).

6. Comparing the Approaches

  • The binary search approach is efficient and only slightly more complex than simple iteration.
  • The mathematical solution is the most efficient in terms of time complexity but might not be immediately obvious.
  • The simple iteration approach is easy to understand but not the most efficient.

In practical applications, the choice between these approaches may depend on the size of n and the optimization requirements.

7. Conclusion

We’ve explored three different approaches to solving the “Arranging Coins” problem on Leetcode using Python. Understanding the different approaches and their time and space complexity helps to build a solid foundation in algorithmic problem solving. The binary search and mathematical solutions are particularly useful as they showcase efficient ways to solve problems which at first might seem to require iterative solutions. Being familiar with mathematical formulae and efficient algorithms is often key to optimizing solutions for programming challenges.

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