Introduction

In this blog post, we will explore how to solve the LeetCode problem "441" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You have n coins and you want to build a staircase with these coins. The staircase consists of k rows where the ith row has exactly i coins. The last row of the staircase may be incomplete. Given the integer n, return the number of complete rows of the staircase you will build. Example 1: Input: n = 5 Output: 2 Explanation: Because the 3rd row is incomplete, we return 2. Example 2: Input: n = 8 Output: 3 Explanation: Because the 4th row is incomplete, we return 3. Constraints: 1 <= n <= 231 - 1

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• Approach:

  • Use binary search to find the largest k such that the sum of the first k natural numbers (k*(k+1)/2) is less than or equal to n.
  • The binary search efficiently narrows down the range of possible k values.
  • The mid value during binary search is checked against the condition, and the range is updated accordingly.

• Complexity:

  • Runtime Complexity: O(log n)
  • Storage Complexity: O(1)

Code

    def arrange_coins(n: int) -> int:
    """
    You have n coins and you want to build a staircase with these coins.
    The staircase consists of k rows where the ith row has exactly i coins.
    The last row of the staircase may be incomplete.
    Given the integer n, return the number of complete rows of the staircase you will build.

    Example 1:
    Input: n = 5
    Output: 2
    Explanation: Because the 3rd row is incomplete, we return 2.

    Example 2:
    Input: n = 8
    Output: 3
    Explanation: Because the 4th row is incomplete, we return 3.

    Constraints:
    1 <= n <= 231 - 1
    """
    left, right = 0, n
    while left <= right:
        mid = left + (right - left) // 2
        coins_needed = mid * (mid + 1) // 2
        if coins_needed == n:
            return mid
        elif coins_needed < n:
            left = mid + 1
        else:
            right = mid - 1
    return right