Introduction

In this blog post, we will explore how to solve the LeetCode problem "2659" 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 are given an integer array nums containing distinct numbers, and you can perform the following operations until the array is empty: If the first element has the smallest value, remove it Otherwise, put the first element at the end of the array. Return an integer denoting the number of operations it takes to make nums empty. Example 1: Input: nums = [3,4,-1] Output: 5 Operation Array 1 [4, -1, 3] 2 [-1, 3, 4] 3 [3, 4] 4 [4] 5 [] Example 2: Input: nums = [1,2,4,3] Output: 5 Operation Array 1 [2, 4, 3] 2 [4, 3] 3 [3, 4] 4 [4] 5 [] Example 3: Input: nums = [1,2,3] Output: 3 Operation Array 1 [2, 3] 2 [3] 3 [] Constraints: 1 <= nums.length <= 105 -109 <= nums[i] <= 109 All values in nums are distinct.

Explanation

Here's a breakdown of the optimal solution:

• Utilize Binary Indexed Tree (BIT) / Fenwick Tree: The core idea is to efficiently track the number of elements smaller than a given element that have already been removed. This helps determine the effective position of an element after several removals. A BIT allows for fast prefix sum queries and updates, crucial for this tracking.

• Rank the Elements: Since the actual values don't matter, but rather their relative order, we first rank the elements. This maps each number to its position in the sorted array. This conversion makes the BIT operations more manageable and intuitive.

• Simulate Operations with BIT: Iterate through the ranked array. For each element, calculate its effective position based on the number of smaller elements removed so far (using the BIT). Update the BIT after processing each element.

• Time & Space Complexity: O(N log N) time complexity, where N is the length of the input array. O(N) space complexity.

Code

    class BIT:
    def __init__(self, n):
        self.n = n
        self.tree = [0] * (n + 1)

    def update(self, i, val):
        i += 1  # 1-based indexing
        while i <= self.n:
            self.tree[i] += val
            i += i & (-i)

    def query(self, i):
        i += 1  # 1-based indexing
        res = 0
        while i > 0:
            res += self.tree[i]
            i -= i & (-i)
        return res

def make_array_empty(nums):
    n = len(nums)
    ranked = sorted((x, i) for i, x in enumerate(nums))
    pos = [0] * n
    for i in range(n):
        pos[ranked[i][1]] = i

    bit = BIT(n)
    ans = 0
    for i in range(n):
        current_pos = pos[i]
        removed_before = bit.query(current_pos)
        effective_pos = current_pos - removed_before

        ans += effective_pos + 1

        if effective_pos == 0:
            ans -=1
        bit.update(current_pos, 1)

    return ans

def make_array_empty(nums):
    n = len(nums)
    ranked = sorted((x, i) for i, x in enumerate(nums))
    pos = [0] * n
    for i in range(n):
        pos[ranked[i][1]] = i

    bit = BIT(n)
    ans = 0
    for i in range(n):
        current_pos = pos[i]
        removed_before = bit.query(current_pos)
        effective_pos = current_pos - removed_before

        ans += effective_pos + 1

        bit.update(current_pos, 1)

    return ans