Introduction

In this blog post, we will explore how to solve the LeetCode problem "1649" 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

Given an integer array instructions, you are asked to create a sorted array from the elements in instructions. You start with an empty container nums. For each element from left to right in instructions, insert it into nums. The cost of each insertion is the minimum of the following: The number of elements currently in nums that are strictly less than instructions[i]. The number of elements currently in nums that are strictly greater than instructions[i]. For example, if inserting element 3 into nums = [1,2,3,5], the cost of insertion is min(2, 1) (elements 1 and 2 are less than 3, element 5 is greater than 3) and nums will become [1,2,3,3,5]. Return the total cost to insert all elements from instructions into nums. Since the answer may be large, return it modulo 109 + 7 Example 1: Input: instructions = [1,5,6,2] Output: 1 Explanation: Begin with nums = []. Insert 1 with cost min(0, 0) = 0, now nums = [1]. Insert 5 with cost min(1, 0) = 0, now nums = [1,5]. Insert 6 with cost min(2, 0) = 0, now nums = [1,5,6]. Insert 2 with cost min(1, 2) = 1, now nums = [1,2,5,6]. The total cost is 0 + 0 + 0 + 1 = 1. Example 2: Input: instructions = [1,2,3,6,5,4] Output: 3 Explanation: Begin with nums = []. Insert 1 with cost min(0, 0) = 0, now nums = [1]. Insert 2 with cost min(1, 0) = 0, now nums = [1,2]. Insert 3 with cost min(2, 0) = 0, now nums = [1,2,3]. Insert 6 with cost min(3, 0) = 0, now nums = [1,2,3,6]. Insert 5 with cost min(3, 1) = 1, now nums = [1,2,3,5,6]. Insert 4 with cost min(3, 2) = 2, now nums = [1,2,3,4,5,6]. The total cost is 0 + 0 + 0 + 0 + 1 + 2 = 3. Example 3: Input: instructions = [1,3,3,3,2,4,2,1,2] Output: 4 Explanation: Begin with nums = []. Insert 1 with cost min(0, 0) = 0, now nums = [1]. Insert 3 with cost min(1, 0) = 0, now nums = [1,3]. Insert 3 with cost min(1, 0) = 0, now nums = [1,3,3]. Insert 3 with cost min(1, 0) = 0, now nums = [1,3,3,3]. Insert 2 with cost min(1, 3) = 1, now nums = [1,2,3,3,3]. Insert 4 with cost min(5, 0) = 0, now nums = [1,2,3,3,3,4]. ​​​​​​​Insert 2 with cost min(1, 4) = 1, now nums = [1,2,2,3,3,3,4]. ​​​​​​​Insert 1 with cost min(0, 6) = 0, now nums = [1,1,2,2,3,3,3,4]. ​​​​​​​Insert 2 with cost min(2, 4) = 2, now nums = [1,1,2,2,2,3,3,3,4]. The total cost is 0 + 0 + 0 + 0 + 1 + 0 + 1 + 0 + 2 = 4. Constraints: 1 <= instructions.length <= 105 1 <= instructions[i] <= 105

Explanation

Here's the solution to the problem, combining an explanation with efficient code:

• Core Idea: Use a data structure that allows efficient counting of elements less than and greater than a given value. A Binary Indexed Tree (BIT), also known as a Fenwick Tree, is ideal for this.
• BIT Operations: The BIT enables us to query the number of elements less than or equal to a value (prefix sum) and update element counts quickly. We calculate "less than" and "greater than" counts using these queries.

• Cost Calculation: For each instruction, calculate the minimum of "less than" and "greater than" counts and add it to the total cost.

• Runtime Complexity: O(N log M), where N is the length of instructions and M is the maximum value in instructions.

• Storage Complexity: O(M), where M is the maximum value in instructions.

Code

    def createSortedArray(instructions):
    """
    Calculates the minimum cost to insert elements from instructions into a sorted array.

    Args:
        instructions: A list of integers representing the instructions.

    Returns:
        The total cost modulo 10^9 + 7.
    """

    MOD = 10**9 + 7
    max_val = max(instructions)  # Determine the size of the BIT
    bit = [0] * (max_val + 1)  # Initialize the Binary Indexed Tree

    def update(index, value):
        """Updates the BIT with the given value at the given index."""
        index += 1  # BIT is 1-indexed
        while index <= max_val:
            bit[index] += value
            index += index & (-index)

    def query(index):
        """Queries the BIT to get the prefix sum up to the given index."""
        index += 1  # BIT is 1-indexed
        total = 0
        while index > 0:
            total += bit[index]
            index -= index & (-index)
        return total

    total_cost = 0
    for num in instructions:
        less = query(num - 1)
        greater = query(max_val) - query(num)  # Count elements > num
        total_cost = (total_cost + min(less, greater)) % MOD
        update(num, 1)  # Increment the count of num

    return total_cost