Introduction

In this blog post, we will explore how to solve the LeetCode problem "2673" 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 n representing the number of nodes in a perfect binary tree consisting of nodes numbered from 1 to n. The root of the tree is node 1 and each node i in the tree has two children where the left child is the node 2 i and the right child is 2 i + 1. Each node in the tree also has a cost represented by a given 0-indexed integer array cost of size n where cost[i] is the cost of node i + 1. You are allowed to increment the cost of any node by 1 any number of times. Return the minimum number of increments you need to make the cost of paths from the root to each leaf node equal. Note: A perfect binary tree is a tree where each node, except the leaf nodes, has exactly 2 children. The cost of a path is the sum of costs of nodes in the path. Example 1: Input: n = 7, cost = [1,5,2,2,3,3,1] Output: 6 Explanation: We can do the following increments: - Increase the cost of node 4 one time. - Increase the cost of node 3 three times. - Increase the cost of node 7 two times. Each path from the root to a leaf will have a total cost of 9. The total increments we did is 1 + 3 + 2 = 6. It can be shown that this is the minimum answer we can achieve. Example 2: Input: n = 3, cost = [5,3,3] Output: 0 Explanation: The two paths already have equal total costs, so no increments are needed. Constraints: 3 <= n <= 105 n + 1 is a power of 2 cost.length == n 1 <= cost[i] <= 104

Explanation

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

• High-Level Approach:

  • The problem requires making all root-to-leaf paths have the same cost in a perfect binary tree. A bottom-up approach is most efficient.
  • Start from the last level (leaves) and work upwards, ensuring that the costs of paths through sibling nodes are equalized at each level.
  • Keep track of the necessary increments and propagate the maximum path cost upwards.

• Complexity:

  • Runtime Complexity: O(n) - We iterate through all nodes of the tree once.
  • Storage Complexity: O(1) - We modify the input array in place, using no significant extra space.

Code

    def minIncrements(n: int, cost: list[int]) -> int:
    """
    Calculates the minimum increments needed to make the cost of paths from the root to each leaf node equal in a perfect binary tree.

    Args:
        n (int): The number of nodes in the perfect binary tree.
        cost (list[int]): A 0-indexed integer array where cost[i] is the cost of node i + 1.

    Returns:
        int: The minimum number of increments needed.
    """
    increments = 0
    for i in range(n // 2 - 1, -1, -1):
        left_child_index = 2 * (i + 1) - 1
        right_child_index = 2 * (i + 1)

        increments += abs(cost[left_child_index] - cost[right_child_index])
        cost[i] += max(cost[left_child_index], cost[right_child_index])

    return increments