Introduction

In this blog post, we will explore how to solve the LeetCode problem "2471" 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 the root of a binary tree with unique values. In one operation, you can choose any two nodes at the same level and swap their values. Return the minimum number of operations needed to make the values at each level sorted in a strictly increasing order. The level of a node is the number of edges along the path between it and the root node. Example 1: Input: root = [1,4,3,7,6,8,5,null,null,null,null,9,null,10] Output: 3 Explanation: - Swap 4 and 3. The 2nd level becomes [3,4]. - Swap 7 and 5. The 3rd level becomes [5,6,8,7]. - Swap 8 and 7. The 3rd level becomes [5,6,7,8]. We used 3 operations so return 3. It can be proven that 3 is the minimum number of operations needed. Example 2: Input: root = [1,3,2,7,6,5,4] Output: 3 Explanation: - Swap 3 and 2. The 2nd level becomes [2,3]. - Swap 7 and 4. The 3rd level becomes [4,6,5,7]. - Swap 6 and 5. The 3rd level becomes [4,5,6,7]. We used 3 operations so return 3. It can be proven that 3 is the minimum number of operations needed. Example 3: Input: root = [1,2,3,4,5,6] Output: 0 Explanation: Each level is already sorted in increasing order so return 0. Constraints: The number of nodes in the tree is in the range [1, 105]. 1 <= Node.val <= 105 All the values of the tree are unique.

Explanation

Here's a breakdown of the solution and the Python code:

• High-Level Approach:

  • Perform a level-order traversal of the binary tree to group nodes by their level.
  • For each level, calculate the minimum number of swaps needed to sort the node values using cycle decomposition.
  • Sum the swap counts across all levels to get the total minimum swaps.

• Complexity:

  • Runtime Complexity: O(N), where N is the number of nodes in the tree. This is because we visit each node once during level order traversal and the cycle decomposition process is linear in the size of each level.
  • Storage Complexity: O(W), where W is the maximum width of the tree. This is primarily due to the queue used in the level order traversal and the lists used to store node values at each level. In the worst case, W can be N (for a complete binary tree), but it can be much smaller for balanced trees.

Code

    from collections import deque

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def minimum_swaps(arr):
    """
    Calculates the minimum number of swaps to sort an array using cycle decomposition.
    """
    n = len(arr)
    arrpos = [*enumerate(arr)]
    arrpos.sort(key=lambda it: it[1])

    visited = [False] * n
    ans = 0
    for i in range(n):
        if visited[i] or arrpos[i][0] == i:
            continue

        cycle_size = 0
        j = i
        while not visited[j]:
            visited[j] = True
            j = arrpos[j][0]
            cycle_size += 1

        if cycle_size > 0:
            ans += (cycle_size - 1)
    return ans

def minimum_operations(root):
    """
    Calculates the minimum number of operations needed to make the values at each level sorted.
    """
    if not root:
        return 0

    queue = deque([root])
    total_swaps = 0

    while queue:
        level_size = len(queue)
        level_values = []

        for _ in range(level_size):
            node = queue.popleft()
            level_values.append(node.val)

            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)

        total_swaps += minimum_swaps(level_values)

    return total_swaps