What is the brute force solution for Minimum Number of Operations to Sort a Binary Tree by Level?

The brute force approach for Minimum Number of Operations to Sort a Binary Tree by Level is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves collecting the values at each level of the binary tree and then sorting them. We can then count how many swaps are needed to transform the current arrangement into the sorted arrangement. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Operations to Sort a Binary Tree by Level?

Minimum Number of Operations to Sort a Binary Tree by Level uses the following concepts: Tree, Breadth-First Search, Binary Tree. The recommended patterns to study are: Cycle Detection, Breadth-First Search.

What are the interview tips for Minimum Number of Operations to Sort a Binary Tree by Level?

Explain your thought process clearly during the traversal. Be prepared to discuss time and space complexities. Practice cycle detection problems to strengthen your understanding.

What are common mistakes when solving Minimum Number of Operations to Sort a Binary Tree by Level?

Not performing level-order traversal correctly. Failing to account for the correct number of swaps needed.

#2471

Minimum Number of Operations to Sort a Binary Tree by Level

Medium

TreeBreadth-First SearchBinary TreeCycle DetectionBreadth-First Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a level-order traversal to gather values at each level and then counts the minimum number of swaps needed to sort each level using a more efficient approach based on the concept of cycles in permutations.

⚙️

Algorithm

3 steps

1Step 1: Perform a level-order traversal (BFS) to collect values at each level of the binary tree.
2Step 2: For each level, create a sorted version of the level values and map the original indices to their sorted positions.
3Step 3: Count the number of swaps needed to sort the level using the cycle detection method.

solution.py42 lines

1from collections import deque
2
3def minOperations(root):
4    if not root:
5        return 0
6    levels = []
7    queue = deque([root])
8    while queue:
9        level_size = len(queue)
10        level = []
11        for _ in range(level_size):
12            node = queue.popleft()
13            level.append(node.val)
14            if node.left:
15                queue.append(node.left)
16            if node.right:
17                queue.append(node.right)
18        levels.append(level)
19
20    def count_swaps(arr):
21        sorted_arr = sorted(arr)
22        index_map = {value: idx for idx, value in enumerate(arr)}
23        visited = [False] * len(arr)
24        swaps = 0
25
26        for i in range(len(arr)):
27            if visited[i] or arr[i] == sorted_arr[i]:
28                continue
29            cycle_size = 0
30            j = i
31            while not visited[j]:
32                visited[j] = True
33                j = index_map[sorted_arr[j]]
34                cycle_size += 1
35            if cycle_size > 0:
36                swaps += (cycle_size - 1)
37        return swaps
38
39    total_swaps = 0
40    for level in levels:
41        total_swaps += count_swaps(level)
42    return total_swaps

ℹ

Complexity note: The time complexity is O(n) because we traverse each level once and count swaps using cycle detection, which is linear. The space complexity is O(n) due to the storage of level values and the index map.

1Each level can be sorted independently.
2Swaps can be minimized by using cycle detection in permutations.

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