What is the time complexity of Minimum Number of Operations to Sort a Binary Tree by Level?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Minimum Number of Operations to Sort a Binary Tree by Level?

This problem uses Tree, Breadth-First Search, Binary Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Number of Operations to Sort a Binary Tree by Level?

Insight 1: Level-order traversal is necessary to process nodes at each level independently. Insight 2: The core problem boils down to finding the minimum number of swaps to sort an array, which can be solved by cycle decomposition.

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Minimum Number of Operations to Sort a Binary Tree by Level - LeetCode 2471 Solution

Q: Which companies ask LeetCode 2471. Minimum Number of Operations to Sort a Binary Tree by Level in interviews?

This problem is frequently asked at Guidewire.

Minimum Number of Operations to Sort a Binary Tree by Level - Complete Solution Guide

Minimum Number of Operations to Sort a Binary Tree by Level is LeetCode problem 2471, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 2 nd level becomes [3,4]. -

Detailed Explanation

The problem requires you to find the minimum number of swaps needed to sort each level of a binary tree. The input is the root of a binary tree where all node values are unique. For each level of the tree, you can swap the values of any two nodes. The goal is to sort the values at each level in strictly increasing order and return the total number of swaps required across all levels.

Solution Approach

The provided solutions use a Breadth-First Search (BFS) to traverse the binary tree level by level. For each level, the values of the nodes are extracted into a list. Then, the `_min_swaps_to_sort` function calculates the minimum number of swaps required to sort this list. This is achieved by identifying cycles in the array's correct placement. The algorithm then sums up the swaps needed for each level to get the total minimum swaps for the entire tree.

Step-by-Step Algorithm

Step 1: Perform a level-order (BFS) traversal of the binary tree.
Step 2: For each level, extract the node values into a list.
Step 3: Implement the `_min_swaps_to_sort` function: Create a dictionary mapping values to their original indices. Also, create a sorted version of the array.
Step 4: In `_min_swaps_to_sort`, iterate through the array. If an element is not yet visited, start a cycle.
Step 5: Traverse the cycle using the dictionary to find the correct index for each element. Mark the visited elements.
Step 6: The number of swaps needed for a cycle is its length minus 1. The minimum number of swaps for the array is the total number of elements minus the number of cycles.
Step 7: Sum the minimum swaps for each level to obtain the final result.

Key Insights

Insight 1: Level-order traversal is necessary to process nodes at each level independently.
Insight 2: The core problem boils down to finding the minimum number of swaps to sort an array, which can be solved by cycle decomposition.
Insight 3: Representing the array as a graph where an edge exists from index i to j if the element at i should be at j allows identification of cycles.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Tree, Breadth-First Search, Binary Tree.

Companies

Asked at: Guidewire.

Minimum Number of Operations to Sort a Binary Tree by Level | LeetCode 2471

Medium

Tree Breadth-First Search Binary Tree

Time: O(n log n)Space: O(n)

Solution Code

Available in:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
import java.util.*;

class Solution {
    public int minimumOperations(TreeNode root) {
        if (root == null) {
            return 0;
        }

        int total_operations = 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while (!queue.isEmpty()) {
            int level_size = queue.size();
            List<Integer> level_values = new ArrayList<>();
            for (int i = 0; i < level_size; i++) {
                TreeNode node = queue.poll();
                level_values.add(node.val);
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }

            total_operations += _min_swaps_to_sort(level_values);
        }

        return total_operations;
    }

    private int _min_swaps_to_sort(List<Integer> arr) {
        int n = arr.size();
        if (n <= 1) {
            return 0;
        }

        Map<Integer, Integer> val_to_current_idx = new HashMap<>();
        for (int i = 0; i < n; i++) {
            val_to_current_idx.put(arr.get(i), i);
        }

        List<Integer> sorted_arr = new ArrayList<>(arr);
        Collections.sort(sorted_arr);

        boolean[] visited = new boolean[n];
        int num_cycles = 0;

        for (int i = 0; i < n; i++) {
            if (visited[i]) {
                continue;
            }

            num_cycles += 1;
            int j = i;
            while (!visited[j]) {
                visited[j] = true;
                int correct_val_for_j = sorted_arr.get(j);
                j = val_to_current_idx.get(correct_val_for_j);
            }
        }

        return n - num_cycles;
    }
}

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Problem Info

Description

Problem 2471: Minimum Number of Operations to Sort a Binary Tree by Level - Detailed explanation and solution approaches.

Topics

Tree Breadth-First Search Binary Tree

Companies

Guidewire

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2471

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Perform a level-order (BFS) traversal of the binary tree.
2
Step 2: For each level, extract the node values into a list.
3
Step 3: Implement the `_min_swaps_to_sort` function: Create a dictionary mapping values to their original indices. Also, create a sorted version of the array.
4
Step 4: In `_min_swaps_to_sort`, iterate through the array. If an element is not yet visited, start a cycle.
5
Step 5: Traverse the cycle using the dictionary to find the correct index for each element. Mark the visited elements.
6
Step 6: The number of swaps needed for a cycle is its length minus 1. The minimum number of swaps for the array is the total number of elements minus the number of cycles.
7
Step 7: Sum the minimum swaps for each level to obtain the final result.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The BFS visits each node once. The `_min_swaps_to_sort` function involves sorting, which takes O(k log k) time for a level of size k. Finding the cycles takes O(k) for each level. In the worst case, we may have to store all N nodes in the queue during BFS or in the level values list.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using Inversions Counting

Instead of cycle decomposition, you could use an inversion counting algorithm to determine the number of swaps needed. An inversion occurs when two elements are out of order. The minimum number of swaps is equal to the number of inversions.

Time: O(N log N), where N is the total number of nodes. Counting inversions typically requires O(n log n) time for an array of size n, and we do it for each level.Space: O(N), auxiliary space might be needed for merge sort based inversion counting

Trade-offs: Inversion counting can be slightly more complex to implement compared to cycle decomposition, but it might be more intuitive for some.

✨

Key Insights

1Insight 1: Level-order traversal is necessary to process nodes at each level independently.
2Insight 2: The core problem boils down to finding the minimum number of swaps to sort an array, which can be solved by cycle decomposition.
3Insight 3: Representing the array as a graph where an edge exists from index i to j if the element at i should be at j allows identification of cycles.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree (root is None). The solution handles this by returning 0 immediately.
Edge case 2: Tree with only one node. The solution handles this correctly as BFS processes only the root, and `_min_swaps_to_sort` returns 0 for an array of size 1.
Edge case 3: A level is already sorted. The _min_swaps_to_sort will correctly return 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the empty tree case. Forgetting to return 0 when the root is None will cause errors.

✗Common mistake 2: Failing to correctly calculate the number of cycles in the `_min_swaps_to_sort` function. Incorrect cycle detection will lead to an incorrect swap count.

✗Common mistake 3: Modifying the original array while sorting instead of creating a copy, which would corrupt the mapping from value to index.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing data arrangement in databases or file systems to minimize the number of data movements required.

🏦Real-world scenario 2: Task scheduling, where the goal is to minimize the number of task re-arrangements to meet dependencies or deadlines.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Most tree problems can be solved recursively. The base case is usually a null node. Think about what information needs to flow up vs down the tree.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Not handling null nodes properly. Always check if the current node exists before accessing its children.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 10, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Number of Operations to Sort a Binary Tree by Level"?

For this medium-level Tree problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?