What is the time complexity of Make Costs of Paths Equal in a Binary Tree?

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

What data structures are used to solve Make Costs of Paths Equal in a Binary Tree?

This problem uses Array, Dynamic Programming, Greedy, Tree, Binary Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Make Costs of Paths Equal in a Binary Tree?

Insight 1: The optimal solution can be found by working backwards from the leaves towards the root. This is because the increments made at a node affect all paths originating from that node. Insight 2: At each internal node, the cost of its left and right subtrees must be equalized. The difference in costs represents the number of increments needed. This increment value will be propagated upwards.

Which companies ask LeetCode 2673. Make Costs of Paths Equal in a Binary Tree in interviews?

This problem is frequently asked at DE Shaw.

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Make Costs of Paths Equal in a Binary Tree - LeetCode 2673 Solution

Make Costs of Paths Equal in a Binary Tree - Complete Solution Guide

Make Costs of Paths Equal in a Binary Tree is LeetCode problem 2673, 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 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. Retu

Detailed Explanation

The problem asks us to minimize the number of increments needed to make the costs of all root-to-leaf paths in a perfect binary tree equal. The tree is represented implicitly by an array `cost`, where `cost[i]` is the cost of node `i+1`. We can increment the cost of any node by 1 as many times as we want. A perfect binary tree means that every non-leaf node has exactly two children, and all leaves are at the same level. We need to find the minimum total increments across all nodes to ensure that the sum of node costs along any path from the root to a leaf is the same for all paths.

Solution Approach

The provided solution uses a bottom-up approach. Starting from the last non-leaf node, it compares the costs of its two children. The absolute difference between these costs is added to the total increments. The cost of the current node is then increased by the maximum cost of its two children. This process is repeated until the root node is reached. The final value of `total_increments` represents the minimum number of increments needed.

Step-by-Step Algorithm

Step 1: Iterate from the last non-leaf node to the root. The index of the last non-leaf node is `n // 2 - 1`.
Step 2: For each non-leaf node at index `i`, calculate the indices of its left and right children: `left_child_idx = 2 * i + 1` and `right_child_idx = 2 * i + 2`.
Step 3: Calculate the absolute difference between the costs of the left and right children: `abs(cost[left_child_idx] - cost[right_child_idx])`. Add this difference to the `total_increments`.
Step 4: Update the cost of the current node by adding the maximum of the costs of its two children: `cost[i] += max(cost[left_child_idx], cost[right_child_idx])`.
Step 5: Repeat steps 2-4 until the root node (index 0) is reached.
Step 6: Return the `total_increments`.

Key Insights

Insight 1: The optimal solution can be found by working backwards from the leaves towards the root. This is because the increments made at a node affect all paths originating from that node.
Insight 2: At each internal node, the cost of its left and right subtrees must be equalized. The difference in costs represents the number of increments needed. This increment value will be propagated upwards.
Insight 3: The target cost for a given path is not known initially, but the equalization process ensures all paths become equal as we move towards the root.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming, Greedy, Tree, Binary Tree.

Companies

Asked at: DE Shaw.

Make Costs of Paths Equal in a Binary Tree | LeetCode 2673

Medium

Array Dynamic Programming Greedy Tree Binary Tree

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minIncrements(int n, int[] cost) {
        int total_increments = 0;
        for (int i = n / 2 - 1; i >= 0; i--) {
            int left_child_idx = 2 * i + 1;
            int right_child_idx = 2 * i + 2;

            int left_path_cost = cost[left_child_idx];
            int right_path_cost = cost[right_child_idx];

            total_increments += Math.abs(left_path_cost - right_path_cost);

            cost[i] += Math.max(left_path_cost, right_path_cost);
        }
        return total_increments;
    }
}

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Problem Info

Description

Problem 2673: Make Costs of Paths Equal in a Binary Tree - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Greedy Tree Binary Tree

Companies

DE Shaw

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2673

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

6 steps

1
Step 1: Iterate from the last non-leaf node to the root. The index of the last non-leaf node is `n // 2 - 1`.
2
Step 2: For each non-leaf node at index `i`, calculate the indices of its left and right children: `left_child_idx = 2 * i + 1` and `right_child_idx = 2 * i + 2`.
3
Step 3: Calculate the absolute difference between the costs of the left and right children: `abs(cost[left_child_idx] - cost[right_child_idx])`. Add this difference to the `total_increments`.
4
Step 4: Update the cost of the current node by adding the maximum of the costs of its two children: `cost[i] += max(cost[left_child_idx], cost[right_child_idx])`.
5
Step 5: Repeat steps 2-4 until the root node (index 0) is reached.
6
Step 6: Return the `total_increments`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates through the first half of the `cost` array once (from `n // 2 - 1` down to 0), performing a constant number of operations at each index. Therefore, the time complexity is O(n). The algorithm only uses a constant amount of extra space, regardless of the input size. The provided solutions modify the input array `cost` in-place, avoiding the creation of new data structures that scale with the input size.

📖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

Top-Down Approach (Recursion)

A recursive approach could traverse the tree from the root down to the leaves, keeping track of the current path cost. At each leaf, the maximum path cost encountered so far would be updated. Then, during the backtracking phase, each node's cost would be adjusted to match the maximum path cost encountered.

Time: O(n)Space: O(log n) due to the recursion stack

Trade-offs: The iterative bottom-up approach is generally preferred due to its constant space complexity compared to the logarithmic space complexity of the recursive top-down approach. The recursive approach might be more intuitive for some, but the iterative approach is more efficient in terms of memory usage.

✨

Key Insights

1Insight 1: The optimal solution can be found by working backwards from the leaves towards the root. This is because the increments made at a node affect all paths originating from that node.
2Insight 2: At each internal node, the cost of its left and right subtrees must be equalized. The difference in costs represents the number of increments needed. This increment value will be propagated upwards.
3Insight 3: The target cost for a given path is not known initially, but the equalization process ensures all paths become equal as we move towards the root.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Tree

Hierarchical structure with nodes. Like a family tree!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: n = 3, cost = [5, 3, 3]. The algorithm correctly calculates 0 increments, as the two leaf nodes already have equal costs.
Edge case 2: When all node costs are equal initially. The algorithm correctly determines that 0 increments are needed.
Edge case 3: Large cost values. The algorithm handles large cost values appropriately as the addition and subtraction operations are within the integer range.
Edge case 4: When n is small (e.g., n=3, n=7). The loop condition `i >= 0` and the calculations within the loop correctly handle the small values of n

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Starting the iteration from the beginning of the cost array (index 0) instead of the last non-leaf node (index `n // 2 - 1`). This leads to incorrect results as it violates the bottom-up approach.

✗Common mistake 2: Not taking the absolute difference between the costs of the children, which can lead to negative increments and incorrect calculations.

✗Common mistake 3: Forgetting to update the cost of the parent node after equalizing the costs of its children, which causes the path costs to be unequal further up the tree.

✅

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 resource allocation in a hierarchical organizational structure where the goal is to balance resource consumption across different departments or teams.

🏦Real-world scenario 2: Balancing network traffic load across different network paths in a tree-like network topology to ensure optimal performance and minimize congestion.

🎮Real-world scenario 3: Financial planning to ensure that different investment portfolios yield similar returns over time, minimizing discrepancies and maintaining a consistent financial outcome.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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: November 1, 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

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

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

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 "Make Costs of Paths Equal in a Binary Tree"?

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

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