What is the time complexity of Minimum Cost Tree From Leaf Values?

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

What data structures are used to solve Minimum Cost Tree From Leaf Values?

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

What are the key insights for solving Minimum Cost Tree From Leaf Values?

Insight 1: The optimal tree construction involves combining smaller leaf values first to minimize the product at higher levels of the tree. Insight 2: A stack-based approach is suitable for efficiently tracking and combining leaf values to minimize the overall sum of non-leaf nodes.

Which companies ask LeetCode 1130. Minimum Cost Tree From Leaf Values in interviews?

This problem is frequently asked at MathWorks, PhonePe.

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Minimum Cost Tree From Leaf Values - LeetCode 1130 Solution

Minimum Cost Tree From Leaf Values - Complete Solution Guide

Minimum Cost Tree From Leaf Values is LeetCode problem 1130, 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

Given an array arr of positive integers, consider all binary trees such that: Each node has either 0 or 2 children; The values of arr correspond to the values of each leaf in an in-order traversal of the tree. The value of each non-leaf node is equal to the product of the largest leaf value in its left and right subtree, respectively. Among all possible binary trees considered, return the smallest possible sum of the values of each non-leaf node . It is guaranteed this sum fits into a 32-bit int

Detailed Explanation

The problem asks us to find the minimum sum of non-leaf node values for all possible binary trees constructed from a given array of positive integers. These integers represent the leaf nodes of the binary tree in an in-order traversal. The key constraints are: each non-leaf node must have exactly two children, and the value of a non-leaf node is the product of the largest leaf values in its left and right subtrees. We need to return the minimum possible sum of all non-leaf node values across all possible binary trees.

Solution Approach

The solution utilizes a stack to maintain a monotonically increasing sequence of leaf values. As we iterate through the input array, if we encounter a value smaller than the top of the stack, we push it onto the stack. If we encounter a value greater than or equal to the top of the stack, we pop elements from the stack until the top of the stack is greater than the current value. Each time we pop an element (mid), we calculate the cost of merging it with its adjacent smaller element (either the new top of the stack or the current value). This cost is added to the overall result. Finally, after iterating through the array, there may be remaining elements in the stack. These elements represent the 'top' part of the tree and we continue to merge them together until only one element remains.

Step-by-Step Algorithm

Step 1: Initialize an empty stack and push a large sentinel value (infinity) onto it. This ensures the algorithm works correctly even if the input array is monotonically increasing.
Step 2: Iterate through the input array 'arr'.
Step 3: While the top of the stack is less than or equal to the current element 'num' from 'arr', pop the top element 'mid' from the stack.
Step 4: Calculate the cost of merging 'mid' which is 'mid * min(stack[-1], num)'. Add this cost to the 'res' (total sum). 'stack[-1]' is the new top of stack and 'num' is the current element.
Step 5: Push the current element 'num' onto the stack.
Step 6: After iterating through the array, if there are more than two elements in the stack, pop the top element and multiply it with the new top of the stack, adding the product to 'res'. Repeat until only two elements (sentinel value and last element) remain.
Step 7: Return the 'res' (total sum).

Key Insights

Insight 1: The optimal tree construction involves combining smaller leaf values first to minimize the product at higher levels of the tree.
Insight 2: A stack-based approach is suitable for efficiently tracking and combining leaf values to minimize the overall sum of non-leaf nodes.
Insight 3: The problem has a greedy nature, where merging the smallest adjacent values yields the optimal solution.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Stack, Greedy, Monotonic Stack.

Companies

Asked at: MathWorks, PhonePe.

Minimum Cost Tree From Leaf Values | LeetCode 1130

Medium

Array Dynamic Programming Stack Greedy Monotonic Stack

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

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public int mctFromLeafValues(int[] arr) {
        int res = 0;
        Stack<Integer> stack = new Stack<>();
        stack.push(Integer.MAX_VALUE);

        for (int num : arr) {
            while (stack.peek() <= num) {
                int mid = stack.pop();
                res += mid * Math.min(stack.peek(), num);
            }
            stack.push(num);
        }

        while (stack.size() > 2) {
            res += stack.pop() * stack.peek();
        }

        return res;
    }
}

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Problem Info

Description

Problem 1130: Minimum Cost Tree From Leaf Values - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Stack Greedy Monotonic Stack

Companies

MathWorksPhonePe

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1130

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: Initialize an empty stack and push a large sentinel value (infinity) onto it. This ensures the algorithm works correctly even if the input array is monotonically increasing.
2
Step 2: Iterate through the input array 'arr'.
3
Step 3: While the top of the stack is less than or equal to the current element 'num' from 'arr', pop the top element 'mid' from the stack.
4
Step 4: Calculate the cost of merging 'mid' which is 'mid * min(stack[-1], num)'. Add this cost to the 'res' (total sum). 'stack[-1]' is the new top of stack and 'num' is the current element.
5
Step 5: Push the current element 'num' onto the stack.
6
Step 6: After iterating through the array, if there are more than two elements in the stack, pop the top element and multiply it with the new top of the stack, adding the product to 'res'. Repeat until only two elements (sentinel value and last element) remain.
7
Step 7: Return the 'res' (total sum).

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because each element in the input array is pushed and popped from the stack at most once. The space complexity is O(n) because, in the worst-case scenario (e.g., a monotonically increasing input array), the stack will store all the elements of the input array.

📖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

Dynamic Programming

Dynamic programming can be used to solve this problem by calculating the minimum cost for all possible subranges of the input array. dp[i][j] represents the minimum cost to construct a tree from arr[i] to arr[j].

Time: O(n^3), where n is the length of the input array.Space: O(n^2), where n is the length of the input array.

Trade-offs: The dynamic programming approach is less efficient than the stack-based approach due to its higher time complexity. However, it can be useful if a bottom-up approach is preferred or if additional constraints are introduced that make the stack-based approach less suitable.

✨

Key Insights

1Insight 1: The optimal tree construction involves combining smaller leaf values first to minimize the product at higher levels of the tree.
2Insight 2: A stack-based approach is suitable for efficiently tracking and combining leaf values to minimize the overall sum of non-leaf nodes.
3Insight 3: The problem has a greedy nature, where merging the smallest adjacent values yields the optimal solution.

📚

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.

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: Monotonically increasing array (e.g., [1, 2, 3, 4, 5]). The stack-based approach correctly merges the smallest adjacent pairs first, leading to the optimal solution.
Edge case 2: Monotonically decreasing array (e.g., [5, 4, 3, 2, 1]). The sentinel value (infinity) in the stack ensures the initial elements are handled correctly and pushed to the stack.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not realizing the greedy nature of the problem and trying to brute-force all possible tree constructions, which leads to an exponential time complexity.

✗Common mistake 2: Incorrectly implementing the stack-based approach, such as not using a sentinel value or not correctly calculating the cost of merging elements.

✅

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: This algorithm is applicable in optimizing the order of merging operations in compiler design, where the cost of merging intermediate results needs to be minimized.

🏦Real-world scenario 2: It can also be used in resource allocation problems where merging smaller tasks with minimal cost is crucial.

💡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

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

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: July 21, 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Cost Tree From Leaf Values"?

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.

🤖 Need help with this problem?