What is the time complexity of Divide an Array Into Subarrays With Minimum Cost I?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Divide an Array Into Subarrays With Minimum Cost I?

This problem uses Array, Sorting, Enumeration. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Divide an Array Into Subarrays With Minimum Cost I?

Insight 1: The cost of each subarray is determined solely by its first element. Therefore, we only need to consider the first element of each subarray when calculating the total cost. Insight 2: A brute-force approach is feasible due to the small constraints on the input array size. We can iterate through all possible ways to divide the array into three subarrays and find the minimum cost.

Which companies ask LeetCode 3010. Divide an Array Into Subarrays With Minimum Cost I in interviews?

This problem is frequently asked at American Express.

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Divide an Array Into Subarrays With Minimum Cost I - LeetCode 3010 Solution

Divide an Array Into Subarrays With Minimum Cost I - Complete Solution Guide

Divide an Array Into Subarrays With Minimum Cost I is LeetCode problem 3010, a Easy 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 array of integers nums of length n . The cost of an array is the value of its first element. For example, the cost of [1,2,3] is 1 while the cost of [3,4,1] is 3 . You need to divide nums into 3 disjoint contiguous subarrays . Return the minimum possible sum of the cost of these subarrays . Example 1: Input: nums = [1,2,3,12] Output: 6 Explanation: The best possible way to form 3 subarrays is: [1], [2], and [3,12] at a total cost of 1 + 2 + 3 = 6. The other possible ways to form

Detailed Explanation

The problem asks you to divide an input array of integers into three disjoint, contiguous subarrays. The cost of each subarray is the value of its first element. The goal is to find the minimum possible sum of the costs of these three subarrays. The input is an array `nums` containing integers, and the output is a single integer representing the minimum total cost. Constraints limit the array size (3 to 50 elements) and the value range of elements (1 to 50).

Solution Approach

The provided code uses a brute-force approach. It iterates through all possible combinations of dividing the array into three subarrays. For each combination, it calculates the total cost (sum of the first elements of each subarray) and updates the minimum cost found so far. This approach works because it exhaustively checks all possibilities, guaranteeing the discovery of the minimum cost.

Step-by-Step Algorithm

Step 1: Initialize `minCost` (or `ans`) to a very large value (infinity in Python, `Integer.MAX_VALUE` in Java, `INT_MAX` in C++). This variable will store the minimum cost found.
Step 2: Iterate through all possible positions `i` for the end of the first subarray (from index 1 to `n-2`, where `n` is the array length).
Step 3: For each `i`, iterate through all possible positions `j` for the end of the second subarray (from `i+1` to `n-1`).
Step 4: Calculate the total cost as `nums[0] + nums[i] + nums[j]`. This represents the cost of the three subarrays: [nums[0]...nums[i-1]], [nums[i]...nums[j-1]], [nums[j]...nums[n-1]]
Step 5: Update `minCost` if the current cost is less than the current minimum cost.
Step 6: After iterating through all combinations, return `minCost`.

Key Insights

Insight 1: The cost of each subarray is determined solely by its first element. Therefore, we only need to consider the first element of each subarray when calculating the total cost.
Insight 2: A brute-force approach is feasible due to the small constraints on the input array size. We can iterate through all possible ways to divide the array into three subarrays and find the minimum cost.
Insight 3: The first element of the array is always part of the cost calculation, as it's always the first element of the first subarray. This simplifies the calculation.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Array, Sorting, Enumeration.

Companies

Asked at: American Express.

Divide an Array Into Subarrays With Minimum Cost I | LeetCode 3010

Easy

Array Sorting Enumeration

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minimumCost(int[] nums) {
        int n = nums.length;
        int minCost = Integer.MAX_VALUE;

        for (int i = 1; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                minCost = Math.min(minCost, nums[0] + nums[i] + nums[j]);
            }
        }

        return minCost;
    }
}

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Problem Info

Description

Problem 3010: Divide an Array Into Subarrays With Minimum Cost I - Detailed explanation and solution approaches.

Topics

Array Sorting Enumeration

Companies

American Express

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3010

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: Initialize `minCost` (or `ans`) to a very large value (infinity in Python, `Integer.MAX_VALUE` in Java, `INT_MAX` in C++). This variable will store the minimum cost found.
2
Step 2: Iterate through all possible positions `i` for the end of the first subarray (from index 1 to `n-2`, where `n` is the array length).
3
Step 3: For each `i`, iterate through all possible positions `j` for the end of the second subarray (from `i+1` to `n-1`).
4
Step 4: Calculate the total cost as `nums[0] + nums[i] + nums[j]`. This represents the cost of the three subarrays: [nums[0]...nums[i-1]], [nums[i]...nums[j-1]], [nums[j]...nums[n-1]]
5
Step 5: Update `minCost` if the current cost is less than the current minimum cost.
6
Step 6: After iterating through all combinations, return `minCost`.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The nested loops iterate through all possible pairs of indices (i, j), resulting in O(n^2) time complexity. The algorithm uses a constant amount of extra space to store the minimum cost and loop counters, resulting in O(1) space complexity.

📖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

✨

Key Insights

1Insight 1: The cost of each subarray is determined solely by its first element. Therefore, we only need to consider the first element of each subarray when calculating the total cost.
2Insight 2: A brute-force approach is feasible due to the small constraints on the input array size. We can iterate through all possible ways to divide the array into three subarrays and find the minimum cost.
3Insight 3: The first element of the array is always part of the cost calculation, as it's always the first element of the first subarray. This simplifies the calculation.

📚

Key Terms Explained

For Students

Array

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Arrays with only three elements. The algorithm correctly handles this case, as the outer loop only runs once, and the inner loop runs once.
Edge case 2: Arrays where the minimum cost involves large numbers. The algorithm correctly handles this because it considers all possible combinations and always updates the minimum cost appropriately.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Off-by-one errors in loop bounds. Incorrectly setting the loop limits might lead to missing some combinations or accessing elements out of bounds.

✗Common mistake 2: Failing to initialize `minCost` to a sufficiently large value. If it's not initialized correctly, the algorithm may not correctly find the minimum cost.

✅

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: Resource allocation. Imagine dividing a project into three tasks, with each task having a setup cost. This algorithm could be used to find the minimum total setup cost.

🏦Real-world scenario 2: Network routing. Dividing a network path into segments with costs associated with each segment can use a similar approach to minimize the overall cost.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

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 9, 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 should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Divide an Array Into Subarrays With Minimum Cost I"?

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

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