What is the time complexity of Collecting Chocolates?

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

What data structures are used to solve Collecting Chocolates?

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

What are the key insights for solving Collecting Chocolates?

Insight 1: The problem requires finding the minimum cost across all possible numbers of shift operations (from 0 to n-1). Insight 2: For each chocolate type, we only need to consider the minimum cost among its initial cost and its costs after each shift. This is because we can always choose the cheapest option for each type of chocolate.

Which companies ask LeetCode 2735. Collecting Chocolates in interviews?

This problem is frequently asked at Deutsche Bank.

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Collecting Chocolates - LeetCode 2735 Solution

Collecting Chocolates - Complete Solution Guide

Collecting Chocolates is LeetCode problem 2735, 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 a 0-indexed integer array nums of size n representing the cost of collecting different chocolates. The cost of collecting the chocolate at the index i is nums[i] . Each chocolate is of a different type, and initially, the chocolate at the index i is of i th type. In one operation, you can do the following with an incurred cost of x : Simultaneously change the chocolate of i th type to ((i + 1) mod n) th type for all chocolates. Return the minimum cost to collect chocolates of all t

Detailed Explanation

The problem asks us to find the minimum cost to collect all types of chocolates. We are given an array `nums` where `nums[i]` is the initial cost of the chocolate of type `i`. We can perform an operation where we shift the type of each chocolate by one (i.e., the chocolate of type `i` becomes type `(i+1) mod n`) at a cost of `x`. We need to determine the minimum total cost considering the initial costs and the costs of shifting chocolate types before buying them. The output should be the minimal total cost achieved after considering all possible shift operations (0 to n-1 shifts).

Solution Approach

The solution iterates through all possible numbers of shifts (from 0 to n-1). For each number of shifts `k`, it calculates the cost of performing `k` shifts (k * x). Then, for each chocolate type `j`, it finds the minimum cost to obtain that type of chocolate considering all previous shifts. This is done by comparing the current minimum cost for chocolate type `j` with the cost of the chocolate at the `(j-k)%n` index in the original `nums` array. After finding the minimum cost for each chocolate type, the solution calculates the total cost (shift cost + sum of minimum costs for all chocolate types) and updates the overall minimum cost.

Step-by-Step Algorithm

Step 1: Initialize `min_costs_per_type` array with the original costs from the `nums` array. This array stores the minimum cost found so far for each chocolate type.
Step 2: Initialize `overall_min_cost` with the sum of the initial costs in `nums` (corresponding to zero shifts).
Step 3: Iterate from `k = 1` to `n-1` (number of shifts).
Step 4: Calculate the cost of `k` shifts: `op_cost = k * x`.
Step 5: For each chocolate type `j` (from 0 to `n-1`):
Step 6: Calculate the index of the chocolate type after `k` shifts: `(j - k + n) % n`. The `+ n` ensures a positive result before applying the modulo operator.
Step 7: Update `min_costs_per_type[j]` with the minimum between its current value and the cost of the chocolate at the shifted index in the original `nums` array.
Step 8: Calculate the total cost for the current number of shifts: `current_total_cost = op_cost + sum(min_costs_per_type)`.
Step 9: Update `overall_min_cost` with the minimum between its current value and `current_total_cost`.
Step 10: Return `overall_min_cost`.

Key Insights

Insight 1: The problem requires finding the minimum cost across all possible numbers of shift operations (from 0 to n-1).
Insight 2: For each chocolate type, we only need to consider the minimum cost among its initial cost and its costs after each shift. This is because we can always choose the cheapest option for each type of chocolate.
Insight 3: Modulo operator `%` is crucial for handling the circular shifts of chocolate types.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Enumeration.

Companies

Asked at: Deutsche Bank.

Collecting Chocolates | LeetCode 2735

Medium

Array Enumeration

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

Solution Code

Available in:

class Solution {
    public long minCost(int[] nums, int x) {
        int n = nums.length;
        int[] min_costs_per_type = new int[n];
        for (int i = 0; i < n; i++) {
            min_costs_per_type[i] = nums[i];
        }

        long overall_min_cost = 0;
        for (int cost : min_costs_per_type) {
            overall_min_cost += cost;
        }

        for (int k = 1; k < n; k++) {
            long op_cost = (long)k * x;

            for (int j = 0; j < n; j++) {
                int cost_at_k_shifts = nums[(j - k + n) % n];
                min_costs_per_type[j] = Math.min(min_costs_per_type[j], cost_at_k_shifts);
            }

            long current_total_cost = op_cost;
            for (int cost : min_costs_per_type) {
                current_total_cost += cost;
            }
            overall_min_cost = Math.min(overall_min_cost, current_total_cost);
        }

        return overall_min_cost;
    }
}

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Problem Info

Description

Problem 2735: Collecting Chocolates - Detailed explanation and solution approaches.

Topics

Array Enumeration

Companies

Deutsche Bank

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2735

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

10 steps

1
Step 1: Initialize `min_costs_per_type` array with the original costs from the `nums` array. This array stores the minimum cost found so far for each chocolate type.
2
Step 2: Initialize `overall_min_cost` with the sum of the initial costs in `nums` (corresponding to zero shifts).
3
Step 3: Iterate from `k = 1` to `n-1` (number of shifts).
4
Step 4: Calculate the cost of `k` shifts: `op_cost = k * x`.
5
Step 5: For each chocolate type `j` (from 0 to `n-1`):
6
Step 6: Calculate the index of the chocolate type after `k` shifts: `(j - k + n) % n`. The `+ n` ensures a positive result before applying the modulo operator.
7
Step 7: Update `min_costs_per_type[j]` with the minimum between its current value and the cost of the chocolate at the shifted index in the original `nums` array.
8
Step 8: Calculate the total cost for the current number of shifts: `current_total_cost = op_cost + sum(min_costs_per_type)`.
9
Step 9: Update `overall_min_cost` with the minimum between its current value and `current_total_cost`.
10
Step 10: Return `overall_min_cost`.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The nested loops (shift amount and chocolate type index) result in O(n*n) complexity. The space complexity stems from the 'min_costs_per_type' array which requires O(n) space.

📖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 with Memoization

Instead of recomputing the minimum cost for each shift, we can store the minimum costs for each chocolate type after each shift in a 2D array. This might provide marginal optimization depending on the cost function, but the core time complexity remains similar.

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

Trade-offs: This approach uses more space, potentially without significant time complexity improvement, making the provided solution generally better.

✨

Key Insights

1Insight 1: The problem requires finding the minimum cost across all possible numbers of shift operations (from 0 to n-1).
2Insight 2: For each chocolate type, we only need to consider the minimum cost among its initial cost and its costs after each shift. This is because we can always choose the cheapest option for each type of chocolate.
3Insight 3: Modulo operator `%` is crucial for handling the circular shifts of chocolate types.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: When the cost of shifting `x` is very high, it might be optimal to not perform any shifts and just buy the chocolates at their original prices. The solution correctly handles this by initializing `overall_min_cost` with the sum of original costs and comparing it with the cost after each possible number of shifts.
Edge case 2: When `n` is small, the solution still works correctly because the loops iterate the appropriate number of times. When `n` is 1, it correctly returns `nums[0]`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to use the modulo operator `%` when calculating the shifted index, which will lead to incorrect results and potential array out-of-bounds errors.

✗Common mistake 2: Not considering the case when no shifts are performed (i.e., just buying the chocolates at their original prices). The initial `overall_min_cost` must be set correctly for this scenario.

✅

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: Inventory management where the cost of items varies with time and there is a fixed cost to re-arrange the items. The goal is to minimize the total cost of acquiring the necessary items.

🏦Real-world scenario 2: Resource allocation in a circular economy where the cost of accessing resources at different locations is different and there is a fixed cost to transport resources between locations. The problem can be adapted to find the optimal strategy to obtain the necessary resources at the lowest possible 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

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 6, 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

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

@InterviewProJun 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.

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 "Collecting Chocolates"?

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 enumeration techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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