What is the time complexity of Minimum Cost to Split an Array?

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

What data structures are used to solve Minimum Cost to Split an Array?

This problem uses Array, Hash Table, Dynamic Programming, Counting. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Cost to Split an Array?

Insight 1: Dynamic programming is suitable for solving this optimization problem because the optimal solution for a larger array can be built from the optimal solutions of smaller subarrays. Insight 2: The importance value of a subarray can be efficiently computed by keeping track of the frequency of each number within the subarray and calculating the trimmed length accordingly.

Which companies ask LeetCode 2547. Minimum Cost to Split an Array in interviews?

This problem is frequently asked at Indeed.

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Minimum Cost to Split an Array - LeetCode 2547 Solution

Minimum Cost to Split an Array - Complete Solution Guide

Minimum Cost to Split an Array is LeetCode problem 2547, a Hard 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 array nums and an integer k . Split the array into some number of non-empty subarrays. The cost of a split is the sum of the importance value of each subarray in the split. Let trimmed(subarray) be the version of the subarray where all numbers which appear only once are removed. For example, trimmed([3,1,2,4,3,4]) = [3,4,3,4]. The importance value of a subarray is k + trimmed(subarray).length . For example, if a subarray is [1,2,3,3,3,4,4] , then trimmed( [1,2,3,3,3,4,4]

Detailed Explanation

The problem asks us to split an array `nums` into subarrays and minimize the total cost of the split. The cost is calculated by summing the importance value of each subarray. The importance value of a subarray is `k + trimmed(subarray).length`, where `trimmed(subarray)` is the subarray after removing all numbers that appear only once. The goal is to find the minimum possible cost among all possible splits of the array.

Solution Approach

The solution uses dynamic programming to determine the minimum cost to split the array. `dp[i]` represents the minimum cost to split the array `nums[0...i-1]`. The algorithm iterates from `i = 1` to `n` (the length of `nums`), calculating `dp[i]` by considering all possible split positions `j` (from `0` to `i-1`). For each split position `j`, the cost of the split is `dp[j] + k + trimmed(nums[j...i-1]).length`. The algorithm maintains the minimum cost found so far for `dp[i]`.

Step-by-Step Algorithm

Step 1: Initialize a DP array `dp` of size `n+1` with `dp[0] = 0` and all other values as infinity. `dp[i]` stores the minimum cost to split `nums[0...i-1]`.
Step 2: Iterate through the array from `i = 1` to `n` (outer loop), representing the ending index of the subarray.
Step 3: For each `i`, iterate backwards from `j = i-1` to `0` (inner loop), representing the starting index of a potential subarray.
Step 4: Calculate the trimmed length of the subarray `nums[j...i-1]` by keeping track of the frequency of each element in the subarray. Increment `trimmed_len` by 2 when a number appears for the second time and by 1 when a number appears for the third or more times.
Step 5: Calculate the cost of splitting the array at index `j` as `dp[j] + k + trimmed_len`.
Step 6: Update `dp[i]` with the minimum of its current value and the calculated cost.
Step 7: After the outer loop completes, `dp[n]` will contain the minimum cost to split the entire array `nums`.

Key Insights

Insight 1: Dynamic programming is suitable for solving this optimization problem because the optimal solution for a larger array can be built from the optimal solutions of smaller subarrays.
Insight 2: The importance value of a subarray can be efficiently computed by keeping track of the frequency of each number within the subarray and calculating the trimmed length accordingly.
Insight 3: Consider all possible split positions to determine the split with minimal cost.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Dynamic Programming, Counting.

Companies

Asked at: Indeed.

Minimum Cost to Split an Array | LeetCode 2547

Hard

Array Hash Table Dynamic Programming Counting

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;

class Solution {
    public int minCost(int[] nums, int k) {
        int n = nums.length;
        int[] dp = new int[n + 1];
        Arrays.fill(dp, Integer.MAX_VALUE);
        dp[0] = 0;

        for (int i = 1; i <= n; i++) {
            Map<Integer, Integer> freq = new HashMap<>();
            int trimmed_len = 0;
            for (int j = i - 1; j >= 0; j--) {
                int num = nums[j];
                freq.put(num, freq.getOrDefault(num, 0) + 1);

                if (freq.get(num) == 2) {
                    trimmed_len += 2;
                } else if (freq.get(num) > 2) {
                    trimmed_len += 1;
                }

                int cost = k + trimmed_len;
                if (dp[j] != Integer.MAX_VALUE) {
                    dp[i] = Math.min(dp[i], dp[j] + cost);
                }
            }
        }

        return dp[n];
    }
}

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Problem Info

Description

Problem 2547: Minimum Cost to Split an Array - Detailed explanation and solution approaches.

Topics

Array Hash Table Dynamic Programming Counting

Companies

Indeed

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2547

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 a DP array `dp` of size `n+1` with `dp[0] = 0` and all other values as infinity. `dp[i]` stores the minimum cost to split `nums[0...i-1]`.
2
Step 2: Iterate through the array from `i = 1` to `n` (outer loop), representing the ending index of the subarray.
3
Step 3: For each `i`, iterate backwards from `j = i-1` to `0` (inner loop), representing the starting index of a potential subarray.
4
Step 4: Calculate the trimmed length of the subarray `nums[j...i-1]` by keeping track of the frequency of each element in the subarray. Increment `trimmed_len` by 2 when a number appears for the second time and by 1 when a number appears for the third or more times.
5
Step 5: Calculate the cost of splitting the array at index `j` as `dp[j] + k + trimmed_len`.
6
Step 6: Update `dp[i]` with the minimum of its current value and the calculated cost.
7
Step 7: After the outer loop completes, `dp[n]` will contain the minimum cost to split the entire array `nums`.

💡

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 dominate the time complexity. The outer loop runs `n` times. The inner loop also runs up to `n` times for each iteration of the outer loop. The space complexity is determined by the `dp` array which stores the minimal costs for subarrays of lengths from 0 to n, which gives us O(n) 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

Recursion with Memoization

This approach involves recursively exploring all possible splits and using memoization to avoid recomputing the minimum cost for the same subarrays. The base case is when the array is empty (cost is 0).

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

Trade-offs: The recursive approach with memoization can be more intuitive to understand for some, but may incur higher overhead due to function calls. Also, the space complexity can be higher than the iterative DP solution because of recursion stack and memoization table.

✨

Key Insights

1Insight 1: Dynamic programming is suitable for solving this optimization problem because the optimal solution for a larger array can be built from the optimal solutions of smaller subarrays.
2Insight 2: The importance value of a subarray can be efficiently computed by keeping track of the frequency of each number within the subarray and calculating the trimmed length accordingly.
3Insight 3: Consider all possible split positions to determine the split with minimal cost.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty array (not explicitly handled but works correctly because the loop conditions prevent it). The `dp` array would be size 1, and `dp[0] = 0` would be returned.
Edge case 2: `k` being very large compared to the trimmed length. The algorithm correctly handles this, since it chooses the minimum cost among all possible splits, including splitting the array into one subarray.
Edge case 3: Array with all distinct elements. The trimmed length will always be 0. The solution still correctly computes the minimum cost.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not correctly calculating the trimmed length of a subarray. For instance, not accounting for the fact that the count needs to be greater than 1 to be included in the trimmed length.

✗Common mistake 2: Failing to initialize the `dp` array with appropriate initial values (e.g., infinity) to ensure the `min` function works correctly.

✗Common mistake 3: Not considering all possible split positions, leading to a suboptimal solution.

✗Common mistake 4: Forgetting to add `k` when calculating the cost of each subarray.

✅

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: Data compression - splitting data into segments to minimize the total size after applying a trimming/compression algorithm to each segment. The cost of compression may be similar to 'k'.

🏦Real-world scenario 2: Resource allocation - splitting tasks across multiple workers. The 'trimmed length' could represent redundant work that needs to be minimized. 'k' represents the cost of distributing the tasks.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: June 28, 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

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

@InterviewProJul 14

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

@CodeNinjaJul 7

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

@TechLeadJun 30

❓

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 do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 Cost to Split an Array"?

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

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