What is the time complexity of Minimum Cost to Cut a Stick?

The optimal solution has a time complexity of O(n^3).

What is the space complexity of Minimum Cost to Cut a Stick?

The space complexity is O(n^2).

What data structures are used to solve Minimum Cost to Cut a Stick?

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

What are the key insights for solving Minimum Cost to Cut a Stick?

Insight 1: The optimal cutting order is crucial for minimizing the total cost. Different cutting orders lead to different costs. Insight 2: Dynamic programming (specifically, memoization) can be used to efficiently explore all possible cutting orders by breaking the problem into overlapping subproblems.

Which companies ask LeetCode 1547. Minimum Cost to Cut a Stick in interviews?

This problem is frequently asked at LINE, PhonePe.

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Minimum Cost to Cut a Stick - LeetCode 1547 Solution

Minimum Cost to Cut a Stick - Complete Solution Guide

Minimum Cost to Cut a Stick is LeetCode problem 1547, 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

Given a wooden stick of length n units. The stick is labelled from 0 to n . For example, a stick of length 6 is labelled as follows: Given an integer array cuts where cuts[i] denotes a position you should perform a cut at. You should perform the cuts in order, you can change the order of the cuts as you wish. The cost of one cut is the length of the stick to be cut, the total cost is the sum of costs of all cuts. When you cut a stick, it will be split into two smaller sticks (i.e. the sum of the

Detailed Explanation

The problem asks us to find the minimum cost to cut a wooden stick of length 'n' at specified positions given in the 'cuts' array. We can cut the stick in any order. The cost of each cut is the length of the stick being cut. The total cost is the sum of the costs of all cuts. We need to find the cutting order that minimizes this total cost. The stick is labeled from 0 to n, and the cuts array specifies the positions where the stick should be cut.

Solution Approach

The solution uses a top-down dynamic programming approach with memoization. We consider all possible cutting orders recursively. For each possible cut, we compute the cost (which is the length of the current stick) and recursively compute the minimum cost to cut the remaining segments of the stick. Memoization is used to avoid recomputing the solutions to overlapping subproblems, significantly improving efficiency.

Step-by-Step Algorithm

Step 1: Sort the 'cuts' array and add the endpoints 0 and 'n' to create a new array called 'points'. This represents all possible cut positions.
Step 2: Define a recursive function `solve(i, j)` that calculates the minimum cost to cut the segment of the stick between 'points[i]' and 'points[j]'.
Step 3: Base case for `solve(i, j)`: If 'j - i <= 1', it means there are no cuts to be made between points[i] and points[j], so return 0.
Step 4: Calculate the length of the stick segment between 'points[i]' and 'points[j]' as 'points[j] - points[i]'. This will be the cost of the first cut we make on this segment.
Step 5: Iterate through all possible cut positions 'k' between 'i + 1' and 'j - 1'. For each 'k', recursively calculate the cost of cutting the stick at 'k', which is `solve(i, k) + solve(k, j)`.
Step 6: Find the minimum cost among all possible 'k' values. Add this minimum cost to the cost of the first cut (stick length).
Step 7: Memoize the result for `solve(i, j)` to avoid recomputation.
Step 8: Call `solve(0, len(points) - 1)` to find the minimum cost to cut the entire stick.

Key Insights

Insight 1: The optimal cutting order is crucial for minimizing the total cost. Different cutting orders lead to different costs.
Insight 2: Dynamic programming (specifically, memoization) can be used to efficiently explore all possible cutting orders by breaking the problem into overlapping subproblems.
Insight 3: Adding the endpoints (0 and n) to the cuts array simplifies the implementation by uniformly handling all stick segments.

Complexity Analysis

Time Complexity: O(n^3)

Space Complexity: O(n^2)

Topics

This problem involves: Array, Dynamic Programming, Sorting.

Companies

Asked at: LINE, PhonePe.

Minimum Cost to Cut a Stick | LeetCode 1547

Hard

Array Dynamic Programming Sorting

Time: O(n^3)Space: O(n^2)

Solution Code

Available in:

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

class Solution {
    private int[] points;
    private Map<String, Integer> memo;

    public int minCost(int n, int[] cuts) {
        // Add the endpoints of the stick to the cuts array to handle all segments uniformly.
        // This new array, let's call it `points`, represents all potential endpoints of smaller sticks.
        points = new int[cuts.length + 2];
        points[0] = 0;
        for (int i = 0; i < cuts.length; i++) {
            points[i + 1] = cuts[i];
        }
        points[cuts.length + 1] = n;
        Arrays.sort(points);

        // Use memoization to store results of subproblems.
        memo = new HashMap<>();

        // The initial problem is to find the cost for the entire stick,
        // which corresponds to the segment from points[0] to the last point.
        return solve(0, points.length - 1);
    }

    private int solve(int i, int j) {
        String key = i + "," + j;
        if (memo.containsKey(key)) {
            return memo.get(key);
        }

        /**
         * Calculates the minimum cost to cut the stick segment from points[i] to points[j].
         * i and j are indices into the `points` array.
         */

        // Base case: If there are no cuts to be made within this segment.
        // This happens when j is adjacent to i (j = i + 1), meaning the segment
        // is between two consecutive points and cannot be cut further.
        if (j - i <= 1) {
            return 0;
        }

        // The cost of the first cut on the segment [points[i], points[j]] is its length.
        // This cost is incurred regardless of where we make the first cut within this segment.
        int current_stick_length = points[j] - points[i];

        // We try every possible position `k` for the first cut within the current segment.
        // The available cut points are from points[i+1] to points[j-1].
        // The total cost for a choice of `k` is the cost of the subproblems it creates.
        int min_subproblem_cost = Integer.MAX_VALUE;
        for (int k = i + 1; k < j; k++) {
            min_subproblem_cost = Math.min(min_subproblem_cost, solve(i, k) + solve(k, j));
        }

        // The total cost for this segment is the cost of the first cut (its length)
        // plus the minimum cost to solve the resulting subproblems.
        int totalCost = current_stick_length + min_subproblem_cost;
        memo.put(key, totalCost);
        return totalCost;
    }
}

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Problem Info

Description

Problem 1547: Minimum Cost to Cut a Stick - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Sorting

Companies

LINEPhonePe

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:1547

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

8 steps

1
Step 1: Sort the 'cuts' array and add the endpoints 0 and 'n' to create a new array called 'points'. This represents all possible cut positions.
2
Step 2: Define a recursive function `solve(i, j)` that calculates the minimum cost to cut the segment of the stick between 'points[i]' and 'points[j]'.
3
Step 3: Base case for `solve(i, j)`: If 'j - i <= 1', it means there are no cuts to be made between points[i] and points[j], so return 0.
4
Step 4: Calculate the length of the stick segment between 'points[i]' and 'points[j]' as 'points[j] - points[i]'. This will be the cost of the first cut we make on this segment.
5
Step 5: Iterate through all possible cut positions 'k' between 'i + 1' and 'j - 1'. For each 'k', recursively calculate the cost of cutting the stick at 'k', which is `solve(i, k) + solve(k, j)`.
6
Step 6: Find the minimum cost among all possible 'k' values. Add this minimum cost to the cost of the first cut (stick length).
7
Step 7: Memoize the result for `solve(i, j)` to avoid recomputation.
8
Step 8: Call `solve(0, len(points) - 1)` to find the minimum cost to cut the entire stick.

💡

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^3)

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is determined by the number of subproblems (O(n^2)) multiplied by the time taken to solve each subproblem (O(n) due to the inner loop). The space complexity is dominated by the memoization table, which stores results for each subproblem which gives O(n^2).

📖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

Bottom-up Dynamic Programming

Instead of recursion and memoization, we can build a DP table bottom-up. `dp[i][j]` would store the minimum cost to cut the stick from points[i] to points[j].

Time: O(n^3)Space: O(n^2)

Trade-offs: Bottom-up DP can sometimes be slightly faster due to avoiding recursive function call overhead. It might also be easier to reason about the order of computation. However, top-down memoization can be more intuitive for some and only computes the necessary subproblems.

✨

Key Insights

1Insight 1: The optimal cutting order is crucial for minimizing the total cost. Different cutting orders lead to different costs.
2Insight 2: Dynamic programming (specifically, memoization) can be used to efficiently explore all possible cutting orders by breaking the problem into overlapping subproblems.
3Insight 3: Adding the endpoints (0 and n) to the cuts array simplifies the implementation by uniformly handling all stick segments.

📚

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.

Sorting

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

Edge Cases

Edge case 1: No cuts provided (cuts array is empty). The algorithm correctly handles this by adding 0 and n, resulting in points = [0, n] and thus a base case return of 0.
Edge case 2: Only one cut provided. The algorithm will still work since the points will be [0, cut, n], and the recursion will correctly calculate the costs.
Edge case 3: Input cuts contain duplicates. The problem states that all integers in the cuts array are distinct, so it is guaranteed this will not happen.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the cuts array. The points need to be sorted to properly determine the length of the stick to cut.

✗Common mistake 2: Not including the endpoints 0 and n in the sorted cuts array. This leads to incorrect calculation of stick lengths at the boundaries.

✗Common mistake 3: Forgetting to memoize the results of subproblems. This results in exponential time complexity and causes the algorithm to time out for larger inputs.

✗Common mistake 4: Incorrect base case for the recursive function. The base case should be when there are no more cuts to make between two points.

✅

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 optimizing the cutting of a roll of fabric to fulfill orders of various sizes, minimizing waste.

🏦Real-world scenario 2: Task scheduling. Consider a task that needs to be split into subtasks at specific points. Optimizing the order of splitting minimizes overhead costs.

🎮Real-world scenario 3: DNA sequencing. When breaking DNA strands into smaller fragments for sequencing, minimizing the cost/time based on cutting positions is relevant.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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: September 16, 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 13

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

@CodeNinjaJul 6

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

@TechLeadJun 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.

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 Cut a Stick"?

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

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