What is the time complexity of Minimum Total Space Wasted With K Resizing Operations?

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

What is the space complexity of Minimum Total Space Wasted With K Resizing Operations?

The space complexity is O(n^2).

What data structures are used to solve Minimum Total Space Wasted With K Resizing Operations?

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

What are the key insights for solving Minimum Total Space Wasted With K Resizing Operations?

Insight 1: Dynamic programming is suitable because the optimal solution for a given number of resizes and a subarray depends on the optimal solutions for smaller subarrays and fewer resizes. Insight 2: The core idea is to calculate the cost (wasted space) for all possible subarrays and then use DP to find the minimum total cost for the entire array, given the constraint on the number of resizes.

Which companies ask LeetCode 1959. Minimum Total Space Wasted With K Resizing Operations in interviews?

This problem is frequently asked at Media.net.

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Minimum Total Space Wasted With K Resizing Operations - LeetCode 1959 Solution

Minimum Total Space Wasted With K Resizing Operations - Complete Solution Guide

Minimum Total Space Wasted With K Resizing Operations is LeetCode problem 1959, 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 currently designing a dynamic array. You are given a 0-indexed integer array nums , where nums[i] is the number of elements that will be in the array at time i . In addition, you are given an integer k , the maximum number of times you can resize the array (to any size). The size of the array at time t , size t , must be at least nums[t] because there needs to be enough space in the array to hold all the elements. The space wasted at time t is defined as size t - nums[t] , and the total

Detailed Explanation

The problem asks us to find the minimum total wasted space when resizing a dynamic array. We're given an array `nums` representing the number of elements at each time step, and `k`, the maximum number of resizes allowed. The wasted space at a given time is the difference between the array's size and the number of elements at that time. The goal is to minimize the sum of wasted space across all time steps, given the constraint on the number of resizes. The initial size of the array does not count towards the number of resizing operations.

Solution Approach

The solution uses dynamic programming to determine the minimum total wasted space. It first pre-calculates the 'cost' of keeping a subarray at a single size (the maximum element within that subarray). Then, it uses a DP table `dp[i][j]` to store the minimum wasted space for the first `i` elements with `j` resizes. The base case `dp[0][j] = 0` for all `j` because no space is wasted when the array has zero length. The transitions consider all possible points `p` where a resize could occur before position `i`. Therefore, the solution considers a range of previous best solutions `dp[p][j-1]` plus the cost of not resizing between index `p` and `i-1`. The DP formula is `dp[i][j] = min(dp[i][j], dp[p][j-1] + cost[p][i-1])` where `p` is from 0 to `i-1`.

Step-by-Step Algorithm

Step 1: Pre-calculate the `cost` matrix. `cost[i][j]` stores the wasted space if the subarray from index `i` to `j` is kept at the same size (determined by the maximum value within that subarray).
Step 2: Initialize the `dp` table. `dp[i][j]` represents the minimum wasted space for the first `i` elements using at most `j` resizes. Initialize all values in `dp` to infinity (or a sufficiently large number) except for `dp[0][j] = 0` for all `j` from 0 to k.
Step 3: Iterate through the `dp` table. The outer loop iterates from `i = 1` to `n` (number of elements). The inner loop iterates from `j = 0` to `k` (number of resizes).
Step 4: Fill the `dp` table. For each `dp[i][j]`, consider all possible resize positions `p` from 0 to `i - 1`. Calculate the wasted space if we resize at position `p` and use the best solution for the subarray up to `p` with `j - 1` resizes (`dp[p][j - 1]`). Then, add the cost of not resizing between `p` and `i - 1` (using `cost[p][i - 1]`). Take the minimum of all these possibilities to update `dp[i][j]`.
Step 5: Return `dp[n][k]`. This value represents the minimum wasted space for the entire array using at most `k` resizes.

Key Insights

Insight 1: Dynamic programming is suitable because the optimal solution for a given number of resizes and a subarray depends on the optimal solutions for smaller subarrays and fewer resizes.
Insight 2: The core idea is to calculate the cost (wasted space) for all possible subarrays and then use DP to find the minimum total cost for the entire array, given the constraint on the number of resizes.
Insight 3: Pre-calculating the cost of each possible subarray optimizes the inner loop of the DP, preventing redundant computations.

Complexity Analysis

Time Complexity: O(k*n^2)

Space Complexity: O(n^2)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Media.net.

Minimum Total Space Wasted With K Resizing Operations | LeetCode 1959

Medium

Array Dynamic Programming

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int minSpaceWastedKResizing(int[] nums, int k) {
        int n = nums.length;
        
        int[][] cost = new int[n][n];
        for (int i = 0; i < n; i++) {
            int current_max = 0;
            int current_sum = 0;
            for (int j = i; j < n; j++) {
                current_max = Math.max(current_max, nums[j]);
                current_sum += nums[j];
                cost[i][j] = (j - i + 1) * current_max - current_sum;
            }
        }
        
        int[][] dp = new int[n + 1][k + 1];
        for (int i = 0; i < n + 1; i++) {
            Arrays.fill(dp[i], Integer.MAX_VALUE);
        }
        
        for (int j = 0; j < k + 1; j++) {
            dp[0][j] = 0;
        }
        
        for (int i = 1; i < n + 1; i++) {
            dp[i][0] = cost[0][i - 1];
            
            for (int j = 1; j < k + 1; j++) {
                dp[i][j] = dp[i][j - 1];
                
                for (int p = 0; p < i; p++) {
                    dp[i][j] = Math.min(dp[i][j], dp[p][j - 1] + cost[p][i - 1]);
                }
            }
        }

        return dp[n][k];
    }
}

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Problem Info

Description

Problem 1959: Minimum Total Space Wasted With K Resizing Operations - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

Media.net

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1959

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

5 steps

1
Step 1: Pre-calculate the `cost` matrix. `cost[i][j]` stores the wasted space if the subarray from index `i` to `j` is kept at the same size (determined by the maximum value within that subarray).
2
Step 2: Initialize the `dp` table. `dp[i][j]` represents the minimum wasted space for the first `i` elements using at most `j` resizes. Initialize all values in `dp` to infinity (or a sufficiently large number) except for `dp[0][j] = 0` for all `j` from 0 to k.
3
Step 3: Iterate through the `dp` table. The outer loop iterates from `i = 1` to `n` (number of elements). The inner loop iterates from `j = 0` to `k` (number of resizes).
4
Step 4: Fill the `dp` table. For each `dp[i][j]`, consider all possible resize positions `p` from 0 to `i - 1`. Calculate the wasted space if we resize at position `p` and use the best solution for the subarray up to `p` with `j - 1` resizes (`dp[p][j - 1]`). Then, add the cost of not resizing between `p` and `i - 1` (using `cost[p][i - 1]`). Take the minimum of all these possibilities to update `dp[i][j]`.
5
Step 5: Return `dp[n][k]`. This value represents the minimum wasted space for the entire array using at most `k` resizes.

💡

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(k*n^2)

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is determined by the nested loops in the dynamic programming algorithm. The outermost loop runs `n` times. The middle loop runs `k` times. The innermost loop runs `n` times. This leads to O(n * k * n) = O(k*n^2) complexity. The space complexity is derived from the two 2D arrays: `cost[n][n]` and `dp[n+1][k+1]`, leading to O(n^2) 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

Greedy Approach

A greedy approach might involve resizing whenever the current array size is significantly smaller than the next element. However, this will not guarantee the optimal solution as local optima may lead to a greater overall waste.

Time: O(n*log(n)) - depends on sorting or priority queue usageSpace: O(n)

Trade-offs: Greedy approaches are generally faster, but less accurate. This problem requires an exact optimal solution, making the DP approach necessary. The greedy approach would be significantly harder to guarantee optimality with.

✨

Key Insights

1Insight 1: Dynamic programming is suitable because the optimal solution for a given number of resizes and a subarray depends on the optimal solutions for smaller subarrays and fewer resizes.
2Insight 2: The core idea is to calculate the cost (wasted space) for all possible subarrays and then use DP to find the minimum total cost for the entire array, given the constraint on the number of resizes.
3Insight 3: Pre-calculating the cost of each possible subarray optimizes the inner loop of the DP, preventing redundant computations.

📚

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.

Edge Cases

Edge case 1: `k = 0`: In this case, we cannot resize at all. The solution handles this by initializing `dp[i][0]` with the cost of not resizing from the beginning (index 0) to `i - 1`.
Edge case 2: `k >= n - 1`: In this case, we can resize at every index. However, the solution still works because it considers all possible combinations of resizes and finds the minimum wasted space, even if some resizes are unnecessary.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not pre-calculating the cost matrix: Calculating the wasted space for each subarray within the DP loop would lead to a higher time complexity (O(n^3*k)), and exceeding time limits.

✗Common mistake 2: Incorrect DP transition: Failing to consider all possible resize positions `p` or incorrectly calculating the cost for the subarray between `p` and `i - 1`.

✅

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: Dynamic memory allocation in systems programming. This algorithm can guide decisions on when to reallocate memory to minimize wasted space while balancing the cost of reallocation.

🏦Real-world scenario 2: Cloud resource scaling. This approach can be adapted to determine when to scale up or down cloud resources (e.g., CPU, memory) based on predicted demand, aiming to minimize resource wastage while ensuring adequate performance.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: November 7, 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

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

@TechLeadJul 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 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.

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 "Minimum Total Space Wasted With K Resizing Operations"?

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.

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