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

The optimal solution has a time complexity of O(n*dist*log(dist)).

What is the space complexity of Divide an Array Into Subarrays With Minimum Cost II?

The space complexity is O(dist).

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

This problem uses Array, Hash Table, Sliding Window, Heap (Priority Queue). Understanding these concepts is essential for solving this Hard problem efficiently.

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

Insight 1: The first subarray always starts at index 0, reducing the problem to selecting the best `k-1` starting indices from the rest of the array. Insight 2: The constraint `i_{k-1} - i_1 <= dist` implies that all the selected `k-1` indices must fall within a sliding window of size `dist + 1` in the subarray `nums[1:]`.

Which companies ask LeetCode 3013. Divide an Array Into Subarrays With Minimum Cost II in interviews?

This problem is frequently asked at American Express, jio.

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Divide an Array Into Subarrays With Minimum Cost II - LeetCode 3013 Solution

Q: How to solve Divide an Array Into Subarrays With Minimum Cost II?

The provided code uses a sliding window approach combined with two heaps (priority queues) to efficiently find the `k-1` smallest elements within each window of size `dist + 1` in the subarray `nums[1:]`. The algorithm maintains a max-heap (`small_heap`) containing the `k-1` smallest elements seen so far within the current window, and a min-heap (`large_heap`) containing the remaining elements of the window. The code uses a `to_remove` counter to handle cases where elements that are no longer within the window need to be removed from the heaps efficiently. By maintaining these heaps, we can easily track the sum of the `k-1` smallest elements in the current window, and update them efficiently as the window slides.

Q: Which companies ask LeetCode 3013. Divide an Array Into Subarrays With Minimum Cost II in interviews?

This problem is frequently asked at American Express, jio.

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

Divide an Array Into Subarrays With Minimum Cost II is LeetCode problem 3013, 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 a 0-indexed array of integers nums of length n , and two positive integers k and dist . 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 k disjoint contiguous subarrays , such that the difference between the starting index of the second subarray and the starting index of the kth subarray should be less than or equal to dist . In other words, if you divide nums into the subarra

Detailed Explanation

The problem requires dividing a given array `nums` of length `n` into `k` disjoint contiguous subarrays. The goal is to minimize the sum of the costs of these subarrays, where the cost of each subarray is defined as its first element. A crucial constraint is that the difference between the starting indices of the second and last (k-th) subarrays must be less than or equal to `dist`. The first subarray always starts at index 0. The inputs are the array `nums`, the number of subarrays `k`, and the maximum distance `dist`. The output is the minimum possible sum of the costs of the `k` subarrays.

Solution Approach

The provided code uses a sliding window approach combined with two heaps (priority queues) to efficiently find the `k-1` smallest elements within each window of size `dist + 1` in the subarray `nums[1:]`. The algorithm maintains a max-heap (`small_heap`) containing the `k-1` smallest elements seen so far within the current window, and a min-heap (`large_heap`) containing the remaining elements of the window. The code uses a `to_remove` counter to handle cases where elements that are no longer within the window need to be removed from the heaps efficiently. By maintaining these heaps, we can easily track the sum of the `k-1` smallest elements in the current window, and update them efficiently as the window slides.

Step-by-Step Algorithm

Step 1: Initialize variables: Calculate `m = k - 1` (the number of additional subarrays to create) and `window_size = dist + 1`.
Step 2: Create subarrays: Extract the subarray `v = nums[1:]` for processing the sliding window.
Step 3: Initialize heaps and counter: Create a max-heap `small_heap` to store the `m` smallest elements in the current window, a min-heap `large_heap` to store the remaining elements, and a counter `to_remove` to track the counts of elements that need to be removed from the heaps.
Step 4: Initialize the first window: Populate the heaps and `current_sum` using the first window of size `window_size` in the array `v`. `current_sum` stores the sum of the elements in the `small_heap` (the m smallest elements in current window).
Step 5: Slide the window: Iterate from index `window_size` to the end of the array `v`. In each iteration, process the element being removed from the window (`rem_val`) and the element being added to the window (`add_val`).
Step 6: Update heaps: Remove `rem_val` if it is in either heap (using `to_remove` counter) and add `add_val` to the appropriate heap based on its value relative to the maximum value in `small_heap`.
Step 7: Rebalance heaps: Ensure that `small_heap` contains exactly `m` elements. If it contains fewer, transfer elements from `large_heap`. If it contains more, transfer elements to `large_heap`. Update `current_sum` accordingly.
Step 8: Update minimum sum: After each window slide and heap rebalancing, update `min_sum` with the minimum of the current `min_sum` and `current_sum`.
Step 9: Return result: Finally, return the sum of `nums[0]` and `min_sum`.

Key Insights

Insight 1: The first subarray always starts at index 0, reducing the problem to selecting the best `k-1` starting indices from the rest of the array.
Insight 2: The constraint `i_{k-1} - i_1 <= dist` implies that all the selected `k-1` indices must fall within a sliding window of size `dist + 1` in the subarray `nums[1:]`.
Insight 3: We need to efficiently find the `k-1` smallest elements within the sliding window. This is a classic problem for heaps (priority queues).

Complexity Analysis

Time Complexity: O(n*dist*log(dist))

Space Complexity: O(dist)

Topics

This problem involves: Array, Hash Table, Sliding Window, Heap (Priority Queue).

Companies

Asked at: American Express, jio.

Divide an Array Into Subarrays With Minimum Cost II | LeetCode 3013

Hard

Array Hash Table Sliding Window Heap (Priority Queue)

Time: O(n*dist*log(dist))Space: O(dist)

Solution Code

Available in:

import java.util.*;

class Solution {
    public long minimumCost(int[] nums, int k, int dist) {
        // The first subarray always starts at index 0, so its cost is nums[0].
        // We need to choose k-1 other start indices i_1, ..., i_{k-1}
        // from nums[1:] such that i_{k-1} - i_1 <= dist.
        // This means all k-1 indices must fall within a window of size dist + 1.
        // We want to find a window of this size in nums[1:] that contains
        // k-1 elements with the minimum possible sum.

        int m = k - 1;
        int window_size = dist + 1;

        if (m == 0) {
            return nums[0];
        }

        int[] v = Arrays.copyOfRange(nums, 1, nums.length);
        int n_v = v.length;

        PriorityQueue<Integer> small_heap = new PriorityQueue<>(Comparator.reverseOrder());  // max-heap for the m smallest elements
        PriorityQueue<Integer> large_heap = new PriorityQueue<>();  // min-heap for the rest
        Map<Integer, Integer> to_remove = new HashMap<>();

        Runnable clean_small = () -> {
            while (!small_heap.isEmpty() && to_remove.getOrDefault(-small_heap.peek(), 0) > 0) {
                int val = -small_heap.poll();
                to_remove.put(val, to_remove.getOrDefault(val, 0) - 1);
                if (to_remove.get(val) == 0) {
                    to_remove.remove(val);
                }
            }
        };

        Runnable clean_large = () -> {
            while (!large_heap.isEmpty() && to_remove.getOrDefault(large_heap.peek(), 0) > 0) {
                int val = large_heap.poll();
                to_remove.put(val, to_remove.getOrDefault(val, 0) - 1);
                if (to_remove.get(val) == 0) {
                    to_remove.remove(val);
                }
            }
        };

        // Initialize with the first window
        int[] initial_window = Arrays.copyOfRange(v, 0, Math.min(window_size, n_v));
        Arrays.sort(initial_window);

        long current_sum = 0;
        for (int i = 0; i < Math.min(m, initial_window.length); i++) {
            int val = initial_window[i];
            small_heap.add(-val);
            current_sum += val;
        }

        for (int i = m; i < initial_window.length; i++) {
            large_heap.add(initial_window[i]);
        }

        long min_sum = current_sum;
        int small_valid_count = Math.min(m, initial_window.length);

        // Slide the window over v
        for (int i = window_size; i < n_v; i++) {
            int rem_val = v[i - window_size];
            int add_val = v[i];

            clean_small.run();
            int boundary = small_heap.isEmpty() ? Integer.MAX_VALUE : -small_heap.peek();

            // Process rem_val
            to_remove.put(rem_val, to_remove.getOrDefault(rem_val, 0) + 1);
            if (rem_val <= boundary) {
                small_valid_count--;
                current_sum -= rem_val;
            }

            // Process add_val
            if (add_val < boundary) {
                small_valid_count++;
                current_sum += add_val;
                small_heap.add(-add_val);
            } else {
                large_heap.add(add_val);
            }

            // Rebalance heaps to maintain m valid elements in small_heap
            while (small_valid_count < m) {
                clean_large.run();
                if (large_heap.isEmpty()) break;
                int val = large_heap.poll();
                current_sum += val;
                small_valid_count++;
                small_heap.add(-val);
            }

            while (small_valid_count > m) {
                clean_small.run();
                if (small_heap.isEmpty()) break;
                int val = -small_heap.poll();
                current_sum -= val;
                small_valid_count--;
                large_heap.add(val);
            }

            min_sum = Math.min(min_sum, current_sum);
        }

        return nums[0] + min_sum;
    }
}

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Problem Info

Description

Problem 3013: Divide an Array Into Subarrays With Minimum Cost II - Detailed explanation and solution approaches.

Topics

Array Hash Table Sliding Window Heap (Priority Queue)

Companies

American Expressjio

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:3013

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

9 steps

1
Step 1: Initialize variables: Calculate `m = k - 1` (the number of additional subarrays to create) and `window_size = dist + 1`.
2
Step 2: Create subarrays: Extract the subarray `v = nums[1:]` for processing the sliding window.
3
Step 3: Initialize heaps and counter: Create a max-heap `small_heap` to store the `m` smallest elements in the current window, a min-heap `large_heap` to store the remaining elements, and a counter `to_remove` to track the counts of elements that need to be removed from the heaps.
4
Step 4: Initialize the first window: Populate the heaps and `current_sum` using the first window of size `window_size` in the array `v`. `current_sum` stores the sum of the elements in the `small_heap` (the m smallest elements in current window).
5
Step 5: Slide the window: Iterate from index `window_size` to the end of the array `v`. In each iteration, process the element being removed from the window (`rem_val`) and the element being added to the window (`add_val`).
6
Step 6: Update heaps: Remove `rem_val` if it is in either heap (using `to_remove` counter) and add `add_val` to the appropriate heap based on its value relative to the maximum value in `small_heap`.
7
Step 7: Rebalance heaps: Ensure that `small_heap` contains exactly `m` elements. If it contains fewer, transfer elements from `large_heap`. If it contains more, transfer elements to `large_heap`. Update `current_sum` accordingly.
8
Step 8: Update minimum sum: After each window slide and heap rebalancing, update `min_sum` with the minimum of the current `min_sum` and `current_sum`.
9
Step 9: Return result: Finally, return the sum of `nums[0]` and `min_sum`.

💡

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*dist*log(dist))

💾

Space Complexity

O(dist)

📝

Why?

The time complexity is dominated by the sliding window and the heap operations. The space complexity is determined by the size of the heaps and the counter which are proportional to `dist`.

📖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

Sorting within each window

Instead of using heaps, we could sort the elements within each sliding window of size `dist + 1` and then sum the first `k-1` elements. However, sort operation takes O(dist*log(dist)) time.

Time: O(n * dist * log(dist))Space: O(dist)

Trade-offs: This approach is simpler to implement but has a higher time complexity than the heap-based solution. Therefore, the heap based solution is preferable.

Using Quickselect within each window

Instead of full sorting, we could use Quickselect to find the k-1 smallest elements. Quickselect can achieve average O(dist) but worst case O(dist^2).

Time: Average O(n * dist), Worst case O(n * dist^2)Space: O(dist)

Trade-offs: This can potentially perform better than the sort solution but has same/worse complexity than Heap approach and has worst case O(n * dist^2), thus making heap based approach preferable due to consistent time complexity.

✨

Key Insights

1Insight 1: The first subarray always starts at index 0, reducing the problem to selecting the best `k-1` starting indices from the rest of the array.
2Insight 2: The constraint `i_{k-1} - i_1 <= dist` implies that all the selected `k-1` indices must fall within a sliding window of size `dist + 1` in the subarray `nums[1:]`.
3Insight 3: We need to efficiently find the `k-1` smallest elements within the sliding window. This is a classic problem for heaps (priority queues).

📚

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!

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: `k = 1`. In this case, the solution returns `nums[0]` directly because only the first element matters.
Edge case 2: `dist` is very small. The solution still works correctly as long as the heaps can maintain the `k-1` smallest numbers. The space complexity will reduce in this scenario.
Edge case 3: `nums` contains duplicate values. The `to_remove` counter handles duplicates correctly by keeping track of the number of times each value has been removed and adjusting the heaps accordingly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to rebalance the heaps after adding or removing elements, leading to incorrect `current_sum` values.

✗Common mistake 2: Not handling duplicates correctly, which can lead to incorrect counts of elements in the `small_heap` and `large_heap`.

✗Common mistake 3: Incorrectly calculating the boundary between the `small_heap` and `large_heap` during the sliding window process.

✗Common mistake 4: Integer overflow when calculating the sum of costs in long arrays. It is necessary to use `long long` variables.

✅

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 allocating resources to `k` projects from a stream of potential projects. The 'cost' could be the resource requirement, and the 'distance' could be a time constraint. The goal would be to minimize the total resource cost while meeting the time constraints.

🏦Real-world scenario 2: Data stream analysis. Identifying the `k` smallest values within a sliding window of a data stream, such as finding the `k` lowest latency servers in a network over a specific time period.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: October 12, 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

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

@CodeNinjaJul 13

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

@TechLeadJul 6

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

@DevExpert92Jun 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 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 "Divide an Array Into Subarrays With Minimum Cost II"?

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.

🤖 Need help with this problem?