What is the time complexity of House Robber IV?

The optimal solution has a time complexity of O(nlog(m)).

What data structures are used to solve House Robber IV?

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

What are the key insights for solving House Robber IV?

Insight 1: The problem can be solved using binary search because we are looking for a minimum value (capability) within a defined range (minimum to maximum value in nums). Insight 2: We need to check if a given 'capability' is feasible, i.e., whether we can rob at least 'k' houses with that capability without robbing adjacent houses.

Which companies ask LeetCode 2560. House Robber IV in interviews?

This problem is frequently asked at Cashfree, LinkedIn.

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House Robber IV - LeetCode 2560 Solution

House Robber IV - Complete Solution Guide

House Robber IV is LeetCode problem 2560, 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

There are several consecutive houses along a street, each of which has some money inside. There is also a robber, who wants to steal money from the homes, but he refuses to steal from adjacent homes . The capability of the robber is the maximum amount of money he steals from one house of all the houses he robbed. You are given an integer array nums representing how much money is stashed in each house. More formally, the i th house from the left has nums[i] dollars. You are also given an integer

Detailed Explanation

The problem asks us to find the minimum capability required for a robber to steal from at least 'k' houses in a row of houses, such that the robber never steals from adjacent houses. The 'capability' is defined as the maximum amount of money the robber steals from any single house he robbed. The input is an array 'nums' representing the amount of money in each house, and an integer 'k' representing the minimum number of houses to steal from. The output is the minimum capability the robber needs to achieve this.

Solution Approach

The solution uses binary search to find the minimum capability. First, we determine the range for binary search, which is the minimum and maximum values in the 'nums' array. Then, for each 'mid' value within this range (our current guess for the minimum capability), we use a 'check' function to determine if it's possible to rob at least 'k' houses with that capability. If it is, we try a lower capability (move 'high' to 'mid - 1'); otherwise, we try a higher capability (move 'low' to 'mid + 1'). The binary search continues until 'low' becomes greater than 'high'.

Step-by-Step Algorithm

Step 1: Find the minimum and maximum values in the 'nums' array. These will be the initial 'low' and 'high' for the binary search.
Step 2: Initialize 'ans' to 'high'. This will hold the current best answer.
Step 3: Start a while loop that continues as long as 'low' is less than or equal to 'high'.
Step 4: Calculate the 'mid' value as the average of 'low' and 'high' (using the formula `low + (high - low) / 2` to avoid overflow).
Step 5: Call the 'check' function with 'mid' as the capability. The 'check' function determines if we can rob at least 'k' houses with the given capability.
Step 6: If 'check(mid)' returns true, it means we can rob at least 'k' houses with the current capability. Update 'ans' to 'mid', and set 'high' to 'mid - 1' to try to find a smaller capability.
Step 7: If 'check(mid)' returns false, it means we cannot rob at least 'k' houses with the current capability. Set 'low' to 'mid + 1' to try a larger capability.
Step 8: After the while loop finishes, 'ans' will hold the minimum capability that allows us to rob at least 'k' houses. Return 'ans'.

Key Insights

Insight 1: The problem can be solved using binary search because we are looking for a minimum value (capability) within a defined range (minimum to maximum value in nums).
Insight 2: We need to check if a given 'capability' is feasible, i.e., whether we can rob at least 'k' houses with that capability without robbing adjacent houses.
Insight 3: The check function can be implemented using a greedy approach. If the current house's value is less than or equal to the 'capability', we rob it and skip the next house. Otherwise, we skip the current house.

Complexity Analysis

Time Complexity: O(nlog(m))

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Dynamic Programming, Greedy.

Companies

Asked at: Cashfree, LinkedIn.

House Robber IV | LeetCode 2560

Medium

Array Binary Search Dynamic Programming Greedy

Time: O(nlog(m))Space: O(1)

Solution Code

Available in:

class Solution {
    public int minCapability(int[] nums, int k) {
        int low = Integer.MAX_VALUE;
        int high = Integer.MIN_VALUE;
        for (int num : nums) {
            low = Math.min(low, num);
            high = Math.max(high, num);
        }

        int ans = high;

        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (check(nums, k, mid)) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }

        return ans;
    }

    private boolean check(int[] nums, int k, int capability) {
        int count = 0;
        int i = 0;
        int n = nums.length;
        while (i < n) {
            if (nums[i] <= capability) {
                count += 1;
                i += 2;
            } else {
                i += 1;
            }
        }
        return count >= k;
    }
}

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Problem Info

Description

Problem 2560: House Robber IV - Detailed explanation and solution approaches.

Topics

Array Binary Search Dynamic Programming Greedy

Companies

CashfreeLinkedIn

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2560

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: Find the minimum and maximum values in the 'nums' array. These will be the initial 'low' and 'high' for the binary search.
2
Step 2: Initialize 'ans' to 'high'. This will hold the current best answer.
3
Step 3: Start a while loop that continues as long as 'low' is less than or equal to 'high'.
4
Step 4: Calculate the 'mid' value as the average of 'low' and 'high' (using the formula `low + (high - low) / 2` to avoid overflow).
5
Step 5: Call the 'check' function with 'mid' as the capability. The 'check' function determines if we can rob at least 'k' houses with the given capability.
6
Step 6: If 'check(mid)' returns true, it means we can rob at least 'k' houses with the current capability. Update 'ans' to 'mid', and set 'high' to 'mid - 1' to try to find a smaller capability.
7
Step 7: If 'check(mid)' returns false, it means we cannot rob at least 'k' houses with the current capability. Set 'low' to 'mid + 1' to try a larger capability.
8
Step 8: After the while loop finishes, 'ans' will hold the minimum capability that allows us to rob at least 'k' houses. Return 'ans'.

💡

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(nlog(m))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The binary search algorithm's complexity is O(log(m)), where 'm' is the difference between the maximum and minimum element in the array. For each mid-point in the binary search, the `check` function iterates through the `nums` array in O(n) time. Thus the overall time complexity is O(n * log(m)). The space complexity is O(1) because we use only a few integer variables, regardless of the size of the input array.

📖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

A dynamic programming approach could potentially be used, but would be much more complex. It would likely involve iterating through the houses and tracking the minimum capability needed to rob 'i' houses up to the current house.

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

Trade-offs: The DP approach would be less efficient in terms of both time and space complexity compared to the binary search approach. The Binary Search solution is preferable.

✨

Key Insights

1Insight 1: The problem can be solved using binary search because we are looking for a minimum value (capability) within a defined range (minimum to maximum value in nums).
2Insight 2: We need to check if a given 'capability' is feasible, i.e., whether we can rob at least 'k' houses with that capability without robbing adjacent houses.
3Insight 3: The check function can be implemented using a greedy approach. If the current house's value is less than or equal to the 'capability', we rob it and skip the next house. Otherwise, we skip the current house.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: When all houses have the same amount of money. The binary search will still converge correctly to the minimum capability, which is that amount of money.
Edge case 2: When k is equal to (nums.length + 1) / 2 (the maximum possible number of houses to rob). The algorithm will still find the minimum capability necessary to rob that many houses without robbing adjacent ones.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using binary search correctly. Specifically, not handling the 'low' and 'high' pointers correctly during the binary search, which can lead to an infinite loop or incorrect results.

✗Common mistake 2: Incorrectly implementing the 'check' function. Failing to properly account for the constraint that adjacent houses cannot be robbed can lead to an inaccurate assessment of whether a given capability is feasible.

✗Common mistake 3: Integer overflow when calculating the `mid` value during binary search. Using `mid = low + (high - low) / 2` is safer than `mid = (low + high) / 2`.

✅

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 with constraints. Imagine allocating bandwidth to users, where each user needs a minimum bandwidth but adjacent users cannot share bandwidth efficiently. This problem helps determine the minimum bandwidth required to satisfy a certain number of users.

🏦Real-world scenario 2: Selecting optimal locations for cellular towers to provide coverage to a certain number of areas, where adjacent locations might interfere with each other. The 'capability' here would be the power of the tower, and the goal is to minimize the power needed while covering enough areas.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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 11, 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 14

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

@AlgoMasterJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "House Robber IV"?

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

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