LeetCode #2560 — MEDIUM

House Robber IV

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

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 k, representing the minimum number of houses the robber will steal from. It is always possible to steal at least k houses.

Return the minimum capability of the robber out of all the possible ways to steal at least k houses.

Example 1:

Input: nums = [2,3,5,9], k = 2
Output: 5
Explanation: 
There are three ways to rob at least 2 houses:
- Rob the houses at indices 0 and 2. Capability is max(nums[0], nums[2]) = 5.
- Rob the houses at indices 0 and 3. Capability is max(nums[0], nums[3]) = 9.
- Rob the houses at indices 1 and 3. Capability is max(nums[1], nums[3]) = 9.
Therefore, we return min(5, 9, 9) = 5.

Example 2:

Input: nums = [2,7,9,3,1], k = 2
Output: 2
Explanation: There are 7 ways to rob the houses. The way which leads to minimum capability is to rob the house at index 0 and 4. Return max(nums[0], nums[4]) = 2.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
1 <= k <= (nums.length + 1)/2

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: 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 ith house from the left has nums[i] dollars. You are also given an integer k, representing the minimum number of houses the robber will steal from. It is always possible to steal at least k houses. Return the minimum capability of the robber out of all the possible ways to steal at least k houses.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Dynamic Programming · Greedy

Example 1

[2,3,5,9]
2

Example 2

[2,7,9,3,1]
2

Related Problems

Container With Most Water (container-with-most-water)
House Robber (house-robber)

Step 02

Core Insight

What unlocks the optimal approach

Can we use binary search to find the minimum value of a non-contiguous subsequence of a given size k?
Initialize the search range with the minimum and maximum elements of the input array.
Use a check function to determine if it is possible to select k non-consecutive elements that are less than or equal to the current "guess" value.
Adjust the search range based on the outcome of the check function, until the range converges and the minimum value is found.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2560: House Robber IV
class Solution {
    public int minCapability(int[] nums, int k) {
        int left = 0, right = (int) 1e9;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (f(nums, mid) >= k) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

    private int f(int[] nums, int x) {
        int cnt = 0, j = -2;
        for (int i = 0; i < nums.length; ++i) {
            if (nums[i] > x || i == j + 1) {
                continue;
            }
            ++cnt;
            j = i;
        }
        return cnt;
    }
}

// Accepted solution for LeetCode #2560: House Robber IV
func minCapability(nums []int, k int) int {
	return sort.Search(1e9+1, func(x int) bool {
		cnt, j := 0, -2
		for i, v := range nums {
			if v > x || i == j+1 {
				continue
			}
			cnt++
			j = i
		}
		return cnt >= k
	})
}

# Accepted solution for LeetCode #2560: House Robber IV
class Solution:
    def minCapability(self, nums: List[int], k: int) -> int:
        def f(x):
            cnt, j = 0, -2
            for i, v in enumerate(nums):
                if v > x or i == j + 1:
                    continue
                cnt += 1
                j = i
            return cnt >= k

        return bisect_left(range(max(nums) + 1), True, key=f)

// Accepted solution for LeetCode #2560: House Robber IV
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2560: House Robber IV
// class Solution {
//     public int minCapability(int[] nums, int k) {
//         int left = 0, right = (int) 1e9;
//         while (left < right) {
//             int mid = (left + right) >> 1;
//             if (f(nums, mid) >= k) {
//                 right = mid;
//             } else {
//                 left = mid + 1;
//             }
//         }
//         return left;
//     }
// 
//     private int f(int[] nums, int x) {
//         int cnt = 0, j = -2;
//         for (int i = 0; i < nums.length; ++i) {
//             if (nums[i] > x || i == j + 1) {
//                 continue;
//             }
//             ++cnt;
//             j = i;
//         }
//         return cnt;
//     }
// }

// Accepted solution for LeetCode #2560: House Robber IV
function minCapability(nums: number[], k: number): number {
    const f = (mx: number): boolean => {
        let cnt = 0;
        let j = -2;
        for (let i = 0; i < nums.length; ++i) {
            if (nums[i] <= mx && i - j > 1) {
                ++cnt;
                j = i;
            }
        }
        return cnt >= k;
    };

    let left = 1;
    let right = Math.max(...nums);
    while (left < right) {
        const mid = (left + right) >> 1;
        if (f(mid)) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log m)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.