Introduction

In this blog post, we will explore how to solve the LeetCode problem "2560" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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 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. 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

Explanation

Here's the breakdown of the solution:

• Binary Search: The core idea is to use binary search on the possible range of capabilities. The lower bound is the smallest value in nums, and the upper bound is the largest value in nums.
• Feasibility Check: For a given capability value mid, we check if it's possible to rob at least k houses such that the money stolen from each house is less than or equal to mid. This is done greedily by iterating through the houses and robbing if possible, ensuring we don't rob adjacent houses.

• Minimize Capability: We adjust the binary search range based on the feasibility. If it's possible to rob at least k houses with capability mid, we try a smaller capability. Otherwise, we try a larger capability.

• Time Complexity: O(n log(max(nums))), where n is the length of nums.

• Space Complexity: O(1)

Code

    def minCapability(nums, k):
    """
    Finds the minimum capability of the robber to steal from at least k houses.

    Args:
        nums (list[int]): An array of integers representing the money in each house.
        k (int): The minimum number of houses to steal from.

    Returns:
        int: The minimum capability.
    """

    def can_rob(capability):
        """
        Checks if it is possible to rob at least k houses with the given capability.

        Args:
            capability (int): The maximum amount of money the robber can steal from one house.

        Returns:
            bool: True if possible, False otherwise.
        """
        count = 0
        i = 0
        while i < len(nums):
            if nums[i] <= capability:
                count += 1
                i += 2  # Skip the next house as we can't rob adjacent houses
            else:
                i += 1
        return count >= k

    left = min(nums)
    right = max(nums)
    ans = right  # Initialize with the maximum possible capability

    while left <= right:
        mid = left + (right - left) // 2
        if can_rob(mid):
            ans = mid  # Update the answer
            right = mid - 1  # Try a smaller capability
        else:
            left = mid + 1  # Need a larger capability

    return ans