Introduction

In this blog post, we will explore how to solve the LeetCode problem "2172" 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

You are given an integer array nums of length n and an integer numSlots such that 2 numSlots >= n. There are numSlots slots numbered from 1 to numSlots. You have to place all n integers into the slots such that each slot contains at most two numbers. The AND sum of a given placement is the sum of the bitwise AND of every number with its respective slot number. For example, the AND sum of placing the numbers [1, 3] into slot 1 and [4, 6] into slot 2 is equal to (1 AND 1) + (3 AND 1) + (4 AND 2) + (6 AND 2) = 1 + 1 + 0 + 2 = 4. Return the maximum possible AND sum of nums given numSlots slots. Example 1: Input: nums = [1,2,3,4,5,6], numSlots = 3 Output: 9 Explanation: One possible placement is [1, 4] into slot 1, [2, 6] into slot 2, and [3, 5] into slot 3. This gives the maximum AND sum of (1 AND 1) + (4 AND 1) + (2 AND 2) + (6 AND 2) + (3 AND 3) + (5 AND 3) = 1 + 0 + 2 + 2 + 3 + 1 = 9. Example 2: Input: nums = [1,3,10,4,7,1], numSlots = 9 Output: 24 Explanation: One possible placement is [1, 1] into slot 1, [3] into slot 3, [4] into slot 4, [7] into slot 7, and [10] into slot 9. This gives the maximum AND sum of (1 AND 1) + (1 AND 1) + (3 AND 3) + (4 AND 4) + (7 AND 7) + (10 AND 9) = 1 + 1 + 3 + 4 + 7 + 8 = 24. Note that slots 2, 5, 6, and 8 are empty which is permitted. Constraints: n == nums.length 1 <= numSlots <= 9 1 <= n <= 2 numSlots 1 <= nums[i] <= 15

Explanation

Here's a breakdown of the solution:

• Dynamic Programming with Bitmasking: Represent the state of used numbers with a bitmask. The DP state dp[mask] stores the maximum AND sum achievable using the numbers represented by the bits set in mask.
• Iterate Through Slots: For each slot, try placing two, one, or zero available numbers in that slot. The 'available' numbers are determined by the numbers not yet included in the current bitmask.

• Maximize AND Sum: Update the dp table by considering all possible placements and choosing the one that maximizes the AND sum.

• Runtime Complexity: O(numSlots * n^2 * 2^n), Storage Complexity: O(2^n)

Code

    def maxANDSum(nums, numSlots):
    n = len(nums)
    dp = {}

    def solve(mask, slot):
        if slot > numSlots:
            return 0

        if (mask, slot) in dp:
            return dp[(mask, slot)]

        ans = 0
        available_nums = []
        for i in range(n):
            if not (mask & (1 << i)):
                available_nums.append(i)

        # No numbers in this slot
        ans = max(ans, solve(mask, slot + 1))

        # One number in this slot
        for i in range(len(available_nums)):
            num_index = available_nums[i]
            new_mask = mask | (1 << num_index)
            ans = max(ans, (nums[num_index] & slot) + solve(new_mask, slot + 1))

        # Two numbers in this slot
        for i in range(len(available_nums)):
            for j in range(i + 1, len(available_nums)):
                num1_index = available_nums[i]
                num2_index = available_nums[j]
                new_mask = mask | (1 << num1_index) | (1 << num2_index)
                ans = max(ans, (nums[num1_index] & slot) + (nums[num2_index] & slot) + solve(new_mask, slot + 1))

        dp[(mask, slot)] = ans
        return ans

    return solve(0, 1)