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Maximum AND Sum of Array

Difficulty: Hard

Problem Description

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. Return the maximum possible AND sum of nums given numSlots slots.

Key Insights

  • Each slot can hold at most two numbers, hence we need to consider combinations of numbers in different slots.
  • The bitwise AND operation can produce a maximum result when the numbers are aligned with their slot numbers.
  • The problem can be approached using dynamic programming or backtracking combined with bitmasking due to the limited number of slots (maximum of 9).

Space and Time Complexity

Time Complexity: O(n * 2^numSlots)
Space Complexity: O(numSlots)

Solution

The problem can be solved using a backtracking approach that explores all possible placements of numbers into the slots while keeping track of the maximum AND sum. A bitmask can be used to represent the numbers assigned to each slot, allowing us to efficiently manage and check the current state of slot assignments. We will recursively place each number in an available slot and calculate the potential AND sum.

Code Solutions

def maximumANDSum(nums, numSlots):
    from itertools import combinations
    
    n = len(nums)
    max_sum = 0
    
    # Recursive function to place numbers in slots
    def backtrack(slot, used, current_sum):
        nonlocal max_sum
        if slot > numSlots:
            max_sum = max(max_sum, current_sum)
            return
        
        # Try to place numbers in the current slot
        for i in range(n):
            if used[i] < 2:  # Check if the number can be used
                used[i] += 1
                backtrack(slot + 1, used, current_sum + (nums[i] & slot))
                used[i] -= 1
        
        # Move to next slot without using any number
        backtrack(slot + 1, used, current_sum)
    
    used = [0] * n
    backtrack(1, used, 0)
    return max_sum
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