Problem Description

You have an integer array nums with n elements and a number of slots numSlots. The constraint guarantees that 2 * numSlots >= n, meaning you have enough space to place all numbers.

Each slot is numbered from 1 to numSlots, and each slot can hold at most two numbers from the array.

Your task is to distribute all n numbers from nums into these slots to maximize the AND sum.

The AND sum is calculated by:

• Taking each number you've placed
• Performing a bitwise AND operation between that number and its slot number
• Summing all these AND results

For example, if you place numbers [1, 3] in slot 1 and [4, 6] in slot 2:

• 1 AND 1 = 1
• 3 AND 1 = 1
• 4 AND 2 = 0
• 6 AND 2 = 2
• Total AND sum = 1 + 1 + 0 + 2 = 4

The goal is to find the optimal placement of all numbers that gives you the maximum possible AND sum.

The solution uses dynamic programming with bitmask to track which "positions" (considering each slot has 2 positions) are filled. The state f[i] represents the maximum AND sum when the positions indicated by bitmask i are occupied. For each state, it tries removing each number and calculates the best possible sum by considering which slot position that number was in.

Intuition

The key insight is that we need to try different assignments of numbers to slots to find the maximum AND sum. Since each slot can hold up to 2 numbers and we have numSlots slots, we essentially have 2 * numSlots positions available.

We can think of this as having 2 * numSlots distinct positions where we can place our numbers. Position 0 and 1 belong to slot 1, positions 2 and 3 belong to slot 2, and so on. For position j, the corresponding slot number is j // 2 + 1.

The challenge is exploring all possible placements efficiently. This is where bitmask dynamic programming comes in. We can use a bitmask of length 2 * numSlots where each bit represents whether a position is occupied or not.

For a given bitmask i, the number of 1 bits tells us how many numbers from nums we've placed so far. If we've placed cnt numbers, they must be the first cnt numbers from the array (we can process them in order since the final AND sum doesn't depend on which specific number goes where, only on the pairing of numbers with slots).

The dynamic programming approach works backwards: for each state with cnt numbers placed, we try removing the last placed number (nums[cnt-1]) from each possible position. If position j is occupied in the current state (bit j is set), we can compute what the maximum sum would be if we remove that number from position j. The AND contribution of that number would be nums[cnt-1] & (j // 2 + 1).

By iterating through all valid states (where the number of placed items doesn't exceed n) and trying all possible last placements, we build up the maximum AND sum for each configuration. The answer is the maximum value among all states in our DP table.

Learn more about Dynamic Programming and Bitmask patterns.

Solution Approach

The solution uses bitmask dynamic programming to explore all possible placements of numbers into slots.

State Representation:

• We create an array f of size 1 << m where m = numSlots * 2 (total positions available)
• f[i] stores the maximum AND sum achievable when the positions indicated by bitmask i are occupied
• Each bit in the bitmask represents whether a position is filled (1) or empty (0)

Implementation Steps:

1. Initialize variables:

   • n = len(nums) - number of elements to place
   • m = numSlots << 1 - total positions (2 per slot)
   • f = [0] * (1 << m) - DP array initialized with zeros

2. Iterate through all possible states:

   for i in range(1 << m):

   Each i represents a different configuration of filled positions.

3. Count placed numbers:

   cnt = i.bit_count()
   if cnt > n:
       continue

   • bit_count() returns the number of 1s in the bitmask
   • Skip states where we've placed more numbers than available

4. Try removing each placed number:

   for j in range(m):
       if i >> j & 1:

   • Check if position j is occupied in state i
   • If yes, calculate the maximum sum by considering this as the last placement

5. Update DP value:

   f[i] = max(f[i], f[i ^ (1 << j)] + (nums[cnt - 1] & (j // 2 + 1)))

   • i ^ (1 << j) gives the previous state with position j empty
   • nums[cnt - 1] is the last number placed (we place numbers in order)
   • j // 2 + 1 converts position index to slot number (positions 0,1 → slot 1, positions 2,3 → slot 2, etc.)
   • The AND operation nums[cnt - 1] & (j // 2 + 1) calculates the contribution

6. Return the maximum:

   return max(f)

   The answer is the maximum value across all valid states in the DP table.

Time Complexity: O(2^(2*numSlots) * 2*numSlots) - we iterate through all states and for each state, check all positions.

Space Complexity: O(2^(2*numSlots)) - for storing the DP array.

Example Walkthrough

Let's walk through a small example with nums = [3, 5] and numSlots = 2.

Setup:

• We have 2 numbers to place: [3, 5]
• We have 2 slots (slot 1 and slot 2), each can hold up to 2 numbers
• Total positions available: m = 2 * 2 = 4 positions
  • Positions 0,1 belong to slot 1
  • Positions 2,3 belong to slot 2
• DP array size: 2^4 = 16 states

Key States Walkthrough:

Initially, f[0000] = 0 (no numbers placed, AND sum = 0)

Placing 1 number (cnt = 1, placing nums[0] = 3):

• State 0001 (position 0 filled):
  • f[0001] = f[0000] + (3 & 1) = 0 + 1 = 1
• State 0010 (position 1 filled):
  • f[0010] = f[0000] + (3 & 1) = 0 + 1 = 1
• State 0100 (position 2 filled):
  • f[0100] = f[0000] + (3 & 2) = 0 + 2 = 2
• State 1000 (position 3 filled):
  • f[1000] = f[0000] + (3 & 2) = 0 + 2 = 2

Placing 2 numbers (cnt = 2, placing nums[1] = 5):

• State 0011 (positions 0,1 filled - both in slot 1):
  • From 0001: f[0011] = f[0001] + (5 & 1) = 1 + 1 = 2
  • From 0010: f[0011] = f[0010] + (5 & 1) = 1 + 1 = 2
• State 0101 (positions 0,2 filled - one in each slot):
  • From 0001: f[0101] = f[0001] + (5 & 2) = 1 + 0 = 1
  • From 0100: f[0101] = f[0100] + (5 & 1) = 2 + 1 = 3
  • Maximum: f[0101] = 3
• State 0110 (positions 1,2 filled):
  • From 0010: f[0110] = f[0010] + (5 & 2) = 1 + 0 = 1
  • From 0100: f[0110] = f[0100] + (5 & 1) = 2 + 1 = 3
  • Maximum: f[0110] = 3
• State 1001 (positions 0,3 filled):
  • From 0001: f[1001] = f[0001] + (5 & 2) = 1 + 0 = 1
  • From 1000: f[1001] = f[1000] + (5 & 1) = 2 + 1 = 3
  • Maximum: f[1001] = 3
• State 1100 (positions 2,3 filled - both in slot 2):
  • From 0100: f[1100] = f[0100] + (5 & 2) = 2 + 0 = 2
  • From 1000: f[1100] = f[1000] + (5 & 2) = 2 + 0 = 2

Result: The maximum value in the DP array is 3, which occurs when:

• We place 3 in slot 2 (3 & 2 = 2)
• We place 5 in slot 1 (5 & 1 = 1)
• Total AND sum = 2 + 1 = 3

This demonstrates how the algorithm explores all possible placements and finds the optimal assignment by building up from smaller states to larger ones.

Solution Implementation

Time and Space Complexity

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

The algorithm iterates through all possible states represented by a bitmask of size 2 * numSlots. For each state i in the range [0, 2^(2*numSlots)), it performs the following operations:

• bit_count() operation: O(1) in Python 3.10+
• Inner loop iterating through m = 2 * numSlots positions: O(numSlots)
• Inside the inner loop: bit operations and array access are O(1)

Therefore, the total time complexity is O(2^(2*numSlots) * numSlots).

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

The algorithm uses:

• Array f of size 2^(2*numSlots) to store the DP states: O(2^(2*numSlots))
• A few auxiliary variables (n, m, cnt, i, j): O(1)

The dominant factor is the DP array, resulting in a space complexity of O(2^(2*numSlots)).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Incorrect Slot Number Calculation

Problem: A common mistake is incorrectly mapping position indices to slot numbers. Since each slot has 2 positions, developers often confuse the zero-based position indexing with one-based slot numbering.

Incorrect Implementation:

# Wrong: forgetting to add 1 for 1-based slot numbering
slot_number = position // 2  # This gives 0-based slot index

# Wrong: incorrect division for position-to-slot mapping
slot_number = position / 2 + 1  # Using float division instead of integer

Correct Implementation:

slot_number = (position // 2) + 1  # Correctly converts to 1-based slot number

Pitfall 2: Misunderstanding Number Placement Order

Problem: The algorithm assumes numbers are placed in the order they appear in the input array. Some developers mistakenly think they can place numbers in any order, leading to incorrect indexing when accessing nums[filled_positions - 1].

Why This Matters:

• The DP state only tracks which positions are filled, not which specific numbers are in those positions
• By placing numbers in order, we know that if k positions are filled, we've placed nums[0] through nums[k-1]
• Trying to optimize by reordering numbers would require a completely different state representation

Solution: Always place numbers sequentially from the input array. If you want to explore different orderings, you'd need to modify the entire DP approach to track which specific numbers have been placed.

Pitfall 3: Overflow with Large numSlots

Problem: The space complexity is O(2^(2*numSlots)), which grows exponentially. For numSlots = 16, this requires 2^32 array elements, consuming gigabytes of memory.

Symptoms:

• Memory allocation failures
• Program crashes or timeouts
• Python MemoryError exceptions

Solution:

1. Check problem constraints carefully - most problems limit numSlots to reasonable values (typically ≤ 9)
2. If larger values are needed, consider alternative approaches like branch-and-bound or greedy heuristics
3. Add validation:

if numSlots > 15:  # Practical limit for bitmask DP
    raise ValueError("numSlots too large for bitmask DP approach")

Pitfall 4: Confusion About State Representation

Problem: Misunderstanding what each bit in the mask represents. Some developers think each bit represents a slot (not a position), leading to incorrect state transitions.

Incorrect Mental Model:

• Thinking mask bit i represents whether slot i has any numbers
• This would only allow tracking whether slots are used, not how many numbers each contains

Correct Understanding:

• Each slot has exactly 2 positions (even if not fully utilized)
• Positions 0,1 belong to slot 1; positions 2,3 belong to slot 2, etc.
• The mask tracks individual positions, allowing us to know if a slot has 0, 1, or 2 numbers

Debugging Tip: Add helper visualization:

def visualize_mask(mask, total_positions):
    positions = []
    for i in range(total_positions):
        if (mask >> i) & 1:
            slot = (i // 2) + 1
            pos_in_slot = i % 2
            positions.append(f"Slot{slot}_Pos{pos_in_slot}")
    return positions