Cell fusion in Python

Cell fusion is a problem where cells of different sizes interact according to specific rules. Given a list of cell sizes, we simulate iterations where the two largest cells interact: if they're equal, both die; otherwise, they merge into a new cell with size floor((a + b) / 3).

For example, with cells [20, 40, 40, 30], the two largest cells (40, 40) are equal so they die, leaving [20, 30]. These merge to form floor((20 + 30) / 3) = 16.

Algorithm

We use a max heap to efficiently find the two largest cells in each iteration ?

• Convert all values to negative (Python's heapq is a min heap)

• Create a heap from the modified list

• While heap has more than one element:

  • Extract two largest cells (convert back to positive)

  • If sizes differ, merge them and add result back to heap

• Return the last remaining cell size or -1 if none remain

Implementation

from heapq import heapify, heappop, heappush

class Solution:
    def solve(self, cells):
        # Convert to negative values for max heap behavior
        cells = [-x for x in cells]
        heapify(cells)
        
        while len(cells) > 1:
            # Extract two largest cells (convert back to positive)
            first, second = -heappop(cells), -heappop(cells)
            
            # If sizes are different, merge them
            if first != second:
                merged_size = (first + second) // 3
                heappush(cells, -merged_size)
        
        # Return last remaining cell or -1 if none
        return -cells[0] if cells else -1

# Test the solution
ob = Solution()
cells = [20, 40, 40, 30]
result = ob.solve(cells)
print(f"Input: {cells}")
print(f"Output: {result}")

Input: [20, 40, 40, 30]
Output: 16

How It Works

Let's trace through the example step by step ?

from heapq import heapify, heappop, heappush

def solve_with_trace(cells):
    print(f"Initial cells: {cells}")
    cells = [-x for x in cells]
    heapify(cells)
    iteration = 1
    
    while len(cells) > 1:
        first, second = -heappop(cells), -heappop(cells)
        print(f"Iteration {iteration}: Largest cells are {first} and {second}")
        
        if first != second:
            merged = (first + second) // 3
            heappush(cells, -merged)
            print(f"  ? Cells merge into {merged}")
        else:
            print(f"  ? Both cells die (equal sizes)")
        
        remaining = [-x for x in cells]
        print(f"  ? Remaining cells: {remaining}")
        iteration += 1
    
    result = -cells[0] if cells else -1
    print(f"Final result: {result}")
    return result

# Trace the example
solve_with_trace([20, 40, 40, 30])

Initial cells: [20, 40, 40, 30]
Iteration 1: Largest cells are 40 and 40
  ? Both cells die (equal sizes)
  ? Remaining cells: [30, 20]
Iteration 2: Largest cells are 30 and 20
  ? Cells merge into 16
  ? Remaining cells: [16]
Final result: 16

Key Points

• Max Heap Simulation: Python's heapq is a min heap, so we negate values to simulate a max heap

• Merge Formula: New cell size is (a + b) // 3 using integer division

• Equal Cells: When two cells have equal size, both are removed without creating a new cell

• Time Complexity: O(n log n) where n is the number of cells

Conclusion

Cell fusion problems are efficiently solved using heaps to repeatedly find the largest elements. The key insight is simulating a max heap with Python's min heap by negating values.