Solution Explanation:

This problem asks for the minimum number of operations to remove all 1s from a binary matrix. In each operation, you select a cell with a 1, and set all cells in its row and column to 0. The solution uses Breadth-First Search (BFS) to explore the state space efficiently.

Approach:

1. State Representation: Each state of the matrix is represented as a single integer. Each bit in the integer corresponds to a cell in the matrix. If the bit is set (1), the cell contains a 1; otherwise, it contains a 0. This allows for efficient checking and manipulation of the matrix state.

2. BFS: The algorithm starts with the initial state of the matrix (represented as an integer). BFS explores all possible states reachable from the initial state. Each state is added to a queue. The algorithm continues until a state with all 1s removed (represented by the integer 0) is found.

3. Operation Simulation: For each state in the queue, the algorithm iterates through all cells. If a cell contains a 1, it simulates the operation by flipping bits (setting them to 0) in the corresponding row and column. This creates a new state.

4. Visited States: To avoid cycles and redundant exploration, the algorithm keeps track of visited states using a set. A new state is only added to the queue if it hasn't been visited.

5. Minimum Operations: The number of BFS levels required to reach the state with all 1s removed represents the minimum number of operations needed.

Time Complexity Analysis:

• The number of possible states is 2(m*n), where 'm' and 'n' are the dimensions of the matrix. In the worst-case scenario, BFS might explore all possible states.
• Each state transition involves iterating through the matrix (O(m*n) time).
• Therefore, the overall time complexity is approximately O(mn * 2(mn)). However, because m*n <= 15, the actual runtime is relatively fast in practice, even though the theoretical complexity looks exponential.

Space Complexity Analysis:

• The space complexity is dominated by the size of the visited states set and the queue. In the worst case, the set and queue could hold up to 2(m*n) elements.
• Therefore, the space complexity is O(2(mn)), but again, the constraints (mn <= 15) keep the space usage manageable in this problem.

Code Explanation (Python):

class Solution:
    def removeOnes(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        state = sum(1 << (i * n + j) for i in range(m) for j in range(n) if grid[i][j]) #Initial state representation.
        q = deque([state])
        vis = {state}
        ans = 0
        while q:
            for _ in range(len(q)): #Process all states at the current level.
                state = q.popleft()
                if state == 0: #Goal state reached.
                    return ans
                for i in range(m):
                    for j in range(n):
                        if grid[i][j] == 0:
                            continue
                        nxt = state
                        for r in range(m): #Flip row bits.
                            nxt &= ~(1 << (r * n + j))
                        for c in range(n): #Flip column bits.
                            nxt &= ~(1 << (i * n + c))
                        if nxt not in vis: #Add unvisited states to queue.
                            vis.add(nxt)
                            q.append(nxt)
            ans += 1 #Increment operation count after processing each BFS level.
        return -1 #Should not reach here with the given constraints.
 

The code in other languages (Java, C++, Go) follows the same logic and structure as the Python solution, differing only in syntax and data structures used. The core algorithm remains the same BFS approach to explore the state space.