CareerCup

#include <iostream>
#include <queue>
#include <climits>
#include <cstring>
using namespace std;

// M x N matrix
#define M 10
#define N 10

// queue node used in BFS
struct Node
{
	// (x, y) represents matrix cell coordinates
	// dist represent its minimum distance from the source
	int x, y, dist;
};

// Below arrays details all 4 possible movements from a cell
int row[] = { -1, 0, 0, 1 };
int col[] = { 0, -1, 1, 0 };

// Function to check if it is possible to go to position (row, col)
// from current position. The function returns false if (row, col)
// is not a valid position or has value 0 or it is already visited
bool isValid(int mat[][N], bool visited[][N], int row, int col)
{
	return (row >= 0) && (row < M) && (col >= 0) && (col < N)
		&& mat[row][col] && !visited[row][col];
}

// Find Shortest Possible Route in a matrix mat from source
// cell (i, j) to destination cell (x, y)
void BFS(int mat[][N], int i, int j, int x, int y)
{
	// construct a matrix to keep track of visited cells
	bool visited[M][N];
	
	// initially all cells are unvisited
	memset(visited, false, sizeof visited);
	
	// create an empty queue
	queue<Node> q;

	// mark source cell as visited and enqueue the source node
	visited[i][j] = true;
	q.push({i, j, 0});

	// stores length of longest path from source to destination
	int min_dist = INT_MAX;

	// run till queue is not empty	
	while (!q.empty())
	{
		// pop front node from queue and process it
		Node node = q.front();
		q.pop();
		
		// (i, j) represents current cell and dist stores its
		// minimum distance from the source
		int i = node.x, j = node.y, dist = node.dist;

		// if destination is found, update min_dist and stop
		if (i == x && j == y)
		{
			min_dist = dist;
			break;
		}

		// check for all 4 possible movements from current cell
		// and enqueue each valid movement
		for (int k = 0; k < 4; k++)
		{
			// check if it is possible to go to position
			// (i + row[k], j + col[k]) from current position
			if (isValid(mat, visited, i + row[k], j + col[k]))
			{
				// mark next cell as visited and enqueue it
				visited[i + row[k]][j + col[k]] = true;
				q.push({ i + row[k], j + col[k], dist + 1 });
			}
		}
	}

	if (min_dist != INT_MAX)
		cout << "The shortest path from source to destination "
				"has length " << min_dist;
	else
		cout << "Destination can't be reached from given source";
}

// Shortest path in a Maze
int main()
{
	// input maze
	int mat[M][N] =
	{
		{ 1, 1, 1, 1, 1, 0, 0, 1, 1, 1 },
		{ 0, 1, 1, 1, 1, 1, 0, 1, 0, 1 },
		{ 0, 0, 1, 0, 1, 1, 1, 0, 0, 1 },
		{ 1, 0, 1, 1, 1, 0, 1, 1, 0, 1 },
		{ 0, 0, 0, 1, 0, 0, 0, 1, 0, 1 },
		{ 1, 0, 1, 1, 1, 0, 0, 1, 1, 0 },
		{ 0, 0, 0, 0, 1, 0, 0, 1, 0, 1 },
		{ 0, 1, 1, 1, 1, 1, 1, 1, 0, 0 },
		{ 1, 1, 1, 1, 1, 0, 0, 1, 1, 1 },
		{ 0, 0, 1, 0, 0, 1, 1, 0, 0, 1 },
	};

	// Find shortest path from source (0, 0) to
	// destination (7, 5)
	BFS(mat, 0, 0, 7, 5);

	return 0;
}

We need a node of bfs like
Node{
int row, col;
boolean flipped;
}

and visited boolean will be a 3D array
boolean visited[][][] = new boolean[2][rows][cols];

If we are on a node we will visit all the neighbors which are not visited and mark the visited [0][row][col] as done and also visit all the neighbours which are zero and push these node as flipped in queue. When visiting these nodes we need to check if visited[1][row][col] is done or not.

At the end we will reach with the minimum steps to reach the destination.

Note that the same problem can be done for k flips allowed as well, we will just need to increase the first dimension to k instead of 2 and similarly check for the destination.

MAX_FLIPS = 1
START,END = (0,0),(2,1)
INP = ["101",
       "101",
       "101",
       "111"]
N,M = len(INP[0]), len(INP)

cells = [(0,0)]*(N+2)*(M+2) # add border for simpler checks
for i,r in enumerate(INP):
    for j,x in enumerate(r):
        cells[(i+1)*(N+2)+j+1] = (int(x),0) # (cell_val, cell_state)
# i-bit set in cell_state means the cell was visited after i flipps

starti = START[0]+1 + (START[1]+1)*(N+2)
endi = END[0]+1 + (END[1]+1)*(N+2)

q = [(starti, 0, 0)] # (pos, steps, flips)
while q:
    (pos, steps, flips) = q.pop()
    cell_val, cell_state = cells[pos]
    if (pos % (N+2)) in [0, N+1] or (pos / (N+2)) in [0, M+1]: # skip border
        continue
    if cell_state & ((1 << (flips+1)) - 1) or flips == MAX_FLIPS and cell_val == 0:
        continue
    flips += (1-cell_val)
    cell_state |= (1 << flips)
    if pos == endi:
        print (steps)
        break
    cells[pos] = (cell_val, cell_state)
    q = [(pos-1, steps+1, flips),
         (pos+1, steps+1, flips),
         (pos-N-2, steps+1, flips),
         (pos+N+2, steps+1, flips)] + q