Saturday, June 4, 2016

Find maximum length Snake sequence - GeeksforGeeks


Find maximum length Snake sequence - GeeksforGeeks
Given a grid of numbers, find maximum length Snake sequence and print it. If multiple snake sequences exists with the maximum length, print any one of them.
A snake sequence is made up of adjacent numbers in the grid such that for each number, the number on the right or the number below it is +1 or -1 its value. For example, if you are at location (x, y) in the grid, you can either move right i.e. (x, y+1) if that number is ± 1 or move down i.e. (x+1, y) if that number is ± 1.

The idea is to use Dynamic Programming. For each cell of the matrix, we keep maximum length of a snake which ends in current cell. The maximum length snake sequence will have maximum value. The maximum value cell will correspond to tail of the snake. In order to print the snake, we need to backtrack from tail all the way back to snake’s head.
Let T[i][i] represent maximum length of a snake which ends at cell (i, j), then for given matrix M, the DP relation is defined as –
T[0][0] = 0
T[i][j] = max(T[i][j], T[i][j – 1] + 1) if M[i][j] = M[i][j – 1] ± 1
T[i][j] = max(T[i][j], T[i – 1][j] + 1) if M[i][j] = M[i – 1][j] ± 1
Time complexity of above solution is O(M*N). Auxiliary space used by above solution is O(M*N). If we are not required to print the snake, space  can be further reduced to O(N) as we only uses the result from last row.

// Function to find maximum length Snake sequence path
// (i, j) corresponds to tail of the snake
list<Point> findPath(int grid[M][N], int mat[M][N],
                     int i, int j)
{
    list<Point> path;
 
    Point pt = {i, j};
    path.push_front(pt);
 
    while (grid[i][j] != 0)
    {
       if (i > 0 &&
           grid[i][j] - 1 == grid[i - 1][j])
       {
           pt = {i - 1, j};
           path.push_front(pt);
           i--;
       }
       else if (j > 0 &&
                grid[i][j] - 1 == grid[i][j - 1])
       {
           pt = {i, j - 1};
           path.push_front(pt);
           j--;
       }
    }
 
    return path;
}
 
// Function to find maximum length Snake sequence
void findSnakeSequence(int mat[M][N])
{
    // table to store results of subproblems
    int lookup[M][N];
 
    // initialize by 0
    memset(lookup, 0, sizeof lookup);
 
    // stores maximum length of Snake sequence
    int max_len = 0;
 
    // store cordinates to snake's tail
    int max_row = 0;
    int max_col = 0;
 
    // fill the table in bottom-up fashion
    for (int i = 0; i < M; i++)
    {
        for (int j = 0; j < N; j++)
        {
            // do except for (0, 0) cell
            if (i || j)
            {
                // look above
                if (i > 0 &&
                    abs(mat[i - 1][j] - mat[i][j]) == 1)
                {
                    lookup[i][j] = max(lookup[i][j],
                               lookup[i - 1][j] + 1);
 
                    if (max_len < lookup[i][j])
                    {
                        max_len = lookup[i][j];
                        max_row = i, max_col = j;
                    }
                }
 
                // look left
                if (j > 0 &&
                    abs(mat[i][j - 1] - mat[i][j]) == 1)
                {
                    lookup[i][j] = max(lookup[i][j],
                                       lookup[i][j - 1] + 1);
                    if (max_len < lookup[i][j])
                    {
                        max_len = lookup[i][j];
                        max_row = i, max_col = j;
                    }
                }
            }
        }
    }
 
    cout << "Maximum length of Snake sequence is: "
         << max_len << endl;
 
    // find maximum length Snake sequence path
    list<Point> path = findPath(lookup, mat, max_row,
                                             max_col);
 
    cout << "Snake sequence is:";
    for (auto it = path.begin(); it != path.end(); it++)
        cout << endl << mat[it->x][it->y] << " ("
             << it->x << ", " << it->y << ")" ;
}
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