Determine the perimeter of the island

Given a map in form of a two-dimensional integer grid where 1 represents land and 0 represents water. Grid cells are connected horizontally/vertically (not diagonally). The grid is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells). The island doesn't have "lakes" (water inside that isn't connected to the water around the island). One cell is a square with side length 1. The grid is rectangular, width and height don't exceed 100. Determine the perimeter of the island.

Example:

[[0,1,0,0],
 [1,1,1,0],
 [0,1,0,0],
 [1,1,0,0]]

Answer: 16
Explanation: The perimeter is the 16 yellow stripes in the image below:
 

 class Solution {
public:
    int islandPerimeter(vector<vector<int>>& grid) {
        int row = grid.size();
        int col = grid[0].size();
        if (grid.empty()) return 0;
        int sum = 0, sr = 0, sc = 0;
        for (int i =0; i < row; i ++) {
            for (int j =0; j < col; j++) {
                if (grid[i][j] == 1) {
                    sr = i;
                    sc = j;
                    break;
                }              
            }
        }
      
        sum += addPerimeter(grid, sr, sc, row, col);
        return sum;
    }
  
    int addPerimeter (vector<vector<int>>&grid, int i , int j, int row, int col) {
        if (i < 0 || i >= row || j < 0 || j >= col || grid[i][j] == 0)
            return 1;
        if (grid[i][j] == -1)
            return 0;
      
        grid[i][j] = -1;
      
        return addPerimeter(grid, i+1, j, row, col) +
            addPerimeter(grid, i-1, j, row, col) +
            addPerimeter(grid, i, j+1, row, col) +
            addPerimeter(grid, i, j-1, row, col);   
      
    }
  
};

Perform Flood Fill for a given 2D image/matrix using DFS

solution for Flood fill question using DFS approach :

Given a coordinate (sr, sc) representing the starting pixel (row and column) of the flood fill, and a pixel value newColor, "flood fill" the image.

To perform a "flood fill", consider the starting pixel, plus any pixels connected 4-directionally to the starting pixel of the same color as the starting pixel, plus any pixels connected 4-directionally to those pixels (also with the same color as the starting pixel), and so on. Replace the color of all of the aforementioned pixels with the newColor. return the modified image.

Example :

Input: 
image = [[1,1,1],[1,1,0],[1,0,1]]
sr = 1, sc = 1, newColor = 3
Output: [[3,3,3],[3,3,0],[3,0,1]]
Explanation: 
From the center of the image (with position (sr, sc) = (1, 1)), all pixels connected 
by a path of the same color as the starting pixel are colored with the new color.
Note the bottom corner is not colored 3, because it is not 4-directionally connected
to the starting pixel.

class Solution {
public:
    vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int newColor) {
        int row = image.size();
        int col = image[0].size();
        if (sr < 0 || sr >= row || sc < 0 || sc >=col )
            return image;
        int m = image[sr][sc];
        fillcolor(image, sr , sc , row,col, m, newColor);
        return image;
    }
   
    void fillcolor(vector<vector<int>>& image, int r, int c , int row, int col, int m, int newColor) {
        if (r < 0 || r >= row || c < 0 || c >= col || (image[r][c] != m) || (image[r][c] == newColor))
            return;
        image[r][c] = newColor;
        fillcolor(image, r+1, c, row, col, m, newColor);
        fillcolor(image, r-1, c, row, col, m,newColor);
        fillcolor(image, r, c+1, row, col, m,newColor);
        fillcolor(image, r, c-1, row, col, m,newColor);
       
    }
};

Pascal's Triangle

Given numRows, generate the first numRows of Pascal's triangle.

For example, given numRows = 4,
Return

[    [1],
    [1,1],
   [1,2,1],
  [1,3,3,1] ]

C program : 

int** generate(int numRows, int** columnSizes) {
    int i, j;
    int numColumns = 0;
    *columnSizes = malloc(numRows*sizeof(int));
    int** result = (int**)malloc(numRows*sizeof(int*));
   
    for (i = 0; i < numRows; i++)
        result[i] = (int*)malloc((i+1)*sizeof(int));
   
    for (i = 0; i < numRows; i++) {
       
        result[i][0] = result[i][i] = 1;
        for (j = 0; j < i; j++) {
         result[i][j] = result[i-1][j-1] + result[i-1][j];
        }
        numColumns++;
        *(*columnSizes+i) = numColumns;
    }
    return result;
}


C++ program : 

class Solution {
public:
    vector<vector<int>> generate(int numRows) {
        vector<vector<int>>result(numRows);
        for (int i = 0; i < numRows; i++) {
            result[i].resize(i+1);
            result[i][0] = result[i][i] = 1;
            for (int j = 0; j < i ; j++) {
                result[i][j] = result[i-1][j-1] + result[i-1][j];
            }
        }
       
        return result;
    }
};

container_of macro

kernel container_of macro

#define container_of(ptr, type, member) ({ \
    const typeof( ((type *)0)->member ) \
    *__mptr = (ptr);
    (type *)( (char *)__mptr - offsetof(type,member) );})

When you use the cointainer_of macro, you want to retrieve the structure that contains the pointer of a given field. For example:

struct numbers {  
   int one; 
   int two; 
   int three; 
} n;  

int *ptr = &n.two; 
struct numbers *n_ptr; 
n_ptr = container_of(ptr, struct numbers, two);

You have a pointer that points in the middle of a structure (and you know that is a pointer totwo
 [the field name in the structure]), but you want to retrieve the entire structure (numbers). So, you calculate the offset of the filed two in the structure:

offsetof(type,member)

and subtract this offset from the given pointer. The result is the pointer to the start of the structure. Finally, you cast this pointer to the structure type to have a valid variable.