463. Island Perimeter.md

463. Island Perimeter

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leetcode Daily Challenge on July 7th, 2020.

leetcode-cn Daily Challenge on October 30th, 2020.

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Difficulty : Easy

Related Topic : HashTable

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You are 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:

Input:
[[0,1,0,0],
 [1,1,1,0],
 [0,1,0,0],
 [1,1,0,0]]

Output: 16

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

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Solution

• mine
  • Java

    • DFS Runtime: 9 ms, faster than 32.06%, Memory Usage: 40.7 MB, less than 37.06% of Java online submissions

      // O(r * c)time
      // O(1)space
      public int islandPerimeter(int[][] grid) {
          int r = grid.length;
          int c = grid[0].length;
          for(int i = 0; i < r; i++){
              for(int j = 0; j < c; j++){
                  if(grid[i][j] == 1){
                      return dfs(grid, i , j);
                  }
              }
          }
          return 0;
      }
      
      int dfs(int[][] grid, int i, int j) {
          if (i < 0 || i >= grid.length || j < 0 || j >= grid[i].length || grid[i][j] == 0) {
              return 1;
          }
          if (grid[i][j] == -1) {
              return 0;
          }
          grid[i][j] = -1;
          int res = 0;
          res += dfs(grid, i - 1, j);
          res += dfs(grid, i, j - 1);
          res += dfs(grid, i + 1, j);
          res += dfs(grid, i, j + 1);
          return res;
      }
      

    • Runtime: 8 ms, faster than 40.31%, Memory Usage: 57.9 MB, less than 10.05% of Java online submissions

      // O(r * c)time
      // O(1)space
      public int islandPerimeter(int[][] grid) {
          int res = 0;
          for (int i = 0; i < grid.length; i++) {
              for (int j = 0; j < grid[i].length; j++) {
                  if (grid[i][j] == 1) {
                      res += 4;
                      if(i + 1 < grid.length && grid[i + 1][j] == 1)  res -= 2;
                      if(j + 1 < grid[0].length && grid[i][j + 1] == 1) res -= 2;
                  }
              }
          }
      
          return res;
      }
      

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• the most votes
  • Runtime: 5 ms, faster than 99.93%, Memory Usage: 40.5 MB, less than 44.66% of Java online submissions

    // O(r * c)time
    // O(1)space
    public int islandPerimeter(int[][] grid) {
        int islands = 0, neighbours = 0;
    
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[i].length; j++) {
                if (grid[i][j] == 1) {
                    islands++; // count islands
                    if (i < grid.length - 1 && grid[i + 1][j] == 1) neighbours++; // count down neighbours
                    if (j < grid[i].length - 1 && grid[i][j + 1] == 1) neighbours++; // count right neighbours
                }
            }
        }
    
        return islands * 4 - neighbours * 2;
    }
    

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