[20200707] Solve july challenge

Author: iamminji

README.md

@@ -228,3 +228,4 @@
 | [Ugly Number II](https://leetcode.com/explore/challenge/card/july-leetcoding-challenge/544/week-1-july-1st-july-7th/3380/)      | [python](challenge/2020/july/Ugly_Number_II.py)  | [solutions](challenge/2020/july/Ugly_Number_II.md)|
 | [Hamming Distance](https://leetcode.com/explore/challenge/card/july-leetcoding-challenge/544/week-1-july-1st-july-7th/3381/)      | [python](challenge/2020/july/Hamming_Distance.py)  | [solutions](challenge/2020/july/Hamming_Distance.md)|
 | [Plus One](https://leetcode.com/explore/challenge/card/july-leetcoding-challenge/544/week-1-july-1st-july-7th/3382/)      | [python](challenge/2020/july/Plus_One.py)  | [solutions](challenge/2020/july/Plus_One.md)|
+| [Island Perimeter](https://leetcode.com/explore/challenge/card/july-leetcoding-challenge/544/week-1-july-1st-july-7th/3383/)      | [python](challenge/2020/july/Island_Perimeter.py)  | [solutions](challenge/2020/july/Island_Perimeter.md)|

challenge/2020/july/Island_Perimeter.md

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+## 문제
+
+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
+```
+
+### 솔루션
+1과 0의 경계선의 카운트를 구하는 문제다. for loop로 1인 경우에만 확장 시켜서, 위/아래/좌/우 가 0이 나오거나 배열을 벗어나면 1로 카운트 하게 한다.
+단, 그냥 하면 중복된 섬(문제에선 1이 섬이다.) 을 볼 수 있기 때문에 캐시를 두고, 이미 확인한 섬은 패스 한다.
+
+코드는 아래와 같다.
+
+```python3
+class Solution:
+
+    def dfs(self, i, j, grid, cache):
+
+        if i < 0 or j < 0 or i >= len(grid) or j >= len(grid[i]):
+            return 1
+
+        if grid[i][j] == 0:
+            return 1
+
+        if (i, j) in cache:
+            return 0
+
+        cache[(i, j)] = True
+
+        ans = 0
+        ans += self.dfs(i, j - 1, grid, cache)
+        ans += self.dfs(i, j + 1, grid, cache)
+        ans += self.dfs(i + 1, j, grid, cache)
+        ans += self.dfs(i - 1, j, grid, cache)
+
+        return ans
+
+    def islandPerimeter(self, grid):
+        """
+        :type grid: List[List[int]]
+        :rtype: int
+        """
+
+        cache = dict()
+        ans = 0
+        for i in range(len(grid)):
+            for j in range(len(grid[i])):
+                if grid[i][j] == 1:
+                    ans += self.dfs(i, j, grid, cache)
+
+        return ans
+```
+
+### 복잡도
+공간 복잡도는 상수고, 시간 복잡도는 O(n) 이다. 
+O(n) 인 이유는 캐시를 두어서 중복된 섬에서 확장 될 일이 없고, 최초에 for loop 만 있기 때문이다. (n 은 배열 길이 m*m)

challenge/2020/july/Island_Perimeter.py

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+class Solution:
+
+    def dfs(self, i, j, grid, cache):
+
+        if i < 0 or j < 0 or i >= len(grid) or j >= len(grid[i]):
+            return 1
+
+        if grid[i][j] == 0:
+            return 1
+
+        if (i, j) in cache:
+            return 0
+
+        cache[(i, j)] = True
+
+        ans = 0
+        ans += self.dfs(i, j - 1, grid, cache)
+        ans += self.dfs(i, j + 1, grid, cache)
+        ans += self.dfs(i + 1, j, grid, cache)
+        ans += self.dfs(i - 1, j, grid, cache)
+
+        return ans
+
+    def islandPerimeter(self, grid):
+        """
+        :type grid: List[List[int]]
+        :rtype: int
+        """
+
+        cache = dict()
+        ans = 0
+        for i in range(len(grid)):
+            for j in range(len(grid[i])):
+                if grid[i][j] == 1:
+                    ans += self.dfs(i, j, grid, cache)
+
+        return ans