[LeetCode] 827. Making A Large Island
In a 2D grid of 0
s and 1
s, we change at most one 0
to a 1
.
After, what is the size of the largest island? (An island is a 4-directionally connected group of 1
s).
Example 1:
Input: [[1, 0], [0, 1]]
Output: 3
Explanation: Change one 0 to 1 and connect two 1s, then we get an island with area = 3.
Example 2:
Input: [[1, 1], [1, 0]]
Output: 4
Explanation: Change the 0 to 1 and make the island bigger, only one island with area = 4.
Example 3:
Input: [[1, 1], [1, 1]]
Output: 4
Explanation: Can't change any 0 to 1, only one island with area = 4.
Notes:
1 <= grid.length = grid[0].length <= 50
.0 <= grid[i][j] <= 1
.
这个题目我只能想到naive的做法, T: O((m*n)^2), 就是对每个 grid[i][j] == 0 的元素, 将其换为1, 然后用BFS来计算size, 一直循环, 找最大值, 如果没有0 的话返回m*n.
然后参考了这个T: O(m*n) 的做法. 思路就是最开始将grid扫一遍, 每次如果grid[i][j] == 1, 将该点及所有跟它相连的为1 的点, 标记为color, 然后将各个不同的island tag为不同颜色, 并且用d2来存color和这个color的island 的size, 思路和做法跟[LeetCode] 200. Number of Islands_ Medium tag: BFS一样, 只是需要标记color.
这样之后可以得到如上图所示的结果, 然后再将grid扫一遍, 这一次是针对 grid[i][j] == 0 的元素, 然后看四个邻居是否为1, 如果是的话将它的size加入到temp里面,只是要注意的是有可能同样颜色的可以是多个neig, 所以要用set来排除已经加入同样颜色的size了, 最后返回最大值即可.
1. Constraints
1) size of grid, [1*1] ~ [50*50]
2) element will only be 0 or 1
2. Ideas
DFS/BFS 用BFS, T: O(m*n) S: O(m*n)
3. Code
class Solution:
def largestIsland(self, grid):
def bfs(grid, d1, i, j, d2, dirs, color):
lr, lc, d1[(i, j)], queue, size = len(grid), len(grid[0]), color, collections.deque([(i,j)]), 1 while queue:
pr, pc = queue.popleft()
for c1, c2 in dirs:
nr, nc = c1 + pr , c2 + pc
if 0 <= nr < lr and 0 <= nc < lc and (nr, nc) not in d1:
d1[(nr, nc)] = color
queue.append((nr, nc))
size += 1
d2[color] = size
d1, d2, lr, lc , dirs, temp, color = {},{}, len(grid), len(grid[0]), [(0,1), (0,-1), (1, 0), (-1,0)], -1, 1
for i in range(lr):
for j in range(lc):
if grid[i][j] == 1 and (i,j) not in d1:
color += 1
bfs(grid, d1, i, j, d2, dirs, color) for i in range(lr):
for j in range(lc):
if grid[i][j] == 0:
colors = set()
for c1, c2 in dirs:
nr, nc = i+ c1, j + c2
if 0 <= nr < lr and 0 <= nc < lc and grid[nr][nc] == 1:
colors.add(d1[nr][nc])
s = sum(d2[color] for color in colors) + 1
temp = s if temp == -1 else max(temp, s)
return temp if temp != -1 else lr*lc
2)
class Solution(object):
def largestIsland(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
lr, lc, dirs, colorMap, ans, colorSize, queue, curColor = len(grid), len(grid[0]), [(1, 0), (-1, 0), (0, 1), (0, -1)], {}, 0, {}, collections.deque(), 0
# color all the islands and save the pos -> color in colorMap, color -> size in colorSize
for i in range(lr):
for j in range(lc):
if grid[i][j] == 1 and (i, j) not in colorMap:
size = 0
curColor += 1
queue.append((i, j))
colorMap[(i, j)] = curColor
while queue:
r, c = queue.popleft()
size += 1
for d1, d2 in dirs:
nr, nc = r + d1, c + d2
if 0 <= nr < lr and 0 <= nc < lc and (nr, nc) not in colorMap and grid[nr][nc] == 1:
queue.append((nr, nc))
colorMap[(nr, nc)] = curColor
colorSize[curColor] = size
ans = max(ans, size) #### to each i, j that grid[i][j] == 0 , put all neigbours that is 1 and add all the color in a set, sum all color sizes and max(ans, sizeSum)
for i in range(lr):
for j in range(lc):
if grid[i][j] == 0:
colorSet = set()
for d1, d2 in dirs:
nr, nc = i + d1, j + d2
if 0 <= nr < lr and 0 <= nc < lc and grid[nr][nc] == 1:
colorSet.add(colorMap[(nr, nc)])
ans = max(ans, 1 + sum(colorSize[color] for color in colorSet))
return ans
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