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[LeetCode] 221. Maximal Square _ Medium Tag: Dynamic Programming

2018-07-17 06:44  Johnson_强生仔仔  阅读(259)  评论(0)  编辑  收藏  举报

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.

Example:

Input: 

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Output: 4


思路是DP, 3种做法, 通用的T: O(m*n) , S: O(m*n)  和只针对部分情况可以use 滚动数组来reduce space成为O(n).
A[i][j] = min(A[i-1][j-1], left[i][j-1], up[i-1][j]) + 1 为边长  i, j > 0

滚动数组
A[i][j] = min(A[i-1][j-1], A[i][j-1], A[i-1][j]) + 1 为边长  i, j > 0

A[i][j] = min(A[i%2-1][j-1], A[i%2][j-1], A[i%2-1][j]) + 1 为边长  i, j > 0


1. Constraints
1) size >=[0*0]
2) element will be "1" or "0"  # note it will be integer or string

2. Ideas

DP     T: O(m*n)   S: O(n)  optimal
1) edge case, empty, m == 1 or n == 1
2) left, up , ans, init
3)
A[i][j] = min(A[i-1][j-1], left[i][j-1], up[i-1][j]) + 1
4) return res*res

3. codes

1) use left, up , and ans   T: O(m*n)    S: O(m*n)
 1 class Solution:
 2     def maxSquare(self, matrix):
 3         if not matrix: return 0
 4         m, n = len(matrix), len(matrix[0])
 5         left, up, ans, res = [[0]*n for _ in range(m)], [[0]*n for _ in range(m)], [[0]*n for _ in range(m)], 0
 6         for i in range(m):
 7             for j in range(n):
 8                 if matrix[i][j] == "1":
 9                     res = 1   # edge case when m == 1 or n == 1
10                     if j == 0:
11                         left[i][j] = ans[i][j] = 1
12                     if i == 0:
13                         up[i][j] = ans[i][j] = 1
14                     if i >0 and j > 0:
15                         left[i][j] = left[i][j-1] + 1
16                         up[i][j] = up[i-1][j] + 1
17         for i in range(1, m):
18             for j in range(1, n):
19                 if matrix[i][j] == "1":
20                     ans[i][j] = min(ans[i-1][j-1], left[i][j-1], up[i-1][j])+1
21                     res = max(res, ans[i][j])
22         return res*res

 

3.2) skip left and up, just use f array

T: O(m*n)    S: O(m*n)
class Solution:    
    def maxSquare(self, matrix):
        if not matrix or not matrix[0]: return 0
        m, n = len(matrix), len(matrix[0])
        f, ans = [[0] * n for _ in range(m)], 0
        # initial f
        for i in range(m):
            if matrix[i][0] == "1":
                f[i][0] = 1
                ans = 1 # edge case when only edge is 1 
        for j in range(n):
            if matrix[0][j] == "1":
                f[0][j] = 1
                ans = 1
        for i in range(1, m):
            for j in range(1, n):
                if matrix[i][j] == "1":
                    f[i][j] = min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1
                    ans = max(ans, f[i][j])
        return ans * ans

 

3.2.1) 将初始化都放在f赋值的两个for loop中：

T: O(m*n)    S: O(m*n)
class Solution:    
    def maxSquare(self, matrix):
        if not matrix or not matrix[0]: return 0
        m, n = len(matrix), len(matrix[0])
        f, ans = [[0] * n for _ in range(m)], 0
        for i in range(m):
            for j in range(n):
                if matrix[i][j] == "1":
                    if i == 0 or j == 0:
                        f[i][j] = 1
                    else:
                        f[i][j] = min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1
                    ans = max(ans, f[i][j])
        return ans * ans

 

 

3.3) 滚动数组,   T: O(m*n),    S: O(n)

class Solution:    
    def maxSquare(self, matrix):
        if not matrix or not matrix[0]: return 0
        m, n = len(matrix), len(matrix[0])
        f, ans = [[0] * n for _ in range(2)], 0
        for i in range(m):
            for j in range(n):
                if matrix[i][j] == "1":
                    if i == 0 or j == 0:
                        f[i % 2][j] = 1
                    else:
                        f[i % 2][j] = min(f[(i - 1) % 2][j], f[i % 2][j - 1], f[(i - 1) % 2][j - 1]) + 1
                        ans = max(ans, f[i % 2][j])
                else:
                        f[i % 2][j] = 0    #Note: must notice when using rolling array, need to initial         
        return ans * ans

 

 

4. Test cases

1) edge case

2) 

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0