forked from kamyu104/LeetCode-Solutions
-
Notifications
You must be signed in to change notification settings - Fork 1
/
maximal-square.cpp
115 lines (107 loc) · 3.69 KB
/
maximal-square.cpp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
// Time: O(n^2)
// Space: O(n)
// DP with rolling window.
class Solution {
public:
int maximalSquare(vector<vector<char>>& A) {
if (A.empty()) {
return 0;
}
const int m = A.size(), n = A[0].size();
vector<vector<int>> size(2, vector<int>(n, 0));
int max_size = 0;
for (int j = 0; j < n; ++j) {
size[0][j] = A[0][j] - '0';
max_size = max(max_size, size[0][j]);
}
for (int i = 1; i < m; ++i) {
size[i % 2][0] = A[i][0] - '0';
for (int j = 1; j < n; ++j) {
if (A[i][j] == '1') {
size[i % 2][j] = min(size[i % 2][j - 1],
min(size[(i - 1) % 2][j],
size[(i - 1) % 2][j - 1])) + 1;
max_size = max(max_size, size[i % 2][j]);
} else {
size[i % 2][j] = 0;
}
}
}
return max_size * max_size;
}
};
// Time: O(n^2)
// Space: O(n^2)
// DP.
class Solution2 {
public:
int maximalSquare(vector<vector<char>>& A) {
if (A.empty()) {
return 0;
}
const int m = A.size(), n = A[0].size();
vector<vector<int>> size(m, vector<int>(n, 0));
int max_size = 0;
for (int j = 0; j < n; ++j) {
size[0][j] = A[0][j] - '0';
max_size = max(max_size, size[0][j]);
}
for (int i = 1; i < m; ++i) {
size[i][0] = A[i][0] - '0';
for (int j = 1; j < n; ++j) {
if (A[i][j] == '1') {
size[i][j] = min(size[i][j - 1],
min(size[i - 1][j],
size[i - 1][j - 1])) + 1;
max_size = max(max_size, size[i][j]);
} else {
size[i][j] = 0;
}
}
}
return max_size * max_size;
}
};
// Time: O(n^2)
// Space: O(n^2)
// DP.
class Solution3 {
public:
struct MaxHW {
int h, w;
};
int maximalSquare(vector<vector<char>>& A) {
if (A.empty()) {
return 0;
}
// DP table stores (h, w) for each (i, j).
vector<vector<MaxHW>> table(A.size(), vector<MaxHW>(A.front().size()));
for (int i = A.size() - 1; i >= 0; --i) {
for (int j = A[i].size() - 1; j >= 0; --j) {
// Find the largest h such that (i, j) to (i + h - 1, j) are feasible.
// Find the largest w such that (i, j) to (i, j + w - 1) are feasible.
table[i][j] = A[i][j] == '1'
? MaxHW{i + 1 < A.size() ? table[i + 1][j].h + 1 : 1,
j + 1 < A[i].size() ? table[i][j + 1].w + 1 : 1}
: MaxHW{0, 0};
}
}
// A table stores the length of largest square for each (i, j).
vector<vector<int>> s(A.size(), vector<int>(A.front().size(), 0));
int max_square_area = 0;
for (int i = A.size() - 1; i >= 0; --i) {
for (int j = A[i].size() - 1; j >= 0; --j) {
int side = min(table[i][j].h, table[i][j].w);
if (A[i][j]) {
// Get the length of largest square with bottom-left corner (i, j).
if (i + 1 < A.size() && j + 1 < A[i + 1].size()) {
side = min(s[i + 1][j + 1] + 1, side);
}
s[i][j] = side;
max_square_area = max(max_square_area, side * side);
}
}
}
return max_square_area;
}
};