3459. Length of Longest V-Shaped Diagonal Segment
You are given a 2D integer matrix grid of size n x m, where each element is either 0, 1, or 2.
A V-shaped diagonal segment is defined as:
- The segment starts with
1.
- The subsequent elements follow this infinite sequence:
2, 0, 2, 0, ....
- The segment:
- Starts along a diagonal direction (top-left to bottom-right, bottom-right to top-left, top-right to bottom-left, or bottom-left to top-right).
- Continues the sequence in the same diagonal direction.
- Makes at most one clockwise 90-degree turn to another diagonal direction while maintaining the sequence.
img
Return the length of the longest V-shaped diagonal segment. If no valid segment exists, return 0.
c++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
|
int DP[2][500][2][2];
class Solution {
vector<vector<int>> rotate(vector<vector<int>>& A) {
int n = A.size(), m = A[0].size();
vector<vector<int>> res(m, vector<int>(n));
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
res[j][n-i-1] = A[i][j];
}
}
return res;
}
int helper(vector<vector<int>>& A) {
int n = A.size(), m = A[0].size(), res = 0;
memset(DP,0,sizeof DP);
for(int i = 0; i < n; i++) {
memset(DP[!(i&1)], 0, sizeof DP[!(i&1)]);
for(int j = 0; j < m; j++) {
if(A[i][j] == 1) {
res = max(res, 1);
if(i + 1 < n) {
if(j + 1 < m and A[i+1][j+1] == 2) DP[!(i&1)][j+1][0][1] = max(DP[!(i&1)][j+1][0][1], 2);
}
}
if(A[i][j] == 2) {
for(int dir = 0; dir < 2; dir++) {
for(int dig = 0; dig < 2; dig++) {
if(!DP[i&1][j][dig][dir]) continue;
res = max(res, DP[i&1][j][dig][dir]);
if(i + 1 == n) continue;
int nj = j + (dir ? 1 : -1);
if(0 <= nj and nj < m and A[i+1][nj] == 0) {
DP[!(i&1)][nj][dig][dir] = max(DP[!(i&1)][nj][dig][dir], DP[i&1][j][dig][dir] + 1);
}
int nnj = j + (dir ? -1 : 1);
if(!dig and 0 <= nnj and nnj < m and A[i+1][nnj] == 0) {
DP[!(i&1)][nnj][!dig][!dir] = max(DP[!(i&1)][nnj][!dig][!dir], DP[i&1][j][dig][dir] + 1);
}
}
}
}
if(A[i][j] == 0) {
for(int dir = 0; dir < 2; dir++) {
for(int dig = 0; dig < 2; dig++) {
if(!DP[i&1][j][dig][dir]) continue;
res = max(res, DP[i&1][j][dig][dir]);
if(i + 1 == n) continue;
int nj = j + (dir ? 1 : -1);
if(0 <= nj and nj < m and A[i+1][nj] == 2) {
DP[!(i&1)][nj][dig][dir] = max(DP[!(i&1)][nj][dig][dir], DP[i&1][j][dig][dir] + 1);
}
int nnj = j + (dir ? -1 : 1);
if(!dig and 0 <= nnj and nnj < m and A[i+1][nnj] == 2) {
DP[!(i&1)][nnj][!dig][!dir] = max(DP[!(i&1)][nnj][!dig][!dir], DP[i&1][j][dig][dir] + 1);
}
}
}
}
}
}
return res;
}
public:
int lenOfVDiagonal(vector<vector<int>>& grid) {
int res = 0;
for(int i = 0; i < 4; i++) {
if(res < grid.size()) {
res = max(res, helper(grid));
}
grid = rotate(grid);
}
return res;
}
};
|
