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Longest Line of Consecutive One in Matrix in C++
Suppose we have one binary matrix M, we have to find the longest line of consecutive one in that matrix. The line could be horizontal, vertical, diagonal or anti-diagonal.
So, if the input is like
| 0 | 1 | 1 | 0 |
| 0 | 1 | 1 | 0 |
| 0 | 0 | 0 | 1 |
then the output will be 3
To solve this, we will follow these steps −
ret := 0
n := row of M
m := column of M
Define one 3D array dp of order n x m x 4
for initialize i := 0, when i < m, update (increase i by 1), do −
for initialize j := 0, when j < 4, update (increase j by 1), do −
dp[0, i, j] := M[0, i]
ret := maximum of ret and dp[0, i, j]
for initialize j := 0, when j < m, update (increase j by 1), do −
if M[0, j] is non-zero and j > 0, then −
dp[0, j, 1] := 1 + dp[0, j - 1, 1]
ret := maximum of ret and dp[0, j, 1]
for initialize i := 1, when i < n, update (increase i by 1), do −
for initialize j := 0, when j < m, update (increase j by 1), do −
dp[i, j, 0] := (if M[i, j] is non-zero, then 1 + dp[i - 1, j, 0], otherwise 0)
if j > 0, then −
dp[i, j, 1] := (if M[i, j] is non-zero, then dp[i, j - 1, 1] + 1, otherwise 0)
dp[i, j, 2] := (if M[i, j] is non-zero, then dp[i - 1, j - 1, 2] + 1, otherwise 0)
Otherwise
dp[i, j, 1] := M[i, j]
dp[i, j, 2] := M[i, j]
if j + 1 < m, then −
dp[i, j, 3] := (if M[i, j] is non-zero, then dp[i - 1, j + 1, 3] + 1, otherwise 0)
Otherwise
dp[i, j, 3] := M[i, j]
for initialize k := 0, when k < 4, update (increase k by 1), do −
ret := maximum of ret and dp[i, j, k]
return ret
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int longestLine(vector<vector<int>>& M) {
int ret = 0;
int n = M.size();
int m = !n ? 0 : M[0].size();
vector<vector<vector<int> > > dp(n, vector<vector<int> >(m, vector<int>(4)));
for (int i = 0; i < m; i++) {
for (int j = 0; j < 4; j++) {
dp[0][i][j] = M[0][i];
ret = max(ret, dp[0][i][j]);
}
}
for (int j = 0; j < m; j++) {
if (M[0][j] && j > 0) {
dp[0][j][1] = 1 + dp[0][j - 1][1];
ret = max(ret, dp[0][j][1]);
}
}
for (int i = 1; i < n; i++) {
for (int j = 0; j < m; j++) {
dp[i][j][0] = M[i][j] ? 1 + dp[i - 1][j][0] : 0;
if (j > 0) {
dp[i][j][1] = M[i][j] ? dp[i][j - 1][1] + 1 : 0;
dp[i][j][2] = M[i][j] ? dp[i - 1][j - 1][2] + 1 : 0;
}
else {
dp[i][j][1] = M[i][j];
dp[i][j][2] = M[i][j];
}
if (j + 1 < m) {
dp[i][j][3] = M[i][j] ? dp[i - 1][j + 1][3] + 1 : 0;
}
else {
dp[i][j][3] = M[i][j];
}
for (int k = 0; k < 4; k++) {
ret = max(ret, dp[i][j][k]);
}
}
}
return ret;
}
};
main(){
Solution ob;
vector<vector<int>> v = {{0,1,1,0},{0,1,1,0},{0,0,0,1}};
cout << (ob.longestLine(v));
}Input
{{0,1,1,0},{0,1,1,0},{0,0,0,1}}Output
3
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