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

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 −

 Live Demo

#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