Maximum decimal value path in a binary matrix in C++

Given the task is to find the maximum integer value that can be obtained while travelling in a path from the top left element to the bottom right element of a given square binary array, that is, starting from index [0][0] to index [n - 1][n - 1].

While covering the path we can only move to right ([i][j + 1]) or to the bottom ([i + 1][j])

The integer value will be calculate using the bits of the traversed path.

Let’s now understand what we have to do using an example −

Input

m = {
   {1, 1, 1, 1},
   {0, 0, 1, 0},
   {1, 0, 1, 1},
   {0, 1, 1, 1}
}

Output

Explanation

The path that we took is: [0, 0] →[0, 1] → [0, 2] → [1, 2] → [2, 2] → [3, 2] →[3, 3]

Therefore the decimal value becomes = 1*(2⁰) + 1*(2¹) + 1*(2²) + 1*(2³) + 1*(2⁴) + 1*(2⁵) + 1*(2⁶)

= 1 + 2 + 4 + 8 + 16 + 32 + 64

= 127

Input

m = {
   {1, 0, 1, 1},
   {0, 0, 1, 0},
   {1, 0, 0, 1},
   {0, 1, 1, 1}
}

Output

Approach used in the below program as follows

First define the size of side of the square matrix on the top using #define.
In main() function create a 2D array int m[][4] to store the matrix and call Max(m, 0, 0, 0)
In max() function first check if (i >= side || j >= side ). If so, then it means the we are out of the matrix boundary and return 0.
Create a new variable int ans and put ans = max(Max(m, i, j+1, pw+1), Max(m, i+1, j, pw+1)).
Then check if (m[i][j] == 1). If so, then return pow(2, pw) + ans.
Else simply return ans.

Example

Live Demo

#include<bits/stdc++.h>
using namespace std;
#define side 4
// pw is power of 2
int Max(int m[][side], int i, int j, int pw){
   // Out of boundary
   if (i >= side || j >= side )
      return 0;
   int ans = max(Max(m, i, j+1, pw+1), Max(m, i+1, j, pw+1));
   if (m[i][j] == 1)
      return pow(2, pw) + ans;
   else
      return ans;
}
//main function
int main(){
   int m[][4] = {{1, 1, 1, 1},{0, 0, 1, 0},{1, 0, 1, 1},{0, 1, 1, 1}};
   cout << Max(m, 0, 0, 0);
   return 0;
}