searching in 2D sorted matrix

/***
Searching in a 2D sorted matrix

Write an efficient algorithm that searches for a value in an n x m table (two-dimensional array). This table is sorted along the rows and columns — that is,

Table[i][j] ≤ Table[i][j + 1],
Table[i][j] ≤ Table[i + 1][j]

Two Solutions:
1) BS in row/column through column/row
       complexity: O(nlogn)
2) walk-wise linear
       complexity: O(n)
***/

#include<iostream>

using namespace std;

#define N 5
#define M 5

// binary search: time complexity O(m*logn)
bool bs(int *in, int start, int end, int n) {
       if (start > end) return false;
      
       int mid = (start + end) / 2;
       if (in[mid] == n) return true;
       else if (in[mid] > n)
       {
              if (bs(in, start, mid-1, n)) return true;
       } else {
              if (bs(in, mid+1, end, n)) return true;
       }
       return false;
}

bool search2D_b(int (*a)[M], int n) {
       for(int i = 0; i<M; i++) {
              if(bs(a[i], 0, N-1, n)) return true;
       }
       return false;
}

bool walk_wise(int (*a)[M], int i, int j, int n) {

       if(i>M-1 || j<0) return false;

       if(a[i][j] == n) return true;
       else if(a[i][j] < n) {
              if(walk_wise(a, i+1, j, n)) return true;
       } else {
              if(walk_wise(a, i, j-1, n)) return true;
       }
       return false;
}

//walk wise: start from upper right
bool search2D(int (*a)[M], int n) {
       if(walk_wise(a, 0, N-1, n)) return true;
       else return false;  
}

void main() {
       int a[][M] = {{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}};

       if (search2D(a, -1))
       {
              cout<<"search2D"<<endl;
       }
       if (search2D_b(a, 22))
       {
              cout<<"search2D_binary"<<endl;
       }

}

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