# HackerEarth Crazy Matrix problem solution

In this HackerEarth Crazy Matrix problem solution Praveen went crazy yesterday and created an arbitrary matrix consisting of 0, 1 or 2. There was no rule observed in forming this matrix. But then he made up some rules himself.

If there are adjacent 1's, they are said to be connected. Similarly, if there are adjacent 2's, they are said to be connected. Here adjacent means all the surrounding positions along with the diagonal positions.

Now given a matrix of 0, 1 or 2, do your computation according to the following rules:
1. If there is a path from any position in top row to any position in bottom row consisting only of 1, then print 1.
2. If there is a path from any position in first column to any position in last column consisting only of 2, then print 2.
3. If both Rule 1 & Rule 2 are true, print AMBIGUOUS.
4. If none of Rule 1, Rule 2 or Rule 3 satisfies, print 0.

## HackerEarth Crazy Matrix problem solution.

`#include <bits/stdc++.h>using namespace std;int n;int arr [100 + 10][100 + 10];int dirc [8][2] = { {0 , 1}, {0 , -1}, {1 , 0}, {-1 , 0}, {1 , 1}, {1 , -1}, {-1 , -1}, {-1 , 1} };     // all the possible directionsbool vis1 [100 + 10][100 + 10];             // for first dfsbool vis2 [100 + 10][100 + 10];             // for second dfsbool dfs(int x, int y, int comp, bool vis [][100 + 10]){    if(comp == 1 && x == n - 1) return true;        // base case in first dfs    if(comp == 2 && y == n - 1) return true;        // base case in second dfs    vis[x][y] = true;    bool ans = false;    for(int i = 0; i < 8 && !ans; i++){        int dx = x + dirc[i][0];                    // transition        int dy = y + dirc[i][1];        if(dx < 0 || dx >= n || dy < 0|| dy >= n) continue;     // out of the board        if(arr[dx][dy] == comp && !vis[dx][dy])            ans |= dfs(dx, dy, comp, vis);              }    return ans;}int main(){    ios::sync_with_stdio(false);cin.tie(0);             // decrease the time for cin, cout    cin >>     n           ;    for(int i = 0; i < n; i++){        for(int j = 0; j < n; j++){            cin >> arr[i][j];        }    }    bool ones = false, twos = false;    for(int i = 0; i < n  && !ones ; i++){        if(arr[0][i] == 1 && !vis1[0][i]) ones = dfs(0 , i , 1, vis1);      // check for a path from the first row    }    for(int i = 0; i < n  && !twos ; i++){        if(arr[i][0] == 2 && !vis2[i][0]) twos = dfs(i , 0 , 2, vis2);      // check for a path from the second row    }    if(ones && twos) cout << "AMBIGUOUS";    else if (ones) cout << "1";    else if (twos) cout << "2";    else cout << "0";    return 0;}`