# HackerEarth Matrix and Volume problem solution

In this HackerEarth Matrix and Volume problem solution You are given a large box of length X, width Y and height Z. This box is entirely made up of small cubic boxes of 1 unit volume each i.e. length of cube is 1 unit. Now some cubes were removed to create empty spaces. All the cubes that were not removed stay fixed in their initial position.

Now you are given the information of the new state of the large box i.e. you are given the status of each cube that was used in making the larger box. If the status is 1 then it means that this cubic block was not removed or else if the status is 0 then it corresponds to the fact that this cubic block was removed.

## HackerEarth Matrix and Volume problem solution.

`#include<bits/stdc++.h>#include <stdint.h>using           namespace           std;#define         ff                  first#define         ss                  second#define         pb                  push_back#define         mp                  make_pair#define         eps                 1e-9#define         inf                 1e18#define         PI                  3.14159265#define         mset(A,num,n);      memset(A,num,n);#define         all(a)              a.begin(),a.end()#define         allr(a)             a.rbegin(),a.rend()#define         nl                  cout << endl#define         vi                  vector<int>#define         vll                 vector<ll>#define         pi                  pair<int,int>//#define         trace1(x)                   cout <<#x<<": "<<x<< endl;#define         trace2(x, y)                cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<< endl;#define         trace3(x, y, z)             cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl;#define         trace4(a, b, c, d)          cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;#define         trace5(a, b, c, d, e)       cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<<": "<<e<<endl;const int       MOD =               1e9+7;const int       N   =               1e6+5;typedef         long long int       ll;typedef         int64_t             lln;ll POWER[65];ll aonb(ll a, ll b) {ll ret=1;while(b) {if(b&1) ret*=a;a*=a;if(ret>=MOD) ret%=MOD;if(a>=MOD) a%=MOD;b>>=1;}return ret;}void precompute() {    POWER[0]=1;    for(int i=1;i<63;i++) POWER[i]=POWER[i-1]<<1LL;}int x,y,z;int A[105][105][105];bool dp[105][105][105];bool check(int ii,int jj, int kk){    if (ii<0 || ii>=x || jj <0 || jj>=y || kk<0 || kk>=z){        return false;    }    return true;}int a[] = {1,-1,0,0,0,0};int b[] = {0,0,1,-1,0,0};int c[] = {0,0,0,0,1,-1};void dfs(int ii, int jj, int kk){    if (!check(ii,jj,kk))        return ;    if (A[ii][jj][kk] == 1)        return ;    A[ii][jj][kk] = 1;    for (int i=0;i<6;i++){        dfs(ii+a[i],jj+b[i],kk+c[i]);    }}int main(){    std::ios_base::sync_with_stdio(false);    freopen("input_format","rt",stdin);    freopen("output_format","wt",stdout);//  std::cout << std::setprecision(10) << std::fixed;    int t;    cin >> t;    while(t--){        cin >> x >> y >> z;        for (int k=0;k<z;k++){            for (int i=0;i<x;i++){                for (int j=0;j<y;j++){                    cin >> A[i][j][k];                }            }        }                for (int i=0;i<x;i++){            for (int j=0;j<y;j++){                if (A[i][j][0] == 0){                    dfs(i,j,0);                }                if (A[i][j][z-1] == 0){                    dfs(i,j,z-1);                }            }        }        for (int i=0;i<x;i++){            for (int k=0;k<z;k++){                if (A[i][0][k] == 0){                    dfs(i,0,k);                }                           if (A[i][y-1][k] == 0){                    dfs(i,y-1,k);                }            }        }        for (int j=0;j<y;j++){            for (int k=0;k<z;k++){                if (A[0][j][k] == 0){                    dfs(0,j,k);                }                if (A[x-1][j][k] == 0){                    dfs(x-1,j,k);                }            }        }                int count=0;        for (int i=0;i<x;i++){            for (int j=0;j<y;j++){                for (int k=0;k<z;k++){                    if (A[i][j][k] == 0){                        count++;                    //  trace4(i,j,k,count);                    }                }            }        }        cout << count << endl;    }    return 0;}/*4 4 31 1 1 11 1 1 11 1 1 11 1 1 11 1 1 11 0 0 11 0 0 11 1 1 11 1 1 11 1 1 11 1 0 11 1 1 1ans = 4;`

### Second solution

`#include<bits/stdc++.h>#define LL long long int#define M 1000000007#define MM 1000000009#define reset(a) memset(a,0,sizeof(a))#define rep(i,j,k) for(i=j;i<=k;++i)#define per(i,j,k) for(i=j;i>=k;--i)#define print(a,start,end) for(i=start;i<=end;++i) cout<<a[i];#define endl "\n"#define inf 100000000000000LL pow(LL a,LL b,LL m){LL x=1,y=a;while(b > 0){if(b%2 == 1){x=(x*y);if(x>m) x%=m;}y = (y*y);if(y>m) y%=m;b /= 2;}return x%m;}LL gcd(LL a,LL b){if(b==0) return a; else return gcd(b,a%b);}LL gen(LL start,LL end){LL diff = end-start;LL temp = rand()%start;return temp+diff;}using namespace std;int p , q , r;set<int> s;int ct = 0;int dp[101][101][101];bool visit[101][101][101];void dfs(int x,int y,int z)    {        if(x < 1 || y < 1 || z < 1 || x > p || y > q|| z > r|| dp[x][y][z] == 1 || visit[x][y][z] == 1)            return;        ++ct;        s.insert(x);        s.insert(y);        s.insert(z);        dfs(x + 1, y , z);        dfs(x - 1, y , z);        dfs(x , y + 1, z);        dfs(x , y - 1, z);        dfs(x , y , z - 1);        dfs(x , y , z + 1);    }int main()    {        ios_base::sync_with_stdio(0);        int t;        cin >> t;        while(t--)            {                int x , y , z;                cin >> x >> y >> z;                                p = x , q = y , r = z;                for(int i = 1; i <= z ; i++)                    {                        for(int j = 1; j <= x; j++)                            {                                for(int k = 1; k <= y ; k++)                                    {                                        visit[i][j][k] = 0;                                        cin >> dp[j][k][i];                                    }                            }                    }                int ans = 0;                for(int i = 1; i <= x ; i++)                    {                        for(int j = 1; j <= y ; j++)                            {                                for(int k = 1; k <= z ; k++)                                    {                                        s.clear();                                        if(dp[i][j][k] == 0 && visit[i][j][k] == 0)                                            {                                                ct = 0;                                                s.insert(i);                                                s.insert(j);                                                s.insert(k);                                                dfs(i , j , k);                                            }                                        if(s.find(1) == s.end() && s.find(x) == s.end() && s.find(y) == s.end() && s.find(z) == s.end())                                            {                                                ans = ans + ct;                                            }                                    }                            }                    }                cout << ans << endl;                            }    }`