HackerEarth Grid and phrase problem solution

YASH PAL,
In this HackerEarth Grid and phrase problem solution, You are given an n*m grid which contains lower case English letters. How many times does the phrase “saba” appear horizontally, vertically, and diagonally in the grid?
HackerEarth Grid and phrase problem solution

HackerEarth Grid and phrase problem solution.

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int M=1e2+10;
char jad[M][M];
int n,m,ans;
bool check(char a,char b,char c,char d)
{
  if(a=='s' && b=='a' && c=='b' && d=='a')
    return true;
  return false;
}
int32_t main()
{
  cin>>n>>m;
  for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
      cin>>jad[i][j];
  for(int i=4;i<=n;i++)
    for(int j=1;j<=m-3;j++)
      if(check(jad[i][j],jad[i-1][j+1],jad[i-2][j+2],jad[i-3][j+3]))
    ans++;
  for(int i=1;i<=n-3;i++)
    for(int j=1;j<=m-3;j++)
      if(check(jad[i][j],jad[i+1][j+1],jad[i+2][j+2],jad[i+3][j+3]))
    ans++;
  for(int i=1;i<=n;i++)
    for(int j=1;j<=m-3;j++)
      if(check(jad[i][j],jad[i][j+1],jad[i][j+2],jad[i][j+3]))
    ans++;
  for(int i=1;i<=n-3;i++)
    for(int j=1;j<=m;j++)
      if(check(jad[i][j],jad[i+1][j],jad[i+2][j],jad[i+3][j]))
    ans++;
  cout<<ans<<endl;
  return 0;
}

Second solution

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int maxN=1e3+123;

string saba="saba";
string s[maxN];

int32_t main(){
  int n,m;
  cin>>n>>m;
  int ans=0;
  for(int i=0;i<n;i++)cin>>s[i];
  for(int i=0;i<n-3;i++){
    for(int j=0;j<m-3;j++){
      bool b=true;
      for(int k=0;k<4;k++){
    if(s[i+k][j+k]!=saba[k])b=false;
      }
      if(b)ans++;
    }
  }
  for(int i=0;i<n-3;i++){
    for(int j=0;j<m;j++){
      bool b=true;
      for(int k=0;k<4;k++){
    if(s[i+k][j]!=saba[k])b=false;
      }
      if(b)ans++;
    }
  }
  for(int i=0;i<n;i++){
    for(int j=0;j<m-3;j++){
      bool b=true;
      for(int k=0;k<4;k++){
    if(s[i][j+k]!=saba[k])b=false;
      }
      if(b)ans++;
    }
  }
  for(int i=3;i<n;i++){
    for(int j=0;j<m-3;j++){
      bool b=true;
      for(int k=0;k<4;k++){
    if(s[i-k][j+k]!=saba[k])b=false;
      }
      if(b)ans++;
    }
  }
  cout<<ans<<endl;
}
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