HackerEarth Printing patterns problem solution

YASH PAL, 31 July 202411 February 2026

In this HackerEarth Printing patterns problem solution You are required to form a matrix of size r x c where r is the number of rows and c is the number of columns. You are required to form the waves of numbers around the provided center, (Ci, Cj) (0-based indexing).

HackerEarth Printing patterns problem solution

HackerEarth Printing patterns problem solution.

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 1e6;

int dx[] = {-1 , -1 , -1, 0 , 0 , 1 , 1 , 1};// xdirection
int dy[] = {-1 , 0 , 1 , -1, 1 , -1, 0, 1};// ydirection

int row , col , x , y;
int pat[1001][1001];
bool visit[1001][1001];  
    
struct  Point{
  int x , y , d;
  Point(int a ,int b , int c){
    x = a , y = b , d = c;
    }
  };

 bool check(int x , int y){
  if(x < 0 || y < 0 || x >= row || y >= col || visit[x][y] == true){
    return false;
  }
  return true;
}

int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
    cin >> row >> col;
    cin >> x >> y;
    queue < Point > Q;
    Point p(x , y , 0) ;
    pat[x][y] = 0;
    Q.push(p);
    visit[x][y] = 1;
    while(Q.size() > 0){
      Point P = Q.front();
      Q.pop();
      int x = P.x;
      int y = P.y;
      int d = P.d;
      for(int i = 0 ; i < 8 ; i++){
        int X = x + dx[i];
        int Y = y + dy[i];
        if(check(X , Y)){    
          Point A(X , Y , d + 1); 
          visit[X][Y] = 1;
          pat[X][Y] = A.d;
          Q.push(A);
        }
      }
    }
    for(int i = 0 ; i < row ; i++){
      for(int j = 0 ; j < col ; j++){
        cout << pat[i][j] << " ";
      }
      cout << endl;
    }

  return 0;

}

Second solution

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
 
const int maxn = 3e2 + 14;
int main(){
    ios::sync_with_stdio(0), cin.tie(0);
    int r, c, x, y;
    cin >> r >> c >> x >> y;
    for(int i = 0; i < r; i++)
        for(int j = 0; j < c; j++)
            cout << max(abs(i - x), abs(j - y)) << " n"[j == c - 1];
}

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