Create A_Unique_Pattern

[MDAFSARALI]

Problem 22 :

5 5 5 5 5 5 5 5 5 
5 4 4 4 4 4 4 4 5 
5 4 3 3 3 3 3 4 5 
5 4 3 2 2 2 3 4 5 
5 4 3 2 1 2 3 4 5 
5 4 3 2 2 2 3 4 5 
5 4 3 3 3 3 3 4 5 
5 4 4 4 4 4 4 4 5 
5 5 5 5 5 5 5 5 5

Practice Link :

CODE SOLUTION

Approach for pattern22 (English Solution) 🚀

• This function prints a square pattern of size 2*n-1 📏.

• For each cell, four distances are calculated based on i and j:

  • 📍 Distance from the top: i
  • 📍 Distance from the bottom: 2*n-2 - i
  • 📍 Distance from the left: j
  • 📍 Distance from the right: 2*n-2 - j

• The minimum of these four distances is taken 🧮, and n - min() is printed in that cell 🖨️.

• This ensures that the value decreases symmetrically 📉 as we move towards the center of the pattern 🎯.

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Approach for pattern22 (Hinglish_Solution)

• Yeh function ek 2*n-1 size ka square pattern print karta hai.
• Har cell ke liye, i aur j ke basis par 4 distances calculate hote hain: top, bottom, left, aur right.
• Inn distances ka minimum leke, n - min() se jo value aati hai, woh us cell mein print hoti hai.
• Har row ke baad endl se new line print hoti hai.

void pattern22(int n) {
        for(int i=0;i<2*n-1;i++){
            for(int j=0;j<2*n-1;j++){
                int top=i,bottom=j;
                int left=(2*n-2)-i,right=(2*n-2)-j;
                cout<<n-(min(min(top,bottom),min(right,left)))<<" ";
            }
            cout<<endl;
        }
    }

Thanks ♥️

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[MDAFSARALI]

A new pattern Problem Added