Beautiful Matrix — CP

Understanding the problem

We have a 5×5 matrix with twenty-four 0s and a single 1. We can swap neighboring rows or columns. Goal: move the 1 to the center cell (row 3, col 3). Find the minimum number of moves.

My thinking

The key insight — each row swap moves the 1 one step closer vertically, each column swap moves it one step closer horizontally. So the minimum moves is just the Manhattan distance from the 1’s position to the center.

Using 0-based indexing, the center is at (2, 2).

So: minSteps = abs(row - 2) + abs(col - 2)

I found the position of 1 by iterating through the matrix and breaking as soon as I found it.

One bug I hit: my first version used break thinking it would exit both loops. It only exited the inner loop — so i kept incrementing and I got wrong output. Fixed it by using return 0 directly after computing the answer.

Code

#include <bits/stdc++.h>
using namespace std;

int main(){
    int i = 0;
    int j = 0;

    for(; i < 5; i++){
        for(; j < 5; j++){
            int val;
            cin >> val;
            if(val == 1){
                int minStep = abs(i - 2) + abs(j - 2);
                cout << minStep << endl;
                return 0;
            }
        }
        j = 0;
    }

    return 0;
}

What I learned

break only exits the innermost loop — if you need to exit multiple loops, use return or a flag
Manhattan distance is the go-to formula for grid movement problems where you can only move horizontally/vertically
Always reset inner loop variables (j = 0) when using non-standard for loop initialization