Find size of the largest '+' formed by all ones in a binary matrix - GeeksforGeeks


Find size of the largest '+' formed by all ones in a binary matrix - GeeksforGeeks
Given a N X N binary matrix, find the size of the largest '+' formed by all 1s.

The idea is to maintain four auxiliary matrices left[][], right[][], top[][], bottom[][] to store consecutive 1’s in every direction.
int findLargestPlus(int mat[N][N])
{
    // left[j][j], right[i][j], top[i][j] and
    // bottom[i][j] store maximum number of
    // consecutive 1's present to the left,
    // right, top and bottom of mat[i][j] including
    // cell(i, j) respectively
    int left[N][N], right[N][N], top[N][N],
        bottom[N][N];
 
    // initialize above four matrix
    for (int i = 0; i < N; i++)
    {
        // initialize first row of top
        top[0][i] = mat[0][i];
 
        // initialize last row of bottom
        bottom[N - 1][i] = mat[N - 1][i];
 
        // initialize first column of left
        left[i][0] = mat[i][0];
 
        // initialize last column of right
        right[i][N - 1] = mat[i][N - 1];
    }
 
    // fill all cells of above four matrix
    for (int i = 0; i < N; i++)
    {
        for (int j = 1; j < N; j++)
        {
            // calculate left matrix (filled left to right)
            if (mat[i][j] == 1)
                left[i][j] = left[i][j - 1] + 1;
            else
                left[i][j] = 0;
 
            // calculate top matrix
            if (mat[j][i] == 1)
                top[j][i] = top[j - 1][i] + 1;
            else
                top[j][i] = 0;
 
            // calculate new value of j to calculate
            // value of bottom(i, j) and right(i, j)
            j = N - 1 - j;
 
            // calculate bottom matrix
            if (mat[j][i] == 1)
                bottom[j][i] = bottom[j + 1][i] + 1;
            else
                bottom[j][i] = 0;
 
            // calculate right matrix
            if (mat[i][j] == 1)
                right[i][j] = right[i][j + 1] + 1;
            else
                right[i][j] = 0;
 
            // revert back to old j
            j = N - 1 - j;
        }
    }
 
    // n stores length of longest + found so far
    int n = 0;
 
    // compute longest +
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < N; j++)
        {
            // find minimum of left(i, j), right(i, j),
            // top(i, j), bottom(i, j)
            int len = min(min(top[i][j], bottom[i][j]),
                          min(left[i][j], right[i][j]));
 
            // largest + would be formed by a cell that
            // has maximum value
            if(len > n)
                n = len;
        }
    }
 
    // 4 directions of length n - 1 and 1 for middle cell
    if (n)
       return 4 * (n - 1) + 1;
 
    // matrix contains all 0's
    return 0;
}
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