27. Anti Diagonal Traversal of Matrix

The problem can be found at the following link: Question Link

My Approach

To traverse the anti-diagonals of the matrix,

• I used a helper function fill that starts at a given position (i, j) and moves diagonally down until it reaches the border of the matrix.

• I called this fill function for the top row and rightmost column to cover all the anti-diagonals.

Time and Auxiliary Space Complexity

• Time Complexity: O(N*N), where N is the total number of rows and columns in the matrix. We visit each element exactly once.

• Auxiliary Space Complexity: O(N*N), where N is the total number of rows and columns in the matrix. This is the space required to store the output vector.

Code (C++)

class Solution {
public:
    void fill(int i, int j, vector<vector<int>>& mat, vector<int>& out){
        while(i < mat.size() && j >= 0){
            out.push_back(mat[i][j]);
            ++i;
            --j;
        }
    }
    vector<int> antiDiagonalPattern(vector<vector<int>>& mat) 
    {
        vector<int> out;
        int n = mat.size();
        for(int j = 0; j < n; ++j)
            fill(0, j, mat, out);
        
        for(int i = 1; i < n; ++i)
            fill(i, n-1, mat, out);
        
        return out;
    }
};

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