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Rahul kumar
Rahul kumar

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Generate a spiral order matrix

#programming #algorithms

Given an integer A, generate a square matrix filled with elements from 1 to A2 in spiral order.

See problem at leetcode

For example:

A = 2
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Then matrix would be

1 2 
4 3
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A = 3
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Then matrix would be

1 2 3
8 9 4
7 6 5
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Idea

To generate the spiral order matrix, you should traverse the matrix layer by layer. First, generate the outer layer, then generate the one layer down, and so on.

Here, I have used 4 variable to track the row and column which we would like to fill.

  1. row-1
  2. column-n
  3. row-n
  4. column-1
  5. row-2
  6. column-(n-1)
  7. row-(n-1) ..... ..... and so on 
vector<vector<int> > Solution::generateMatrix(int A) {
    vector<vector<int>>ans(A,vector<int>(A,0));

    int rowBegin = 0;
    int colBegin = 0;
    int rowEnd = A-1;
    int colEnd = A-1;
    int counter = 1;

    while(rowBegin<=rowEnd && colBegin<=colEnd){
        // top left to top right
        for(int i=colBegin; i<=colEnd; i++)
            ans[rowBegin][i] = counter++; 
        rowBegin++;

        // top right to bottom right
        for(int i=rowBegin; i<=rowEnd; i++)
            ans[i][colEnd] = counter++;
        colEnd--;

        // bottom right to bottom left
        for(int i=colEnd; i>=colBegin; i--)
            ans[rowEnd][i] = counter++;
        rowEnd--;

        // bottom left to top left
        for(int i=rowEnd; i>=rowBegin; i--)
            ans[i][colBegin] = counter++;
        colBegin++;
    }
    return ans;
}
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