# Return an array of anti-diagonals of given N*N square matrix

Given a square matrix of size N*N, return an array of its anti-diagonals. For better understanding let us look at the image given below:
Examples:

```Input :
```

```Output :
1
2  5
3  6  9
4  7  10  13
8  11 14
12 15
16
```

Approach 1:
To solve the problem mentioned above we have two major observations.

• The first one is, some diagonals start from the zeroth row for each column and ends when either start column >= 0 or start row < N.
• While the second observation is that the remaining diagonals start with end column for each row and ends when either start row < N or start column >= 0.

Below is the implementation of the above approach:

 `// CPP implementation to  return` `// an array of its anti-diagonals` `// when an N*N square matrix is given`   `#include ` `using` `namespace` `std;`   `// function to print the diagonals` `void` `diagonal(``int` `A)` `{`   `    ``int` `N = 3;`   `    ``// For each column start row is 0` `    ``for` `(``int` `col = 0; col < N; col++) {`   `        ``int` `startcol = col, startrow = 0;`   `        ``while` `(startcol >= 0 && startrow < N) {` `            ``cout << A[startrow][startcol] << ``" "``;`   `            ``startcol--;`   `            ``startrow++;` `        ``}` `        ``cout << ``"\n"``;` `    ``}`   `    ``// For each row start column is N-1` `    ``for` `(``int` `row = 1; row < N; row++) {` `        ``int` `startrow = row, startcol = N - 1;`   `        ``while` `(startrow < N && startcol >= 0) {` `            ``cout << A[startrow][startcol] << ``" "``;`   `            ``startcol--;`   `            ``startrow++;` `        ``}` `        ``cout << ``"\n"``;` `    ``}` `}`   `// Driver code` `int` `main()` `{`   `    ``// matrix iniliasation` `    ``int` `A = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };`   `    ``diagonal(A);`   `    ``return` `0;` `}`

 `// JAVA implementation to  return` `// an array of its anti-diagonals` `// when an N*N square matrix is given`   `class` `Matrix {`   `    ``// function to print the diagonals` `    ``void` `diagonal(``int` `A[][])` `    ``{`   `        ``int` `N = ``3``;`   `        ``// For each column start row is 0` `        ``for` `(``int` `col = ``0``; col < N; col++) {`   `            ``int` `startcol = col, startrow = ``0``;`   `            ``while` `(startcol >= ``0` `&& startrow < N) {`   `                ``System.out.print(A[startrow][startcol]` `                                 ``+ ``" "``);`   `                ``startcol--;`   `                ``startrow++;` `            ``}` `            ``System.out.println();` `        ``}`   `        ``// For each row start column is N-1` `        ``for` `(``int` `row = ``1``; row < N; row++) {` `            ``int` `startrow = row, startcol = N - ``1``;`   `            ``while` `(startrow < N && startcol >= ``0``) {` `                ``System.out.print(A[startrow][startcol]` `                                 ``+ ``" "``);`   `                ``startcol--;`   `                ``startrow++;` `            ``}` `            ``System.out.println();` `        ``}` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String args[])` `    ``{`   `        ``// matrix initialisation` `        ``int` `A[][]` `            ``= { { ``1``, ``2``, ``3` `}, { ``4``, ``5``, ``6` `}, { ``7``, ``8``, ``9` `} };`   `        ``Matrix m = ``new` `Matrix();`   `        ``m.diagonal(A);` `    ``}` `}`

 `# Python3 implementation to return` `# an array of its anti-diagonals` `# when an N*N square matrix is given`   `# function to print the diagonals`     `def` `diagonal(A):`   `    ``N ``=` `3`   `    ``# For each column start row is 0` `    ``for` `col ``in` `range``(N):`   `        ``startcol ``=` `col` `        ``startrow ``=` `0`   `        ``while``(startcol >``=` `0` `and` `              ``startrow < N):` `            ``print``(A[startrow][startcol],` `                  ``end``=``" "``)`   `            ``startcol ``-``=` `1` `            ``startrow ``+``=` `1`   `        ``print``()`   `    ``# For each row start column is N-1` `    ``for` `row ``in` `range``(``1``, N):` `        ``startrow ``=` `row` `        ``startcol ``=` `N ``-` `1`   `        ``while``(startrow < N ``and` `              ``startcol >``=` `0``):` `            ``print``(A[startrow][startcol],` `                  ``end``=``" "``)`   `            ``startcol ``-``=` `1` `            ``startrow ``+``=` `1`   `        ``print``()`     `# Driver code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``# matrix iniliasation` `    ``A ``=` `[[``1``, ``2``, ``3``],` `         ``[``4``, ``5``, ``6``],` `         ``[``7``, ``8``, ``9``]]`   `    ``diagonal(A)`   `# This code is contributed by AnkitRai01`

 `// C# implementation to return` `// an array of its anti-diagonals` `// when an N*N square matrix is given` `using` `System;`   `class` `GFG {`   `    ``// Function to print the diagonals` `    ``static` `void` `diagonal(``int``[, ] A)` `    ``{` `        ``int` `N = 3;`   `        ``// For each column start row is 0` `        ``for` `(``int` `col = 0; col < N; col++) {` `            ``int` `startcol = col, startrow = 0;`   `            ``while` `(startcol >= 0 && startrow < N) {` `                ``Console.Write(A[startrow, startcol] + ``" "``);` `                ``startcol--;` `                ``startrow++;` `            ``}` `            ``Console.WriteLine();` `        ``}`   `        ``// For each row start column is N-1` `        ``for` `(``int` `row = 1; row < N; row++) {` `            ``int` `startrow = row, startcol = N - 1;`   `            ``while` `(startrow < N && startcol >= 0) {` `                ``Console.Write(A[startrow, startcol] + ``" "``);` `                ``startcol--;` `                ``startrow++;` `            ``}` `            ``Console.WriteLine();` `        ``}` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{`   `        ``// Matrix initialisation` `        ``int``[, ] A` `            ``= { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };`   `        ``diagonal(A);` `    ``}` `}`   `// This code is contributed by AnkitRai01`

Output:

```1
2 4
3 5 7
6 8
9

```

Time Complexity: Time complexity of the above solution is O(N*N).

Approach 2 : Much simpler and concise  ( Same time Complexity)

In this approach we will make the use of sum of indices of any element in a matrix.   Let indices of any element be represented by i (row) an j (column).

If we find the sum of indices of any element in  a N*N matrix, we will observe that the sum of indices for any element lies between 0 (when i = j = 0) and 2*N – 2 (when i = j = N-1).

So we will follow the following steps:

• Declare a vector of vectors of size 2*N – 1 for holding unique sums from sum = 0 to sum = 2*N – 2.
• Now we will loop through the vector and pushback the elements of similar sum to same row in that vector of vectors.

Below is the implementation of the above approach:

 `// CPP program for the above approach` `#include ` `#include ` `using` `namespace` `std;`   `// Function to print diagonals` `void` `diagonal(vector >& A)` `{`   `    ``int` `n = A.size();` `    ``int` `N = 2 * n - 1;`   `    ``vector > result(N);`   `    ``// Push each element in the result vector` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``for` `(``int` `j = 0; j < n; j++)` `            ``result[i + j].push_back(A[i][j]);` `  `  `    ``// Print the diagonals` `    ``for` `(``int` `i = 0; i < result.size(); i++) ` `    ``{` `        ``cout << endl;` `        ``for` `(``int` `j = 0; j < result[i].size(); j++)` `            ``cout << result[i][j] << ``" "``;` `    ``}` `}`   `// Driver Code` `int` `main()` `{`   `    ``vector > A = { { 1, 2, 3, 4 },` `                               ``{ 5, 6, 7, 8 },` `                               ``{ 9, 10, 11, 12 },` `                               ``{ 13, 14, 15, 16 } };` `    `  `    ``// Function Call` `    ``diagonal(A);`   `    ``return` `0;` `}`

Output :

```1
2 5
3 6 9
4 7 10 13
8 11 14
12 15
16
```

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