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Return an array of anti-diagonals of given N*N square matrix
• Difficulty Level : Medium
• Last Updated : 15 Apr, 2021

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:

## C++

 `// C++ 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[3][3])``{` `    ``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[3][3] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };` `    ``diagonal(A);` `    ``return` `0;``}`

## Java

 `// 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

 `# 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#

 `// 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`

## Javascript

 ``
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:

## C++

 `// C++ 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;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;``import` `java.lang.*;` `class` `GFG{`` ` `// Function to print diagonals``static` `void` `diagonal(``int``[][] A)``{``    ``int` `n = A.length;``    ``int` `N = ``2` `* n - ``1``;`` ` `    ``ArrayList> result = ``new` `ArrayList<>();``    ` `    ``for``(``int` `i = ``0``; i < N; i++)``        ``result.add(``new` `ArrayList<>());`` ` `    ``// Push each element in the result vector``    ``for``(``int` `i = ``0``; i < n; i++)``        ``for``(``int` `j = ``0``; j < n; j++)``            ``result.get(i + j).add(A[i][j]);``   ` `    ``// Print the diagonals``    ``for``(``int` `i = ``0``; i < result.size(); i++)``    ``{``        ``System.out.println();``        ``for``(``int` `j = ``0``; j < result.get(i).size(); j++)``            ``System.out.print(result.get(i).get(j) + ``" "``);``    ``}``}``  ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int``[][] A = { { ``1``, ``2``, ``3``, ``4` `},``                  ``{ ``5``, ``6``, ``7``, ``8` `},``                  ``{ ``9``, ``10``, ``11``, ``12` `},``                  ``{ ``13``, ``14``, ``15``, ``16` `} };``     ` `    ``// Function Call``    ``diagonal(A);``}``}` `// This code is contributed by offbeat`

## Python3

 `# Python3 program for the above approach` `# Function to print diagonals``def` `diagonal(A) :` `    ``n ``=` `len``(A)``    ``N ``=` `2` `*` `n ``-` `1` `    ``result ``=` `[]``    ` `    ``for` `i ``in` `range``(N) :``        ``result.append([])``    ` `    ``# Push each element in the result vector``    ``for` `i ``in` `range``(n) :``        ``for` `j ``in` `range``(n) :``            ``result[i ``+` `j].append(A[i][j])` `    ``# Print the diagonals``    ``for` `i ``in` `range``(``len``(result)) :``    ` `        ``for` `j ``in` `range``(``len``(result[i])) :``            ``print``(result[i][j] , end ``=` `" "``)``            ` `        ``print``()` `A ``=` `[ [ ``1``, ``2``, ``3``, ``4` `],``        ``[ ``5``, ``6``, ``7``, ``8` `],``        ``[ ``9``, ``10``, ``11``, ``12` `],``        ``[ ``13``, ``14``, ``15``, ``16` `] ]` `# Function Call``diagonal(A)` `# This code is contributed by divyesh072019`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;``class` `GFG {``    ` `    ``// Function to print diagonals``    ``static` `void` `diagonal(List> A)``    ``{``     ` `        ``int` `n = A.Count;``        ``int` `N = 2 * n - 1;``     ` `        ``List> result = ``new` `List>();``        ` `        ``for` `(``int` `i = 0; i < N; i++)``        ``{``            ``result.Add(``new` `List<``int``>());``        ``}``     ` `        ``// Push each element in the result vector``        ``for` `(``int` `i = 0; i < n; i++)``            ``for` `(``int` `j = 0; j < n; j++)``                ``result[i + j].Add(A[i][j]);``       ` `        ``// Print the diagonals``        ``for` `(``int` `i = 0; i < result.Count; i++)``        ``{``            ``for` `(``int` `j = 0; j < result[i].Count; j++)``                ``Console.Write(result[i][j] + ``" "``);``            ``Console.WriteLine();``        ``}``    ``}``    ` `  ``static` `void` `Main() {``    ``List> A = ``new` `List>();``    ``A.Add(``new` `List<``int``> {1, 2, 3, 4});``    ``A.Add(``new` `List<``int``> {5, 6, 7, 8});``    ``A.Add(``new` `List<``int``> {9, 10, 11, 12});``    ``A.Add(``new` `List<``int``> {13, 14, 15, 16});``      ` `    ``// Function Call``    ``diagonal(A);``  ``}``}`

Output :

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

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