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

11 Apr, 2022
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++
Java
Python3
C#
Javascript
// C++ implementation to  return// an array of its anti-diagonals// when an N*N square matrix is given #include <iostream>using namespace std; // function to print the diagonalsvoid 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 codeint main(){     // matrix initialization    int A[3][3] = { { 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 codeif __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 givenusing 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
<script> // Javascript implementation to  return// an array of its anti-diagonals// when an N*N square matrix is given // Function to print the diagonalsfunction diagonal(A){    let N = 3;     // For each column start row is 0    for(let col = 0; col < N; col++)    {        let startcol = col, startrow = 0;         while (startcol >= 0 && startrow < N)        {            document.write(A[startrow][startcol] + " ");             startcol--;            startrow++;        }        document.write("</br>");    }     // For each row start column is N-1    for(let row = 1; row < N; row++)    {        let startrow = row, startcol = N - 1;         while (startrow < N && startcol >= 0)        {            document.write(A[startrow][startcol] + " ");             startcol--;            startrow++;        }        document.write("</br>");    }} // Driver code // matrix iniliasationlet A = [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]; diagonal(A);     // This code is contributed by suresh07 </script>
1 
2 4 
3 5 7 
6 8 
9
 
Time Complexity: O(N*N), Where N is the number of rows or columns of given matrix.Auxiliary Space: O(1)
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++
Java
Python3
C#
Javascript
// C++ program for the above approach#include <iostream>#include <vector>using namespace std; // Function to print diagonalsvoid diagonal(vector<vector<int> >& A){     int n = A.size();    int N = 2 * n - 1;     vector<vector<int> > 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 Codeint main(){     vector<vector<int> > A = { { 1, 2, 3, 4 },                               { 5, 6, 7, 8 },                               { 9, 10, 11, 12 },                               { 13, 14, 15, 16 } };         // Function Call    diagonal(A);     return 0;}
// Java program for the above approachimport java.util.*;import java.lang.*; class GFG{  // Function to print diagonalsstatic void diagonal(int[][] A){    int n = A.length;    int N = 2 * n - 1;      ArrayList<ArrayList<Integer>> 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 codepublic 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 program for the above approach # Function to print diagonalsdef 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 Calldiagonal(A) # This code is contributed by divyesh072019
// C# program for the above approachusing System;using System.Collections.Generic;class GFG {         // Function to print diagonals    static void diagonal(List<List<int>> A)    {              int n = A.Count;        int N = 2 * n - 1;              List<List<int>> result = new List<List<int>>();                 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<List<int>> A = new List<List<int>>();    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);  }}
<script>    // Javascript program for the above approach         // Function to print diagonals    function diagonal(A)    {               let n = A.length;        let N = 2 * n - 1;               let result = [];        for (let i = 0; i < N; i++)        {            result.push([]);        }               // Push each element in the result vector        for (let i = 0; i < n; i++)            for (let j = 0; j < n; j++)                result[i + j].push(A[i][j]);                 // Print the diagonals        for (let i = 0; i < result.length; i++)        {            for (let j = 0; j < result[i].length; j++)                document.write(result[i][j] + " ");            document.write("</br>");        }    }         let 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 mukesh07.</script>
Output : 
1  
2 5  
3 6 9  
4 7 10 13  
8 11 14  
12 15  
16
Time Complexity: O(N*N), Where N is the number of rows or columns of given matrix.Auxiliary Space: O(N*N)
ankthon
kaustubhdwivedi1729
offbeat
divyeshrabadiya07
divyesh072019
suresh07
mukesh07
prophet1999
Algorithms
Arrays
Java
Matrix
Arrays
Matrix
Java
Algorithms
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