Print all the super diagonal elements of the given square matrix

Given a square matrix mat[][] of size n * n. The task is to print all the elements which lie on the super-diagonal of the given matrix.
Examples:

Input: mat[][] = { 
{1, 2, 3}, 
{3, 3, 4, }, 
{2, 4, 6}} 
Output: 2 4
Input: mat[][] = { 
{1, 2, 3, 4}, 
{3, 3, 4, 4}, 
{2, 4, 6, 3}, 
{1, 1, 1, 3}} 
Output: 2 4 3 

Approach: The super-diagonal of a square matrix is the set of elements that lies directly above the elements comprising the main diagonal. As for main diagonal elements, their indexes are like (i = j), for super-diagonal elements their indexes are as j = i + 1 (i denotes row and j denotes column).
Hence elements arr[0][1], arr[1][2], arr[2][3], arr[3][4], …. are the elements of super-diagonal.
Either traverse all elements of matrix and print only those where j = i + 1 which requires O(n²) time complexity or traverse only column from 1 to columnCount – 1 and print elements as arr[column – 1][column].
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define R 4
#define C 4

// Function to print the super diagonal
// elements of the given matrix
void printSuperDiagonal(int arr[R][C])
{
    for (int i = 1; i < C; i++) {
        cout << arr[i - 1][i] << " ";
    }
}

// Driver code
int main()
{
    int arr[R][C] = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 9, 10, 11, 12 },
                      { 13, 14, 15, 16 } };

    printSuperDiagonal(arr);

    return 0;
}

Java

// Java implementation of the approach
import java.io.*;

class GFG
{

static int R = 4;
static int C = 4;

// Function to print the sub diagonal
// elements of the given matrix
static void printSubDiagonal(int arr[][])
{
    for (int i = 1; i < C; i++)
    {
            System.out.print(arr[i-1][i] + " ");
    }
}

// Driver code
public static void main (String[] args)
{

    int arr[][] = { { 1, 2, 3, 4 },
                    { 5, 6, 7, 8 },
                    { 9, 10, 11, 12 },
                    { 13, 14, 15, 16 } };

    printSubDiagonal(arr);

}
}

// This code is contributed by mohit kumar 29

Python3

# Python3 implementation of the approach

R = 4
C = 4

# Function to print the super diagonal
# elements of the given matrix
def printSuperDiagonal(arr) :

    for i in range(1, C) :
        print(arr[i - 1][i],end= " ");

# Driver code
if __name__ == "__main__" :

    arr = [ [ 1, 2, 3, 4 ],
            [5, 6, 7, 8 ],
            [ 9, 10, 11, 12 ],
            [ 13, 14, 15, 16 ]]
    printSuperDiagonal(arr);

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

lass GFG
{

    static int R = 4;
    static int C = 4;

    // Function to print the sub diagonal
    // elements of the given matrix
    static void printSubDiagonal(int [,]arr)
    {
        for (int i = 1; i < C; i++)
        {
                Console.Write(arr[i-1,i] + " ");
        }
    }

    // Driver code
    public static void Main (String[] args)
    {

        int [,]arr = { { 1, 2, 3, 4 },
                        { 5, 6, 7, 8 },
                        { 9, 10, 11, 12 },
                        { 13, 14, 15, 16 } };

        printSubDiagonal(arr);
    }
}

/* This code is contributed by PrinciRaj1992 */

Javascript

<script>

// Javascript implementation of the approach
var R = 4
var C = 4

// Function to print the super diagonal
// elements of the given matrix
function printSuperDiagonal( arr)
{
    for (var i = 1; i < C; i++) {
        document.write( arr[i - 1][i] + " ");
    }
}

// Driver code
var arr = [ [ 1, 2, 3, 4 ],
                  [ 5, 6, 7, 8 ],
                  [ 9, 10, 11, 12 ],
                  [ 13, 14, 15, 16 ] ];
printSuperDiagonal(arr);

</script>

Output:

2 7 12

Time complexity: O(C) where C is no of columns of given matrix

Auxiliary space: O(1)