Print all the super diagonal elements of the given square matrix

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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)

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