Sum of middle row and column in Matrix

Given an integer matrix of odd dimensions (3 * 3, 5 * 5). then the task is to find the sum of the middle row & column elements. 

Examples: 

Input :  2 5 7
         3 7 2
         5 6 9
Output : Sum of middle row = 12
         Sum of middle column = 18

Input :  1 3 5 6 7
         3 5 3 2 1
         1 2 3 4 5
         7 9 2 1 6
         9 1 5 3 2
Output : Sum of middle row = 15
         Sum of middle column = 18

Implementation:

CPP

// C++ program to find sum of
// middle row and column in matrix
#include <iostream>
using namespace std;
const int MAX = 100;

void middlesum(int mat[][MAX], int n)
{

    int  row_sum = 0, col_sum = 0;

    //loop for sum of row
    for (int i = 0; i < n; i++)
        row_sum += mat[n / 2][i];

    cout << "Sum of middle row = "
                    << row_sum<<endl;

    //loop for sum of column
    for (int i = 0; i < n; i++)
        col_sum += mat[i][n / 2];

    cout << "Sum of middle column = "
                            << col_sum;
}

// Driver function
int main()
{
    int mat[][MAX] = {{2, 5, 7},
                     {3, 7, 2},
                     {5, 6, 9}};

    middlesum(mat, 3);

    return 0;
}

Java

// java  program to find sum of
// middle row and column in matrix

import java.io.*;

class GFG {


static  int MAX = 100;

 static void middlesum(int mat[][], int n)
{

    int row_sum = 0, col_sum = 0;

    // loop for sum of row
    for (int i = 0; i < n; i++)
        row_sum += mat[n / 2][i];

    System.out.println ( "Sum of middle row = "
                  + row_sum);

    // loop for sum of column
    for (int i = 0; i < n; i++)
        col_sum += mat[i][n / 2];

    System.out.println ( "Sum of middle column = "
                 + col_sum);
}

// Driver function
    public static void main (String[] args) {
        int mat[][] = {{2, 5, 7},
                    {3, 7, 2},
                    {5, 6, 9}};

        middlesum(mat, 3);

    }
}

// This code is contributed by vt_m.

Python3

# Python program to find sum of
# middle row and column in matrix


def middlesum(mat,n):

    row_sum = 0
    col_sum = 0

    # loop for sum of row
    for i in range(n):
        row_sum += mat[n // 2][i]

    print("Sum of middle row = ",
                     row_sum)

    # loop for sum of column
    for i in range(n):
        col_sum += mat[i][n // 2]

    print("Sum of middle column = ",
                            col_sum)

# Driver code
mat= [[2, 5, 7],
     [3, 7, 2],
     [5, 6, 9]]

middlesum(mat, 3)

# This code is contributed
# by Anant Agarwal.

C#

// C# program to find sum of
// middle row and column in matrix
using System;

class GFG {


    //static int MAX = 100;

    static void middlesum(int [,]mat, int n)
    {

        int row_sum = 0, col_sum = 0;

        // loop for sum of row
        for (int i = 0; i < n; i++)
            row_sum += mat[n / 2, i];

        Console.WriteLine ( "Sum of middle row = "
                    + row_sum);

        // loop for sum of column
        for (int i = 0; i < n; i++)
            col_sum += mat[i, n / 2];

        Console.WriteLine ( "Sum of middle column = "
                            + col_sum);
    }

    // Driver function
    public static void Main () {
        int [,]mat = {{2, 5, 7},
                    {3, 7, 2},
                    {5, 6, 9}};

        middlesum(mat, 3);

    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to find sum of
// middle row and column in matrix

function middlesum( $mat,  $n)
{

     $row_sum = 0; $col_sum = 0;

    //loop for sum of row
    for ( $i = 0; $i < $n; $i++)
        $row_sum += $mat[$n / 2][$i];

    echo "Sum of middle row = "
                    , $row_sum,"\n";

    //loop for sum of column
    for ( $i = 0; $i < $n; $i++)
        $col_sum += $mat[$i][$n / 2];

    echo "Sum of middle column = "
                            , $col_sum;
}

// Driver function

     $mat = array(array(2, 5, 7),
                 array(3, 7, 2),
                array(5, 6, 9));

    middlesum($mat, 3);

// This code is contributed by anuj_67.
?>

Javascript

<script>

// Javascript  program to find sum of
// middle row and column in matrix
var MAX = 100;

    function middlesum(mat , n)
    {

        var row_sum = 0, col_sum = 0;

        // loop for sum of row
        for (i = 0; i < n; i++)
            row_sum += mat[parseInt(n / 2)][i];

        document.write(
        "Sum of middle row = " + row_sum+"<br/>"
        );

        // loop for sum of column
        for (i = 0; i < n; i++)
            col_sum += mat[i][parseInt(n / 2)];

        document.write(
        "Sum of middle column = " + col_sum
        );
    }

    // Driver function

        var mat = [ [ 2, 5, 7 ],
                    [ 3, 7, 2 ],
                    [ 5, 6, 9 ] ];

        middlesum(mat, 3);

// This code contributed by aashish1995

</script>

Output

Sum of middle row = 12
Sum of middle column = 18

Time Complexity : O(n) 
Auxiliary Space: O(1) using constant space to initialize row_sum and col_sum variables, since no extra space has been taken.

Using Stl:

Here we use the accumulate function to do it.

For JavaScript:

Here we use the reduce function to do it.

C++

#include <iostream>
#include<bits/stdc++.h>
using namespace std;

int main() {
    vector<vector<int>>v{{2, 5, 7},
                         {3, 7, 2},
                         {5, 6, 9}};
    int n=v.size();
    cout<<"The sum of all the element in middle row : "<<endl;
    cout<<accumulate(v[n/2].begin(),v[n/2].end(),0)<<endl;
    for(int i=0;i<n;i++)
        for(int j=i+1;j<n;j++)
            swap(v[i][j],v[j][i]);
   cout<<"The sum of all the element in middle column : "<<endl;
   cout<<accumulate(v[n/2].begin(),v[n/2].end(),0);
    return 0;
}

Java

import java.util.Arrays;

public class Main {
    public static void main(String[] args) {
        int[][] v = {{2, 5, 7},
                     {3, 7, 2},
                     {5, 6, 9}};
        int n = v.length;
        System.out.println("The sum of all the element in middle row : ");
        System.out.println(Arrays.stream(v[n/2]).sum());
        for(int i = 0; i < n; i++)
            for(int j = i + 1; j < n; j++) {
                int temp = v[i][j];
                v[i][j] = v[j][i];
                v[j][i] = temp;
            }
       System.out.println("The sum of all the element in middle column : ");
       System.out.println(Arrays.stream(v[n/2]).sum());
    }
}

Python3

# Python code to implement the above approach
v = [[2, 5, 7],
     [3, 7, 2],
     [5, 6, 9]]
n = len(v)

print("The sum of all the element in middle row : ")
print(sum(v[n//2]))

for i in range(n):
    for j in range(i+1,n):
        v[i][j],v[j][i] = v[j][i],v[i][j]

print("The sum of all the element in middle column : ")
print(sum(v[n//2]))

Javascript

// JavaScript code to implement the above approach
var v = [[2, 5, 7],
         [3, 7, 2],
         [5, 6, 9]];
var n = v.length;

console.log("The sum of all the element in middle row : ");
console.log(v[Math.floor(n/2)].reduce((a,b)=>a+b,0));

for(var i=0;i<n;i++)
    for(var j=i+1;j<n;j++)
        [v[i][j],v[j][i]] = [v[j][i],v[i][j]];

console.log("The sum of all the element in middle column : ");
console.log(v[Math.floor(n/2)].reduce((a,b)=>a+b,0));

// contributed by rishabh

C#

using System;
using System.Collections.Generic;
using System.Linq;

class Program {
    static void Main(string[] args) {
        List<List<int>> v = new List<List<int>> {
            new List<int> { 2, 5, 7 },
            new List<int> { 3, 7, 2 },
            new List<int> { 5, 6, 9 }
        };
        int n = v.Count;
        Console.WriteLine("The sum of all the element in middle row : ");
        Console.WriteLine(v[n / 2].Sum());
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int temp = v[i][j];
                v[i][j] = v[j][i];
                v[j][i] = temp;
            }
        }
        Console.WriteLine("The sum of all the element in middle column : ");
        Console.WriteLine(v[n / 2].Sum());
    }
}
// This code is contributed by user_dtewbxkn77n

Output

The sum of all the element in middle row : 
12
The sum of all the element in middle column : 
18

The accumulated value of f applications. Complexity: O(n×k), where n is the distance from first to last , O(k) is complexity of f function.

Time Complexity: O(n*k).
Auxiliary Space : O(k)