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Sum of non-diagonal parts of a square Matrix

  • Difficulty Level : Basic
  • Last Updated : 08 Apr, 2021

Given a square matrix of size N X N, the task is to find the sum of all elements at each portion when the matrix is divided into four parts along its diagonals. The elements at the diagonals should not be counted in the sum.
Examples: 
 

Input: arr[][] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16}} 
Output: 68 
Explanation: 
 

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From the above image, (1, 6, 11, 16) and (4, 7, 10, 13) are the diagonals. 
The sum of the elements needs to be found are: 
Top: (2 + 3) = 5 
Left: (5 + 9) = 14 
Bottom: (14 + 15) = 29 
Right: (8 + 12) = 20 
Therefore, sum of all parts = 68.
Input: arr[][] = { {1, 3, 1, 5}, {2, 2, 4, 1}, {5, 0, 2, 3}, {1, 3, 1, 5}} 
Output: 19 
 

 

Approach: The idea is to use indexing to identify the elements at the diagonals. 
 

  1. In a 2-dimensional matrix, two diagonals are identified in the following way: 
    1. Principal Diagonal: The first diagonal has the index of the row is equal to the index of the column. 
       
Condition for Principal Diagonal:
The row-column condition is row = column.
  1.  
  2. Secondary Diagonal: The second diagonal has the sum of the index of row and column equal to N(size of the matrix). 
     
Condition for Secondary Diagonal:
The row-column condition is row = numberOfRows - column -1
  1.  
  2. After identifying both the diagonals, the matrix can further be divided into two parts using the diagonal passing through the first element of the last row and the last element of the first row: 
    1. The left part: 
      • If the column index is greater than row index, the element belongs to the top portion of the matrix.
      • If the row index is greater than column index, the element belongs to the left portion of the matrix.
    2. The right part: 
      • If the column index is greater than row index, the element belongs to the right portion of the matrix.
      • If the row index is greater than column index, the element belongs to the bottom portion of the matrix.

  1. So in order to get the sum of the non-diagonal parts of the matrix: 
    • Traverse the matrix rowwise
    • If the element is a part of diagonal, then skip this element
    • If the element is part of the left, right, bottom, or top part (i.e. non-diagonal parts), add the the element in the resultant sum

Below is the implementation of the above approach: 
 

C++




// C++ implementation of the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return a vector which
// consists the sum of
// four portions of the matrix
int sumOfParts(int* arr, int N)
{
    int sum_part1 = 0, sum_part2 = 0,
        sum_part3 = 0, sum_part4 = 0;
    int totalsum = 0;
 
    // Iterating through the matrix
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
 
            // Condition for selecting all values
            // before the second diagonal of metrics
            if (i + j < N - 1) {
 
                // Top portion of the matrix
                if (i < j and i != j and i + j)
                    sum_part1 += (arr + i * N)[j];
 
                // Left portion of the matrix
                else if (i != j)
                    sum_part2 += (arr + i * N)[j];
            }
            else {
 
                // Bottom portion of the matrix
                if (i > j and i + j != N - 1)
                    sum_part3 += (arr + i * N)[j];
 
                // Right portion of the matrix
                else {
                    if (i + j != N - 1 and i != j)
                        sum_part4 += (arr + i * N)[j];
                }
            }
        }
    }
 
    // Adding all the four portions into a vector
    totalsum = sum_part1 + sum_part2
               + sum_part3 + sum_part4;
    return totalsum;
}
 
// Driver code
int main()
{
    int N = 4;
    int arr[N][N] = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 9, 10, 11, 12 },
                      { 13, 14, 15, 16 } };
 
    cout << sumOfParts((int*)arr, N);
}

Java




// Java implementation of the above approach
class GFG
{
  
// Function to return a vector which
// consists the sum of
// four portions of the matrix
static int sumOfParts(int[][] arr, int N)
{
    int sum_part1 = 0, sum_part2 = 0,
        sum_part3 = 0, sum_part4 = 0;
    int totalsum = 0;
  
    // Iterating through the matrix
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
  
            // Condition for selecting all values
            // before the second diagonal of metrics
            if (i + j < N - 1) {
  
                // Top portion of the matrix
                if (i < j && i != j && i + j > 0)
                    sum_part1 += arr[i][j];
  
                // Left portion of the matrix
                else if (i != j)
                    sum_part2 += arr[i][j];
            }
            else {
  
                // Bottom portion of the matrix
                if (i > j && i + j != N - 1)
                    sum_part3 += arr[i][j];
  
                // Right portion of the matrix
                else {
                    if (i + j != N - 1 && i != j)
                        sum_part4 += arr[i][j];
                }
            }
        }
    }
  
    // Adding all the four portions into a vector
    totalsum = sum_part1 + sum_part2
               + sum_part3 + sum_part4;
    return totalsum;
}
  
// Driver code
public static void main(String[] args)
{
    int N = 4;
    int arr[][] = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 9, 10, 11, 12 },
                      { 13, 14, 15, 16 } };
  
    System.out.print(sumOfParts(arr, N));
}
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python3 implementation of the above approach
 
# Function to return a vector which
# consists the sum of
# four portions of the matrix
def sumOfParts(arr,N):
    sum_part1, sum_part2, sum_part3, \
    sum_part4 = 0, 0, 0, 0
    totalsum = 0
 
    # Iterating through the matrix
    for i in range(N):
        for j in range(N):
             
            # Condition for selecting all values
            # before the second diagonal of metrics
            if i + j < N - 1:
                 
                # Top portion of the matrix
                if(i < j and i != j and i + j):
                    sum_part1 += arr[i][j]
                 
                # Left portion of the matrix
                elif i != j:
                    sum_part2 += arr[i][j]
            else:
                 
                # Bottom portion of the matrix
                if i > j and i + j != N - 1:
                    sum_part3 += arr[i][j]
                else:
                     
                # Right portion of the matrix
                    if i + j != N - 1 and i != j:
                        sum_part4 += arr[i][j]
        # Adding all the four portions into a vecto
    return sum_part1 + sum_part2 + sum_part3 + sum_part4
 
# Driver code
N = 4
arr = [[ 1, 2, 3, 4 ],
       [ 5, 6, 7, 8 ],
       [ 9, 10, 11, 12 ],
       [ 13, 14, 15, 16 ]]
 
print(sumOfParts(arr, N))
 
# This code is contributed by mohit kumar 29

C#




// C# implementation of the above approach
using System;
 
class GFG
{
  
    // Function to return a vector which
    // consists the sum of
    // four portions of the matrix
    static int sumOfParts(int[,] arr, int N)
    {
        int sum_part1 = 0, sum_part2 = 0,
            sum_part3 = 0, sum_part4 = 0;
        int totalsum = 0;
      
        // Iterating through the matrix
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
      
                // Condition for selecting all values
                // before the second diagonal of metrics
                if (i + j < N - 1) {
      
                    // Top portion of the matrix
                    if (i < j && i != j && i + j > 0)
                        sum_part1 += arr[i, j];
      
                    // Left portion of the matrix
                    else if (i != j)
                        sum_part2 += arr[i, j];
                }
                else {
      
                    // Bottom portion of the matrix
                    if (i > j && i + j != N - 1)
                        sum_part3 += arr[i, j];
      
                    // Right portion of the matrix
                    else {
                        if (i + j != N - 1 && i != j)
                            sum_part4 += arr[i, j];
                    }
                }
            }
        }
      
        // Adding all the four portions into a vector
        totalsum = sum_part1 + sum_part2
                   + sum_part3 + sum_part4;
        return totalsum;
    }
      
    // Driver code
    public static void Main()
    {
        int N = 4;
        int [,]arr = { { 1, 2, 3, 4 },
                          { 5, 6, 7, 8 },
                          { 9, 10, 11, 12 },
                          { 13, 14, 15, 16 } };
      
        Console.WriteLine(sumOfParts(arr, N));
    }
}
 
// This code is contributed by Yash_R

Javascript




<script>
 
// javascript implementation of the above approach
 
  
// Function to return a vector which
// consists the sum of
// four portions of the matrix
function sumOfParts(arr , N)
{
    var sum_part1 = 0, sum_part2 = 0,
        sum_part3 = 0, sum_part4 = 0;
    var totalsum = 0;
  
    // Iterating through the matrix
    for (i = 0; i < N; i++) {
        for (j = 0; j < N; j++) {
  
            // Condition for selecting all values
            // before the second diagonal of metrics
            if (i + j < N - 1) {
  
                // Top portion of the matrix
                if (i < j && i != j && i + j > 0)
                    sum_part1 += arr[i][j];
  
                // Left portion of the matrix
                else if (i != j)
                    sum_part2 += arr[i][j];
            }
            else {
  
                // Bottom portion of the matrix
                if (i > j && i + j != N - 1)
                    sum_part3 += arr[i][j];
  
                // Right portion of the matrix
                else {
                    if (i + j != N - 1 && i != j)
                        sum_part4 += arr[i][j];
                }
            }
        }
    }
  
    // Adding all the four portions into a vector
    totalsum = sum_part1 + sum_part2
               + sum_part3 + sum_part4;
    return totalsum;
}
  
// Driver code
var N = 4;
var arr = [ [ 1, 2, 3, 4 ],
                  [ 5, 6, 7, 8 ],
                  [ 9, 10, 11, 12 ],
                  [ 13, 14, 15, 16 ] ];
 
document.write(sumOfParts(arr, N));
 
// This code is contributed by 29AjayKumar
 
</script>
Output: 
68

 

Time Complexity: O(N2) as we are traversing the complete matrix rowwise.
 




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