Sum of non-diagonal parts of a square Matrix
\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:
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.
- In a 2-dimensional matrix, two diagonals are identified in the following way:
- Principal Diagonal: The first diagonal has the index of the row is equal to the index of the column.
- 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.
- 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
- 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:
- 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.
- 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.
- The left part:
- 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 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 vector 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> |
68
Time Complexity: O(N2) as we are traversing the complete matrix row-wise.
Auxiliary Space: O(1)
Using Two Nested Loops:
Approach:
Define a function sum_non_diagonal that takes in a 2D matrix as input.
Get the length of the matrix (assuming it’s square) and initialize a variable sum to 0.
Use two nested loops to iterate over the rows and columns of the matrix.
Check if the current element is not on the diagonal (i.e., if i is not equal to j and i is not equal to n – j – 1, where n is the length of the matrix).
If the current element is not on the diagonal, add it to the sum.
Return the sum.
C++
#include <iostream> using namespace std; // Function to calculate the sum of non-diagonal // elements of a matrix int sumNonDiagonal( int matrix[4][4]) { // Get the size of the matrix (assuming a 4x4 matrix) int n = 4; // Initialize a variable to hold the sum int sum = 0; // Use two nested loops to iterate over the rows and columns for ( int i = 0; i < n; i++) { for ( int j = 0; j < n; j++) { // Check if the current element is not on the diagonal if (i != j && i != n - j - 1) { // If the current element is not on the diagonal, add it to the sum sum += matrix[i][j]; } } } // Return the sum return sum; } int main() { int matrix[4][4] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16}}; cout << sumNonDiagonal(matrix) << endl; // Output: 68 return 0; } |
Java
/*package whatever //do not write package name here */ import java.io.*; public class GFG { // Function to calculate the sum of non-diagonal // elements of a matrix public static int sumNonDiagonal( int [][] matrix) { // Get the length of the matrix int n = matrix.length; // Initialize a variable to hold the sum int sum = 0 ; // Use two nested loops to iterate over the rows and // columns for ( int i = 0 ; i < n; i++) { for ( int j = 0 ; j < n; j++) { // Check if the current element is not on // the diagonal if (i != j && i != n - j - 1 ) { // If the current element is not on the // diagonal, add it to the sum sum += matrix[i][j]; } } } // Return the sum return sum; } // Driver code public static void main(String[] args) { int [][] matrix = { { 1 , 2 , 3 , 4 }, { 5 , 6 , 7 , 8 }, { 9 , 10 , 11 , 12 }, { 13 , 14 , 15 , 16 } }; System.out.println( sumNonDiagonal(matrix)); // Output: 68 } } |
Python3
def sum_non_diagonal(matrix): # Get the length of the matrix n = len (matrix) # Initialize a variable to hold the sum sum = 0 # Use two nested loops to iterate over the rows and columns for i in range (n): for j in range (n): # Check if the current element is not on the diagonal if i ! = j and i ! = n - j - 1 : # If the current element is not on the diagonal, add it to the sum sum + = matrix[i][j] # Return the sum return sum # Example usage matrix = [[ 1 , 2 , 3 , 4 ], [ 5 , 6 , 7 , 8 ], [ 9 , 10 , 11 , 12 ], [ 13 , 14 , 15 , 16 ]] print (sum_non_diagonal(matrix)) # Output: 68 |
C#
using System; class Program { // Function to calculate the sum of non-diagonal // elements of a matrix static int SumNonDiagonal( int [, ] matrix) { // Get the size of the matrix (assuming a 4x4 // matrix) int n = 4; // Initialize a variable to hold the sum int sum = 0; // Use two nested loops to iterate over the rows and // columns for ( int i = 0; i < n; i++) { for ( int j = 0; j < n; j++) { // Check if the current element is not on // the diagonal if (i != j && i != n - j - 1) { // If the current element is not on the // diagonal, add it to the sum sum += matrix[i, j]; } } } // Return the sum return sum; } static void Main( string [] args) { int [, ] matrix = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; Console.WriteLine( SumNonDiagonal(matrix)); // Output: 68 } } |
Javascript
function sum_non_diagonal(matrix) { // Get the length of the matrix let n = matrix.length; // Initialize a variable to hold the sum let sum = 0; // Use two nested loops to iterate over the rows and columns for (let i = 0; i < n; i++) { for (let j = 0; j < n; j++) { // Check if the current element is not on the diagonal if (i != j && i != n - j - 1) { // If the current element is not on the diagonal, add it to the sum sum += matrix[i][j]; } } } // Return the sum return sum; } // Example usage let matrix = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]; console.log(sum_non_diagonal(matrix)); // Output: 68 |
68
Time Complexity: O(n^2)
Auxiliary Space: O(1)