\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.
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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.
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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
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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 matrixint 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 codeint 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 approachclass GFG
{ // Function to return a vector which// consists the sum of// four portions of the matrixstatic 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 codepublic 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 matrixdef 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 codeN = 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 approachusing 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 matrixfunction 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 codevar 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 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 matrixint 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 usagematrix = [[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 usagelet 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
|
Output
68
Time Complexity: O(n^2)
Auxiliary Space: O(1)