Given a 2D matrix arr[][] of N*N dimensions, the task is to find the sum of elements of all four parts of the matrix divided by the diagonals without including the diagonal elements in any of the four parts.
Example:
Input: arr[][] = { {1, 2, 3}, {4, 5, 6}, {7, 8, 9} }
Output: 2 4 6 8
Explanation:
(1, 5, 9) and (3, 5, 7) are diagonals of arr[][] matrix. Therefore sum of parts are:
Top = 2
Left = 4
Right = 6
Bottom = 8
Input: arr[][] = { {1, 3, 1, 5}, {2, 2, 4, 1}, {5, 0, 2, 3}, { 1, 3, 3, 5} }
Output: 4 7 4 6
Explanation:
(1, 2, 2, 5) and (5, 4, 0, 1) are diagonals of arr[][] matrix. Therefore sum of parts are:
Top = 3 + 1 = 4
Left = 2 + 5 = 7
Right = 1 + 3 = 4
Bottom = 3 + 3 = 6
Approach:
As shown in the above figure, After the matrix of size NxN is divided by diagonals. We observe the following properties:
- If sum of index of row and column is less than N – 1 then, it belongs to either Top part or Left part.
- If column index is greater than row index it belongs to Top part.
- Else it belongs to Left part.
- Else it belongs to either Right part or Down part.
- If column index is greater than row index it belongs to Right part.
- Else it belongs to Down part.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include "bits/stdc++.h"using namespace std;
// Function to calculate the// sum of all parts of matrixvoid SumOfPartsOfMetrics(int* arr,
int N)
{ // To store the sum of all four
// parts of the diagonals
int top, bottom, left, right;
// Initialise respective sum
// as zero
top = bottom = right = left = 0;
// Traversing the matrix
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
// If i + j < N -1
// then it belongs to
// top or left
if (i + j < N - 1 && i != j) {
// Belongs to top
if (i < j) {
top += (arr + i * N)[j];
}
// Belongs to left
else {
left += (arr + i * N)[j];
}
}
// If i+j > N - 1 then
// it belongs to right
// or bottom
else if (i + j > N - 1 && i != j) {
// Belongs to right
if (i > j) {
bottom += (arr + i * N)[j];
}
// Belongs to bottom
else {
right += (arr + i * N)[j];
}
}
}
}
cout << top << ' ' << left
<< ' ' << right << ' '
<< bottom << endl;
}// Driver Codeint main()
{ int N = 4;
int arr[N][N] = { { 1, 3, 1, 5 },
{ 2, 2, 4, 1 },
{ 5, 0, 2, 3 },
{ 1, 3, 3, 5 } };
// Function call to find print
// sum of al parts
SumOfPartsOfMetrics((int*)arr, N);
return 0;
} |
Java
// Java program for the above approach class GFG {
// Function to calculate the // sum of all parts of matrix static void SumOfPartsOfMetrics(int [][]arr,
int N)
{ // To store the sum of all four
// parts of the diagonals
// Initialise respective sum
// as zero
int top = 0, bottom = 0;
int left = 0, right = 0;
// Traversing the matrix
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
// If i + j < N -1
// then it belongs to
// top or left
if (i + j < N - 1 && i != j)
{
// Belongs to top
if (i < j)
{
top += arr[i][j];
}
// Belongs to left
else
{
left += arr[i][j];
}
}
// If i+j > N - 1 then
// it belongs to right
// or bottom
else if (i + j > N - 1 && i != j)
{
// Belongs to right
if (i > j)
{
bottom += arr[i][j];
}
// Belongs to bottom
else
{
right += arr[i][j];
}
}
}
}
System.out.println(top + " " + left + " " +
right + " " + bottom);
} // Driver Code public static void main (String[] args)
{ int N = 4;
int arr[][] = { { 1, 3, 1, 5 },
{ 2, 2, 4, 1 },
{ 5, 0, 2, 3 },
{ 1, 3, 3, 5 } };
// Function call to find print
// sum of al parts
SumOfPartsOfMetrics(arr, N);
} }// This code is contributed by AnkitRai01 |
Python3
# Python3 program for the above approach # Function to calculate the # sum of all parts of matrix def SumOfPartsOfMetrics(arr, N):
# To store the sum of all four
# parts of the diagonals
# Initialise respective sum
# as zero
top = bottom = right = left = 0;
# Traversing the matrix
for i in range(N):
for j in range(N):
# If i + j < N -1
# then it belongs to
# top or left
if (i + j < N - 1 and i != j):
# Belongs to top
if (i < j):
top += arr[i][j];
# Belongs to left
else:
left += arr[i][j];
# If i+j > N - 1 then
# it belongs to right
# or bottom
elif (i + j > N - 1 and i != j):
# Belongs to right
if (i > j):
bottom += arr[i][j];
# Belongs to bottom
else:
right += arr[i][j];
print(top, left, right, bottom);
# Driver Code if __name__ == "__main__":
N = 4;
arr = [ [ 1, 3, 1, 5 ],
[ 2, 2, 4, 1 ],
[ 5, 0, 2, 3 ],
[ 1, 3, 3, 5 ] ];
# Function call to find print
# sum of al parts
SumOfPartsOfMetrics(arr, N);
# This code is contributed by AnkitRai01 |
C#
// C# program for the above approach using System;
class GFG {
// Function to calculate the // sum of all parts of matrix static void SumOfPartsOfMetrics(int [,]arr,
int N)
{ // To store the sum of all four
// parts of the diagonals
// Initialise respective sum
// as zero
int top = 0, bottom = 0;
int left = 0, right = 0;
// Traversing the matrix
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
// If i + j < N -1
// then it belongs to
// top or left
if (i + j < N - 1 && i != j)
{
// Belongs to top
if (i < j)
{
top += arr[i, j];
}
// Belongs to left
else
{
left += arr[i, j];
}
}
// If i+j > N - 1 then
// it belongs to right
// or bottom
else if (i + j > N - 1 && i != j)
{
// Belongs to right
if (i > j)
{
bottom += arr[i, j];
}
// Belongs to bottom
else
{
right += arr[i, j];
}
}
}
}
Console.WriteLine(top + " " + left + " " +
right + " " + bottom);
} // Driver Code public static void Main (string[] args)
{ int N = 4;
int [,]arr = { { 1, 3, 1, 5 },
{ 2, 2, 4, 1 },
{ 5, 0, 2, 3 },
{ 1, 3, 3, 5 } };
// Function call to find print
// sum of al parts
SumOfPartsOfMetrics(arr, N);
} }// This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript program for the above approach// Function to calculate the// sum of all parts of matrixfunction SumOfPartsOfMetrics(arr, N)
{ // To store the sum of all four
// parts of the diagonals
let top, bottom, left, right;
// Initialise respective sum
// as zero
top = bottom = right = left = 0;
// Traversing the matrix
for (let i = 0; i < N; i++) {
for (let j = 0; j < N; j++) {
// If i + j < N -1
// then it belongs to
// top or left
if (i + j < N - 1 && i != j) {
// Belongs to top
if (i < j) {
top += arr[i][j];
}
// Belongs to left
else {
left += arr[i][j];
}
}
// If i+j > N - 1 then
// it belongs to right
// or bottom
else if (i + j > N - 1 && i != j) {
// Belongs to right
if (i > j) {
bottom += arr[i][j];
}
// Belongs to bottom
else {
right += arr[i][j];
}
}
}
}
document.write(top + ' ' + left
+ ' ' + right + ' '
+ bottom + "<br>");
}// Driver Code let N = 4;
let arr = [ [ 1, 3, 1, 5 ],
[ 2, 2, 4, 1 ],
[ 5, 0, 2, 3 ],
[ 1, 3, 3, 5 ] ];
// Function call to find print
// sum of al parts
SumOfPartsOfMetrics(arr, N);
// This code is contributed by rishavmahato348.</script> |
Output:
4 7 4 6
Time Complexity: O(N2)
Auxiliary Space: O(1) because It is using constant variable
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