Given two numbers M and N denoting the number of rows and columns of a matrix A[] where A[i][j] is the sum of i and j (indices follow 1 based indexing), the task is to find the sum of elements of the matrix.
Examples:
Input: M = 3, N = 3
Output: 36
Explanation: A[]: {{2, 3, 4}, {3, 4, 5}, {4, 5, 6}}. Sum of matrix: 36.Input: M = 3, N = 4
Output: 54
Explanation: A[]: {{2, 3, 4, 5}, {3, 4, 5, 6}, {4, 5, 6, 7}}, Sum of matrix: 54
Naive Approach: To solve the problem follow the below idea:
Create a matrix of size M x N. While creating matrix make element at position (i, j) equal to i + j, where i and j are indices
(1 based indexing) of row and column of matrix. At last traverse on the matrix and return the sum.
Below is the implementation of the above approach:
// C++ Code to Implement the approach #include <bits/stdc++.h> using namespace std;
// Function to find the sum of the matrix int summation( int M, int N)
{ int matrix[M][N], sum = 0;
// Loop to form the matrix and find its sum
for ( int i = 0; i < M; i++) {
for ( int j = 0; j < N; j++) {
matrix[i][j] = (i + 1) + (j + 1);
sum += matrix[i][j];
}
}
// Return the sum of the matrix
return sum;
} // Driver Code int main()
{ int M = 3, N = 4;
// Function Call
cout << summation(M, N);
return 0;
} |
// Java Code to Implement the approach import java.io.*;
class GFG {
// Function to find the sum of the matrix
public static int summation( int M, int N)
{
int matrix[][] = new int [M][N];
int sum = 0 ;
// Loop to form the matrix and find its sum
for ( int i = 0 ; i < M; i++) {
for ( int j = 0 ; j < N; j++) {
matrix[i][j] = (i + 1 ) + (j + 1 );
sum += matrix[i][j];
}
}
// Return the sum of the matrix
return sum;
}
// Driver Code
public static void main(String[] args)
{
int M = 3 , N = 4 ;
// Function Call
System.out.print(summation(M, N));
}
} // This code is contributed by Rohit Pradhan |
# Python code to implement the approach # Function to find the sum of the matrix def summation(M, N):
sum = 0
rows, cols = ( 5 , 5 )
matrix = [[ 0 ] * cols] * rows
# Loop to form the matrix and find its sum
for i in range ( 0 , M):
for j in range ( 0 , N):
matrix[i][j] = (i + 1 ) + (j + 1 )
sum + = matrix[i][j]
# Return the sum of matrix
return sum
M = 3
N = 4
# Function call print (summation(M, N))
# This code is contributed by lokeshmvs21. |
// C# Code to Implement the approach using System;
public class GFG {
// Function to find the sum of the matrix
public static int summation( int M, int N)
{
int [,]matrix = new int [M,N];
int sum = 0;
// Loop to form the matrix and find its sum
for ( int i = 0; i < M; i++) {
for ( int j = 0; j < N; j++) {
matrix[i,j] = (i + 1) + (j + 1);
sum += matrix[i,j];
}
}
// Return the sum of the matrix
return sum;
}
// Driver Code
public static void Main( string [] args)
{
int M = 3, N = 4;
// Function Call
Console.WriteLine(summation(M, N));
}
} // This code is contributed by AnkThon |
<script> // Javascript Code to Implement the approach // Function to find the sum of the matrix function summation(M, N)
{ let matrix = [];
let sum =0;
for (let i=0;i<M;i++)
{matrix[i] = [];
for (let j=0;j<N;j++){
matrix[i][j] = 0
}
}
// Loop to form the matrix and find its sum
for (let i = 0; i < M; i++) {
for (let j = 0; j < N; j++) {
matrix[i][j] = (i + 1) + (j + 1);
sum += matrix[i][j];
}
}
// Return the sum of the matrix
return sum;
} // Driver Code let M = 3;
let N = 4;
// Function Call
console.log(summation(M, N));
// This code is contributed by akashish__
</script> |
54
Time Complexity: O(M*N)
Auxiliary Space: O(M*N)
Efficient Approach: The approach to this problem is based on the following observation.
The elements of the matrix are repeating certain number of times.
If observed carefully, you can see an element with value (i + j) repeats min(i+j-1, N+M – (i+j-1)) times. and the sum of elements lies in the range [2, N+M]. So traverse in the range and find the repetition and find the sum of the matrix.
Illustration:
let M = 3 and N = 4 then matrix will be: {{2, 3, 4, 5}, {3, 4, 5, 6}, {4, 5, 6, 7}}
Occurrence of elements in matrix:
2 -> 1 time
3 -> 2 times
4 -> 3 times
5 -> 3 times
6 -> 2 times
7 -> 1 timeTotal Summation = 2*1 + 3*2 + 4*3 + 5*3 + 6*2 + 7*1 = 54
Follow the steps mentioned below to implement the idea:
- Create a variable sum = 0 and start = M+N.
- Traverse in a loop from i = 1 to M.
- If the current index is greater than or equal to N, increment the sum by (start * N). Otherwise, increment the sum by (start * i).
- For each iteration decrement start.
- Traverse in a loop from i = N – 1 to 0
- If the current index is greater than or equal to M, increment the sum by (start * M). Otherwise, increment the sum by (start * i).
- For each iteration decrement start
- Return the sum as the required answer.
Below is the implementation of the above approach.
// C++ Code to Implement the Idea #include <bits/stdc++.h> using namespace std;
// Function to find the sum of the matrix int summation( int M, int N)
{ int sum = 0, start = M + N;
for ( int i = 1; i <= M; i++) {
if (i >= N) {
sum += start * N;
}
else {
sum += start * i;
}
start--;
}
for ( int i = N - 1; i >= 1; i--) {
if (i >= M) {
sum += start * M;
}
else {
sum += start * i;
}
start--;
}
// Return the sum
return sum;
} // Driver Code int main()
{ int M = 3, N = 4;
// Function Call
cout << summation(M, N);
return 0;
} |
// Java Code to Implement the Idea import java.io.*;
import java.util.*;
class GFG
{ // Function to find the sum of the matrix
public static int summation( int M, int N)
{
int sum = 0 , start = M + N;
for ( int i = 1 ; i <= M; i++) {
if (i >= N) {
sum += start * N;
}
else {
sum += start * i;
}
start--;
}
for ( int i = N - 1 ; i >= 1 ; i--) {
if (i >= M) {
sum += start * M;
}
else {
sum += start * i;
}
start--;
}
// Return the sum
return sum;
}
// Driver program to test above
public static void main(String[] args)
{
int M = 3 , N = 4 ;
// Function Call
System.out.println(summation(M, N));
}
} //this code is contributed by aditya942003patil |
# python3 implementation of above approach # Function to find the sum of the matrix def summation(M, N) :
sum = 0 ; start = M + N;
for i in range ( 1 ,M + 1 ) :
if (i > = N) :
sum + = start * N
else :
sum + = start * i
start - = 1
for i in range (N - 1 , 0 , - 1 ) :
if (i > = M) :
sum + = start * M
else :
sum + = start * i
start - = 1
# Return the sum
return sum ;
# Driver code if __name__ = = "__main__" :
M , N = 3 , 4 ;
# Function Call
print (summation(M, N))
# this code is contributed by aditya942003patil |
// C# code to implement the approach using System;
class GFG
{ // Function to find the sum of the matrix
public static int summation( int M, int N)
{
int sum = 0, start = M + N;
for ( int i = 1; i <= M; i++) {
if (i >= N) {
sum += start * N;
}
else {
sum += start * i;
}
start--;
}
for ( int i = N - 1; i >= 1; i--) {
if (i >= M) {
sum += start * M;
}
else {
sum += start * i;
}
start--;
}
// Return the sum
return sum;
}
// Driver Code public static void Main()
{ int M = 3, N = 4;
// Function Call
Console.WriteLine(summation(M, N));
} } // This code is contributed by aditya942003patil |
<script> // Javascript code to implement the approach. // Function to find the sum of the matrix function summation(M, N)
{ let sum = 0, start = M + N, i;
for (i = 1; i <= M; i++) {
if (i >= N) {
sum += start * N;
}
else {
sum += start * i;
}
start--;
}
for (i = N - 1; i >= 1; i--) {
if (i >= M) {
sum += start * M;
}
else {
sum += start * i;
}
start--;
}
// Return the sum
return sum;
} let M = 3, N = 4;
// Function Call
document.write(summation(M, N));
// This code is contributed by aditya942003patil.
</script>
|
54
Time Complexity: O(M+N)
Auxiliary Space: O(1)
Another Efficient Approach( Time Optimization ): we can do following observation to find sum of all matrix elements in O(1) time .
Steps were to follow this problem:
- We can see that sum of first row is sum of first (n+1) natural number -1.First row sum = (n+1)*(n+2) /2 -1 .
- We can also observe that sum of second row is equal to sum of first row + n .and sum of third row equal to sum of second row + n. and so on for all next rows.
- So , our temp_sum equal to m*sum of first row . because there is atleast ‘sum of first row’ in every row.
- And sum of A.P series we can find by formula .
- SO , our total sum will be sum of temp_sum and sum of A.P series.
- Finally return total sum .
Below is the implementation of the above approach:
// C++ Code to Implement the Idea #include <bits/stdc++.h> using namespace std;
// Function to find the total sum of the elements in the matrix int summation( int M, int N)
{ int N1 = N+1; // N! = N+1 because there is a sum of first N+1
int first_row_sum = (N1*(N1+1))/2-1; // natural numbers
int total_sum = M * first_row_sum ;
//A.P series - N + 2*N + 3*N +......
int a = N; // first term of A.P series
int d = N; // common difference of A.P series
int n = M-1; // n equal to no. of terms in A.P series
// USing standard formula for finding sum of A.P series
// which you have already learned in 9-10th class
int sum_of_AP = (n*(2*a + (n-1)*d)) /2 ;
total_sum += sum_of_AP;
return total_sum; // Return total sum of matrix
} // Drive Code int main()
{ int M = 3, N = 4;
// Function Call
cout << summation(M, N);
return 0;
} // This Approach is contributed by nikhilsainiofficial546 |
# Python Code to Implement the Idea # Function to find the total sum of the elements in the matrix def summation(M, N):
N1 = N + 1 # N! = N+1 because there is a sum of first N+1 natural numbers
first_row_sum = (N1 * (N1 + 1 )) / / 2 - 1
total_sum = M * first_row_sum
# A.P series - N + 2*N + 3*N +......
a = N # first term of A.P series
d = N # common difference of A.P series
n = M - 1 # n equal to no. of terms in A.P series
# Using standard formula for finding sum of A.P series
# which you have already learned in 9-10th class
sum_of_AP = (n * ( 2 * a + (n - 1 ) * d)) / / 2
total_sum + = sum_of_AP
return total_sum # Return total sum of matrix
# Drive Code if __name__ = = '__main__' :
M = 3
N = 4
# Function Call
print (summation(M, N))
|
// JavaScript Code to Implement the Idea // Function to find the total sum of the elements in the matrix function summation(M, N)
{ let N1 = N + 1; // N! = N+1 because there is a sum of first N+1 natural numbers
let first_row_sum = (N1 * (N1 + 1)) / 2 - 1;
let total_sum = M * first_row_sum;
//A.P series - N + 2N + 3N +......
let a = N; // first term of A.P series
let d = N; // common difference of A.P series
let n = M - 1; // n equal to no. of terms in A.P series
// Using standard formula for finding sum of A.P series
// which you have already learned in 9-10th class
let sum_of_AP = (n * (2 * a + (n - 1) * d)) / 2;
total_sum += sum_of_AP;
return total_sum; // Return total sum of matrix
} // Drive Code let M = 3, N = 4; // Function Call console.log(summation(M, N)); |
// C# Code to Implement the Idea using System;
public class GFG{
// Function to find the total sum of the elements in the matrix
public static int Summation( int M, int N)
{
int N1 = N + 1; // N! = N+1 because there is a sum of first N+1
int first_row_sum = (N1 * (N1 + 1)) / 2 - 1; // sum of first N+1 natural numbers
int total_sum = M * first_row_sum;
// A.P series - N + 2*N + 3*N +...
int a = N; // first term of A.P series
int d = N; // common difference of A.P series
int n = M - 1; // n equal to no. of terms in A.P series
// Using standard formula for finding sum of A.P series
// which you have already learned in 9-10th class
int sum_of_AP = (n * (2 * a + (n - 1) * d)) / 2;
total_sum += sum_of_AP;
return total_sum; // Return total sum of matrix
}
// Driver Code
public static void Main( string [] args)
{
int M = 3, N = 4;
// Function Call
Console.WriteLine(Summation(M, N));
}
} |
// Java Code to Implement the Idea public class GFG {
// Function to find the total sum of the elements in the matrix
public static int summation( int M, int N) {
int N1 = N + 1 ; // N! = N+1 because there is a sum of first N+1
int first_row_sum = (N1 * (N1 + 1 )) / 2 - 1 ; // sum of first N+1 natural numbers
int total_sum = M * first_row_sum;
// A.P series - N + 2*N + 3*N +...
int a = N; // first term of A.P series
int d = N; // common difference of A.P series
int n = M - 1 ; // n equal to no. of terms in A.P series
// Using standard formula for finding sum of A.P series
// which you have already learned in 9-10th class
int sum_of_AP = (n * ( 2 * a + (n - 1 ) * d)) / 2 ;
total_sum += sum_of_AP;
return total_sum; // Return total sum of matrix
}
// Driver Code
public static void main(String[] args) {
int M = 3 , N = 4 ;
// Function Call
System.out.println(summation(M, N));
}
} |
54
Time Complexity: O(1)
Auxiliary Space: O(1)