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Sum of Matrix where each element is sum of row and column number

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  • Difficulty Level : Medium
  • Last Updated : 09 Sep, 2022
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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++




// 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




// 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


Python3




# 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#




// 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


Javascript




<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>


Output

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 time

Total 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++




// 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




// 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




# 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#




// 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


Javascript




<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>


Output

54

Time Complexity: O(M+N)
Auxiliary Space: O(1)


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