Find sum of all elements in a matrix except the elements in row and/or column of given cell?

Given a 2D matrix and a set of cell indexes e.g., an array of (i, j) where i indicates row and j column. For every given cell index (i, j), find sums of all matrix elements except the elements present in i’th row and/or j’th column.

Example:

mat[][]  = { {1, 1, 2}
             {3, 4, 6}
             {5, 3, 2} }
Array of Cell Indexes: {(0, 0), (1, 1), (0, 1)}
Output:  15, 10, 16

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A Naive Solution is to one by once consider all given cell indexes. For every cell index (i, j), find the sum of matrix elements that are not present either at i’th row or at j’th column. Below is C++ implementation of the Naive approach.

C++

#include<bits/stdc++.h> 
#define R 3 
#define C 3 
using namespace std; 
  
// A structure to represent a cell index 
struct Cell 
{  
    int r; // r is row, varies from 0 to R-1 
    int c; // c is column, varies from 0 to C-1 
}; 
  
// A simple solution to find sums for a given array of cell indexes 
void printSums(int mat[][C], struct Cell arr[], int n) 
{ 
    // Iterate through all cell indexes 
    for (int i=0; i<n; i++) 
    { 
        int sum = 0, r = arr[i].r, c = arr[i].c; 
  
        // Compute sum for current cell index 
        for (int j=0; j<R; j++) 
            for (int k=0; k<C; k++) 
                if (j != r && k != c) 
                    sum += mat[j][k]; 
        cout << sum << endl; 
    } 
} 
  
// Driver program to test above 
int main() 
{ 
    int mat[][C] = {{1, 1, 2}, {3, 4, 6}, {5, 3, 2}}; 
    struct Cell arr[] = {{0, 0}, {1, 1}, {0, 1}}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    printSums(mat, arr, n); 
    return 0; 
}

Python3

# Python3 implementation of the approach 
  
# A structure to represent a cell index  
class Cell:  
  
    def __init__(self, r, c): 
        self.r = r # r is row, varies from 0 to R-1 
        self.c = c # c is column, varies from 0 to C-1 
  
# A simple solution to find sums  
# for a given array of cell indexes  
def printSums(mat, arr, n): 
  
    # Iterate through all cell indexes  
    for i in range(0, n):  
      
        Sum = 0; r = arr[i].r; c = arr[i].c  
  
        # Compute sum for current cell index  
        for j in range(0, R):  
            for k in range(0, C):  
                if j != r and k != c:  
                    Sum += mat[j][k]  
        print(Sum)  
  
# Driver Code 
if __name__ == "__main__": 
  
    mat = [[1, 1, 2], [3, 4, 6], [5, 3, 2]] 
    R = C = 3
    arr = [Cell(0, 0), Cell(1, 1), Cell(0, 1)]  
    n = len(arr) 
    printSums(mat, arr, n)  
      
# This code is contributed by Rituraj Jain 

Output:

15
10
16

Time complexity of the above solution is O(n * R * C) where n is number of given cell indexes and R x C is matrix size.

An Efficient Solution can compute all sums in O(R x C + n) time. The idea is to precompute total sum, row and column sums before processing the given array of indexes. Below are details
1. Calculate sum of matrix, call it sum.
2. Calculate sum of individual rows and columns. (row[] and col[])
3. For a cell index (i, j), the desired sum will be “sum- row[i] – col[j] + arr[i][j]”

Below is the implementation of above idea.

C++

// An efficient C++ program to compute sum for given array of cell indexes 
#include<bits/stdc++.h> 
#define R 3 
#define C 3 
using namespace std; 
  
// A structure to represent a cell index 
struct Cell 
{ 
    int r; // r is row, varies from 0 to R-1 
    int c; // c is column, varies from 0 to C-1 
}; 
  
void printSums(int mat[][C], struct Cell arr[], int n) 
{ 
    int sum = 0; 
    int row[R] = {}; 
    int col[C] = {}; 
  
    // Compute sum of all elements, sum of every row and sum every column 
    for (int i=0; i<R; i++) 
    { 
      for (int j=0; j<C; j++) 
       { 
             sum += mat[i][j]; 
             col[j] += mat[i][j]; 
             row[i] += mat[i][j]; 
       } 
    } 
  
    // Compute the desired sum for all given cell indexes 
    for (int i=0; i<n; i++) 
    { 
        int ro = arr[i].r, co = arr[i].c; 
        cout << sum - row[ro] - col[co] + mat[ro][co] << endl; 
    } 
} 
  
// Driver program to test above function 
int main() 
{ 
    int mat[][C] = {{1, 1, 2}, {3, 4, 6}, {5, 3, 2}}; 
    struct Cell arr[] = {{0, 0}, {1, 1}, {0, 1}}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    printSums(mat, arr, n); 
    return 0; 
} 

Python3

# Python3 implementation of the approach 
  
# A structure to represent a cell index 
class Cell: 
  
    def __init__(self, r, c): 
        self.r = r # r is row, varies from 0 to R-1 
        self.c = c # c is column, varies from 0 to C-1 
  
# A simple solution to find sums  
# for a given array of cell indexes 
def printSums(mat, arr, n): 
  
    Sum = 0
    row, col = [0] * R, [0] * C 
  
    # Compute sum of all elements, 
    # sum of every row and sum every column 
    for i in range(0, R): 
        for j in range(0, C): 
            Sum += mat[i][j] 
            row[i] += mat[i][j] 
            col[j] += mat[i][j] 
  
    # Compute the desired sum  
    # for all given cell indexes 
    for i in range(0, n): 
        r0, c0 = arr[i].r, arr[i].c 
        print(Sum - row[r0] - col[c0] + mat[r0][c0]) 
  
# Driver Code 
if __name__ == "__main__": 
  
    mat = [[1, 1, 2], [3, 4, 6], [5, 3, 2]] 
    R = C = 3
    arr = [Cell(0, 0), Cell(1, 1), Cell(0, 1)] 
    n = len(arr) 
    printSums(mat, arr, n) 
  
# This code is contributed by Rituraj Jain 

Output:

15
10
16

Time Complexity: O(R x C + n)
Auxiliary Space: O(R + C)

Thanks to Gaurav Ahirwar for suggesting this efficient solution.

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