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
We strongly recommend you to minimize your browser and try this yourself first.
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; } |
chevron_right
filter_none
Java
// Java implementation of the approach class GFG { static int R = 3; static int C = 3; // A structure to represent a cell index static class Cell { int r; // r is row, varies from 0 to R-1 int c; // c is column, varies from 0 to C-1 public Cell(int r, int c) { this.r = r; this.c = c; } }; // A simple solution to find sums for // a given array of cell indexes static void printSums(int mat[][], 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]; } } } System.out.println(sum); } } // Driver code public static void main(String[] args) { int mat[][] = {{1, 1, 2}, {3, 4, 6}, {5, 3, 2}}; Cell arr[] = {new Cell(0, 0), new Cell(1, 1), new Cell(0, 1)}; int n = arr.length; printSums(mat, arr, n); } } // This code is contributed by Princi Singh |
chevron_right
filter_none
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 |
chevron_right
filter_none
C#
// C# implementation of the approach using System; class GFG { static int R = 3; static int C = 3; // A structure to represent a cell index public class Cell { public int r; // r is row, varies from 0 to R-1 public int c; // c is column, varies from 0 to C-1 public Cell(int r, int c) { this.r = r; this.c = c; } }; // A simple solution to find sums for // a given array of cell indexes static void printSums(int [,]mat, 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]; } } } Console.WriteLine(sum); } } // Driver code public static void Main(String[] args) { int [,]mat = {{1, 1, 2}, {3, 4, 6}, {5, 3, 2}}; Cell []arr = {new Cell(0, 0), new Cell(1, 1), new Cell(0, 1)}; int n = arr.Length; printSums(mat, arr, n); } } /* This code is contributed by PrinciRaj1992 */ |
chevron_right
filter_none
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; } |
chevron_right
filter_none
Java
// An efficient Java program to compute // sum for given array of cell indexes class GFG { static int R = 3; static int C = 3; // A structure to represent a cell index static class Cell { int r; // r is row, varies from 0 to R-1 int c; // c is column, varies from 0 to C-1 public Cell(int r, int c) { this.r = r; this.c = c; } }; static void printSums(int mat[][], Cell arr[], int n) { int sum = 0; int []row = new int[R]; int []col = new int[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; System.out.println(sum - row[ro] - col[co] + mat[ro][co]); } } // Driver Code public static void main(String[] args) { int mat[][] = {{1, 1, 2}, {3, 4, 6}, {5, 3, 2}}; Cell arr[] = {new Cell(0, 0), new Cell(1, 1), new Cell(0, 1)}; int n = arr.length; printSums(mat, arr, n); } } // This code is contributed by Princi Singh |
chevron_right
filter_none
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 |
chevron_right
filter_none
C#
// An efficient C# program to compute // sum for given array of cell indexes using System; class GFG { static int R = 3; static int C = 3; // A structure to represent a cell index public class Cell { public int r; // r is row, varies from 0 to R-1 public int c; // c is column, varies from 0 to C-1 public Cell(int r, int c) { this.r = r; this.c = c; } }; static void printSums(int [,]mat, Cell []arr, int n) { int sum = 0; int []row = new int[R]; int []col = new int[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; Console.WriteLine(sum - row[ro] - col[co] + mat[ro, co]); } } // Driver Code public static void Main(String[] args) { int [,]mat = {{1, 1, 2}, {3, 4, 6}, {5, 3, 2}}; Cell []arr = {new Cell(0, 0), new Cell(1, 1), new Cell(0, 1)}; int n = arr.Length; printSums(mat, arr, n); } } // This code is contributed by Rajput-Ji |
chevron_right
filter_none
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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
