Calculate the address of any element in the 2-D array

The 2-dimensional array can be defined as an array of arrays. The 2-Dimensional arrays are organized as matrices which can be represented as the collection of rows and columns as array[M][N] where M is the number of rows and N is the number of columns. 

Example:

Finding address of an element using Row Major Order: 

Given an array, arr[1………10][1………15] with base value 100 and the size of each element is 1 Byte in memory. Find the address of arr[8][6] with the help of row-major order.

Solution:

Given:
Base address B = 100
Storage size of one element store in any array W = 1 Bytes
Row Subset of an element whose address to be found I = 8
Column Subset of an element whose address to be found J = 6
Lower Limit of row/start row index of matrix LR = 1 
Lower Limit of column/start column index of matrix = 1
Number of column given in the matrix N = Upper Bound – Lower Bound + 1
                                                                            = 15 – 1 + 1
                                                                            = 15

Formula:
Address of A[I][J] = B + W * ((I – LR) * N + (J – LC)) 

Solution:
Address of A[8][6] = 100 + 1 * ((8 – 1) * 15 + (6 – 1))
                                   = 100 + 1 * ((7) * 15 + (5))
                                  = 100 + 1 * (110)
Address of A[I][J] = 210

Finding address of an element using Column Major Order: 

Given an array arr[1………10][1………15] with a base value of 100 and the size of each element is 1 Byte in memory find the address of arr[8][6] with the help of column-major order.

Solution:

Given:
Base address B = 100
Storage size of one element store in any array W = 1 Bytes
Row Subset of an element whose address to be found I = 8
Column Subset of an element whose address to be found J = 6
Lower Limit of row/start row index of matrix LR = 1
Lower Limit of column/start column index of matrix = 1
Number of Rows given in the matrix M = Upper Bound – Lower Bound + 1
                                                                            = 10 – 1 + 1
                                                                           = 10

Formula: used
Address of A[I][J] = B + W * ((J – LC) * M + (I – LR))
Address of A[8][6] = 100 + 1 * ((6 – 1) * 10 + (8 – 1))
                                  = 100 + 1 * ((5) * 10 + (7))
                                 = 100 + 1 * (57)
Address of A[I][J] = 157 

From the above examples, it can be observed that for the same position two different address locations are obtained that’s because in row-major order movement is done across the rows and then down to the next row, and in column-major order, first move down to the first column and then next column. So both the answers are right.

Arrays are fundamental data structures in computer science, and mastering them is crucial for success in the GATE exam. This article aims to provide concise yet comprehensive notes on arrays, covering essential concepts and strategies to help you tackle array-related questions in the GATE 2024 exam.