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Emulating a 2-d array using 1-d array in C++
In this problem, we will understand the conversion of 2-D array to 1-D array. We will see how to store the elements of a 2-D array to a 1-D array.
Here, the size of 1-D array is same as the total number of elements in 2-D array which is n*m.
In programming there are two ways to store a 2-D array to 1-D array. They are−
- Row Major
- Column Major
Row Major: In row major, All the elements of a row are stored together then it moves to the next row.
If an element of 2-D array of size nXm has an index (i, j) is stored in 1-D array, its index in 1-D array is
(j) + (i)*m
Column Major: In column major, all elements of a column are stored together then the next column is traversed.
If an element of 2-D array of size nXm has an index (i, j) is stored in 1-D array, its index in 1-D array is
(i) + (j)*n
Lets see an example to understand the problem,
Input: n = 3, m = 5, (i,j) = (0, 2)
Output: row-major =
column-major =
Explanation:
Row-major = 2 + 0*3 = 3
Col-major = 0 + 2*5 = 10
Program to illustrate the conversion of 2-D to 1-D,
Example
#include<iostream>
using namespace std;
int main() {
int n = 3;
int m = 5;
int grid[n][m] = {{1, 2, 3},
{4, 5, 6},
{7, 8, 9},
{10, 11, 12},
{13, 14, 15}};
int i = 0;
int j = 2;
int rowMajorIndex = i*n + j;
cout<<"Index of element at index (0, 2) in 1-D array using row-major is "<<rowMajorIndex<<endl;
int colMajorIndex = i + j*m;
cout<<"Index of element at index (0, 2) in 1-D array using column-major is "<<colMajorIndex<<endl;
return 0;
}Output −
Index of element at index (0, 2) in 1-D array using row-major is 2 Index of element at index (0, 2) in 1-D array using column-major is 10
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