# Emulating a 2-d array using 1-d array in C++

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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

## Example

Live Demo

#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
Updated on 22-Jan-2021 12:48:54