Print a 2D Array or Matrix using single loop
Last Updated :
08 Jun, 2021
Given a matrix mat[][] of N * M dimensions, the task is to print the elements of the matrix using a single for loop.
Examples:
Input: mat[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
Output: 1 2 3 4 5 6 7 8 9
Input: mat[][] = {{7, 9}, {10, 34}, {12, 15}}
Output: 7 9 10 34 12 15
Approach: To traverse the given matrix using a single loop, observe that there are only N * M elements. Therefore, the idea is to use modulus and division to switch the rows and columns while iterating a single loop over the range [0, N * M]. Follow the steps below to solve the given problem:
- Iterate a loop over the range [0, N * M] using the variable i.
- At each iteration, find the index of the current row and column as row = i / M and column = i % M respectively.
- In the above steps, print the value of mat[row][column] to get the value of the matrix at that index.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <iostream>
using namespace std;
// Function to print the element
// of 2D matrix using single loop
void print2DMatrix(int arr[][3],
int rows, int columns)
{
// Iterate over the range
// [0, rows*columns]
for(int i = 0; i < rows * columns; i++)
{
// Find row and column index
int row = i / columns;
int col = i % columns;
// Print the element at
// current index
cout << arr[row][col] << " ";
}
}
// Driver Code
int main()
{
// Given matrix mat[][]
int mat[][3] = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
// Dimensions of the matrix
int N = sizeof(mat) / sizeof(mat[0]);
int M = sizeof(mat[0]) / sizeof(mat[0][0]);
// Function Call
print2DMatrix(mat, N, M);
return 0;
}
// This code is contributed by akhilsaini
// Java Program for the above approach
import java.io.*;
class GFG {
// Function to print the element
// of 2D matrix using single loop
public static void print2DMatrix(
int arr[][], int rows, int columns)
{
// Iterate over the range
// [0, rows*columns]
for (int i = 0;
i < rows * columns; i++) {
// Find row and column index
int row = i / columns;
int col = i % columns;
// Print the element at
// current index
System.out.print(
arr[row][col] + " ");
}
}
// Driver Code
public static void main(String[] args)
{
// Given matrix mat[][]
int[][] mat = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
// Dimensions of the matrix
int N = mat.length;
int M = mat[0].length;
// Function Call
print2DMatrix(mat, N, M);
}
}
# Python3 program for the above approach
# Function to print the element
# of 2D matrix using single loop
def print2DMatrix(arr, rows, columns):
# Iterate over the range
# [0, rows*columns]
for i in range(0, rows * columns):
# Find row and column index
row = i // columns
col = i % columns
# Print the element at
# current index
print(arr[row][col], end = ' ')
# Driver Code
if __name__ == '__main__':
# Given matrix mat[][]
mat = [ [ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ] ]
# Dimensions of the matrix
N = len(mat)
M = len(mat[0])
# Function Call
print2DMatrix(mat, N, M)
# This code is contributed by akhilsaini
// C# program for the above approach
using System;
class GFG{
// Function to print the element
// of 2D matrix using single loop
public static void print2DMatrix(int[, ] arr, int rows,
int columns)
{
// Iterate over the range
// [0, rows*columns]
for(int i = 0; i < rows * columns; i++)
{
// Find row and column index
int row = i / columns;
int col = i % columns;
// Print the element at
// current index
Console.Write(arr[row, col] + " ");
}
}
// Driver Code
public static void Main()
{
// Given matrix mat[][]
int[, ] mat = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
// Dimensions of the matrix
int N = mat.GetLength(0);
int M = mat.GetLength(1);
// Function Call
print2DMatrix(mat, N, M);
}
}
// This code is contributed by akhilsaini
<script>
// JavaScript program for the above approach
// Function to print the element
// of 2D matrix using single loop
function print2DMatrix(arr,rows,columns)
{
// Iterate over the range
// [0, rows*columns]
for(let i = 0; i < rows * columns; i++)
{
// Find row and column index
let row = parseInt(i / columns, 10);
let col = i % columns;
// Print the element at
// current index
document.write(arr[row][col] + " ");
}
}
// Given matrix mat[][]
let mat = [ [ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ] ];
// Dimensions of the matrix
let N = mat.length;
let M = mat[0].length;
// Function Call
print2DMatrix(mat, N, M);
</script>
Output: 1 2 3 4 5 6 7 8 9
Time Complexity: O(N * M)
Auxiliary Space: O(1)
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