# Find maximum element of each row in a matrix

Last Updated : 28 Apr, 2023

Given a matrix, the task is to find the maximum element of each row.

Examples:

```Input :  [1, 2, 3]
[1, 4, 9]
[76, 34, 21]

Output :
3
9
76

Input : [1, 2, 3, 21]
[12, 1, 65, 9]
[1, 56, 34, 2]
Output :
21
65
56```

Approach : Approach is very simple. The idea is to run the loop for no_of_rows. Check each element inside the row and find for the maximum element. Finally, print the element.

Implementation:

## C++

 `#include` `using` `namespace` `std;`   `const` `int` `N = 4;`   `void` `printArray(``int` `result[], ``int` `no_of_rows) {` `    ``for` `(``int` `i = 0; i < no_of_rows; i++) {` `        ``cout<< result[i]<<``"\n"``;` `    ``}` `}`   `void` `maxelement(``int` `no_of_rows, ``int` `arr[][N]) {` `    ``int` `result[no_of_rows];` `    ``for` `(``int` `i = 0; i < no_of_rows; i++) {` `        ``int` `max = *max_element(arr[i], arr[i]+N);` `        ``result[i] = max;` `    ``}` `    ``printArray(result,no_of_rows);` `}`   `int` `main() {` `    ``int` `arr[][N] = { {3, 4, 1, 8},` `                    ``{1, 4, 9, 11},` `                    ``{76, 34, 21, 1},` `                    ``{2, 1, 4, 5} };` `    ``maxelement(4, arr);` `    ``return` `0;` `}`

## Java

 `// Java program to find maximum ` `// element of each row in a matrix` `public` `class` `GFG{`   `    ``// Function to get max element` `    ``public` `static` `void` `maxelement(``int` `no_of_rows, ``int``[][] arr) {` `        ``int` `i = ``0``;` `        `  `        ``// Initialize max to 0 at beginning` `        ``// of finding max element of each row` `        ``int` `max = ``0``;` `        ``int``[] result = ``new` `int``[no_of_rows];` `        ``while` `(i < no_of_rows) {` `            ``for` `(``int` `j = ``0``; j < arr[i].length; j++) {` `                ``if` `(arr[i][j] > max) {` `                    ``max = arr[i][j];` `                ``}` `            ``}` `            ``result[i] = max;` `            ``max =``0``;` `            ``i++;`   `        ``}` `        ``printArray(result);`   `    ``}`   `    ``// Print array element` `    ``private` `static` `void` `printArray(``int``[] result) {` `        ``for` `(``int` `i =``0``; i

## Python

 `# Python program to find maximum ` `# element of each row in a matrix`   `# importing numpy` `import` `numpy`   `# Function to get max element` `def` `maxelement(arr):` `    `  `    ``# get number of rows and columns` `    ``no_of_rows ``=` `len``(arr)` `    ``no_of_column ``=` `len``(arr[``0``])` `    `  `    ``for` `i ``in` `range``(no_of_rows):` `        `  `        ``# Initialize max1 to 0 at beginning` `        ``# of finding max element of each row` `        ``max1 ``=` `0` `        ``for` `j ``in` `range``(no_of_column):` `            ``if` `arr[i][j] > max1 :` `                ``max1 ``=` `arr[i][j]` `                `  `        ``# print maximum element of each row` `        ``print``(max1)`   `# Driver Code` `arr ``=` `[[``3``, ``4``, ``1``, ``8``],` `       ``[``1``, ``4``, ``9``, ``11``],` `       ``[``76``, ``34``, ``21``, ``1``],` `       ``[``2``, ``1``, ``4``, ``5``]]`   `# Calling the function        ` `maxelement(arr)`

## C#

 `// C# program to find maximum ` `// element of each row in a matrix ` `using` `System;`   `class` `GFG` `{`   `// Function to get max element ` `public` `static` `void` `maxelement(``int` `no_of_rows, ` `                              ``int``[][] arr)` `{` `    ``int` `i = 0;`   `    ``// Initialize max to 0 at beginning ` `    ``// of finding max element of each row ` `    ``int` `max = 0;` `    ``int``[] result = ``new` `int``[no_of_rows];` `    ``while` `(i < no_of_rows)` `    ``{` `        ``for` `(``int` `j = 0; ` `                 ``j < arr[i].Length; j++)` `        ``{` `            ``if` `(arr[i][j] > max)` `            ``{` `                ``max = arr[i][j];` `            ``}` `        ``}` `        ``result[i] = max;` `        ``max = 0;` `        ``i++;`   `    ``}` `    ``printArray(result);`   `}`   `// Print array element ` `private` `static` `void` `printArray(``int``[] result)` `{` `    ``for` `(``int` `i = 0; i < result.Length;i++)` `    ``{` `        ``Console.WriteLine(result[i]);` `    ``}`   `}`   `// Driver code ` `public` `static` `void` `Main(``string``[] args)` `{` `    ``int``[][] arr = ``new` `int``[][]` `    ``{` `        ``new` `int``[] {3, 4, 1, 8},` `        ``new` `int``[] {1, 4, 9, 11},` `        ``new` `int``[] {76, 34, 21, 1},` `        ``new` `int``[] {2, 1, 4, 5}` `    ``};` `    `  `    ``// Calling the function ` `    ``maxelement(4, arr);` `}` `}`   `// This code is contributed by Shrikant13`

## PHP

 ` ``\$max``) ` `            ``{` `                ``\$max` `= ``\$arr``[``\$i``][``\$j``];` `            ``}` `        ``}` `        ``\$result``[``\$i``] = ``\$max``;` `        ``\$max` `= 0;` `        ``\$i``++;`   `    ``}` `    ``printArray(``\$result``,``\$no_of_rows``);`   `}`   `// Driver code` `\$arr` `= ``array``(``array``(3, 4, 1, 8),` `                ``array``(1, 4, 9, 11),` `                ``array``(76, 34, 21, 1),` `                ``array``(2, 1, 4, 5));` `// Calling the function ` `maxelement(4, ``\$arr``);`   `// This code is contributed by mits` `?>`

## Javascript

 ``

Output

```8
11
76
5```

Complexity Analysis:

• Time Complexity: O(n*m) (where, n refers to no. of rows and m refers to no. of columns)
• Auxiliary Space: O(n) (where, n refers to no. of rows)

METHOD: Using List Comprehension.

APPROACH:

It is a concise way of creating a list in Python. We iterate through each row in the matrix using list comprehension and find the maximum element in the current row using the max() function. We then append the maximum element to a new list called “output”. Finally, we print the “output” list which contains the maximum element of each row in the matrix.

ALGORITHM:

1. Iterate through each row of the matrix.
2. Find the maximum element in the current row using the max() function.
3. Append the maximum element to the output list.
4. Return the output list.

## Python3

 `matrix ``=` `[[``1``, ``2``, ``3``], [``1``, ``4``, ``9``], [``76``, ``34``, ``21``]]` `output ``=` `[``max``(row) ``for` `row ``in` `matrix]` `print``(output)`

Output

`[3, 9, 76]`

Time complexity: O(nm) where n is the number of rows and m is the number of columns in the matrix.
Auxiliary Space: O(n) where n is the number of rows in the matrix

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