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Find maximum element of each row in a matrix

The problem at hand involves finding the maximum element of each row in a given matrix. A matrix is a two-dimensional array of elements, where rows and columns define its structure. The task is to identify and print the largest element in each row.

Problem Statement

Given a matrix with dimensions ROW x COL, where each element is an integer, we need to find the maximum element in each row and print those maximum values.

Example

Consider the following matrix:

1  0  2  3
4  1  1  4
0 -3  1  0
2  5  1  2

The maximum elements in each row are: 3, 4, 1, and 5.

Idea to Solve

To solve this problem, we need to iterate through each row of the matrix and find the maximum element in that row. We can initialize a variable to hold the maximum element for each row and update it as we traverse the row.

Pseudocode

Here's the pseudocode for the algorithm:

function row_max_element(matrix):
    for i from 0 to ROW-1:
        max_element = matrix[i][0] // Initialize max_element with the first element of the row
        for j from 1 to COL-1:
            if matrix[i][j] > max_element:
                max_element = matrix[i][j] // Update max_element if a larger element is found
        print max_element

Algorithm Explanation

  1. Start by defining a function row_max_element that takes a 2D matrix matrix as its input.
  2. Initialize a loop that iterates through each row of the matrix from 0 to ROW-1 (inclusive).
  3. Inside the outer loop, initialize a variable max_element with the value of the first element in the current row (matrix[i][0]).
  4. Implement an inner loop that starts from 1 and iterates through each column of the current row (from 1 to COL-1).
  5. Within the inner loop, compare the current element (matrix[i][j]) with the max_element. If the current element is greater than max_element, update max_element with the value of the current element.
  6. After the inner loop completes, print the value of max_element, which represents the maximum element in the current row.
  7. The outer loop will continue to the next row, and the process repeats until all rows have been processed.

Program List

//C Program 
//Find maximum element of each row in a matrix
#include<stdio.h>

//Matrix size
#define ROW 4
#define COL 4

//print max element in each row
void row_max_element(int arr[][COL])
{
  int result=0;

  for (int i = 0; i < ROW; ++i)
  {
    //Get first element of i row
    result=arr[i][0];

    for (int j = 1; j < COL; ++j)
    {
      if(result<arr[i][j])
      {
        //Get new max element
        result=arr[i][j];

      }
    }
    printf(" %d\n",result);
  }
}
int main()
{


  int arr[][COL]= { 
    {1, 0, 2, 3},  
    {4, 1, 1, 4},  
    {0, -3, 1, 0},  
    {2, 5, 1, 2}
  };  
  
  row_max_element(arr);

  return 0;
}


Output

 3
 4
 1
 5
/*
  C++ Program
  Find maximum element of each row in a matrix
*/
#include<iostream>
#define ROW 4
#define COL 5
using namespace std;

class MyMatrix {
  public:
   
  //print max element in each row
  void row_max_element(int matrix[][COL])
  {
    int result=0;

    for (int i = 0; i < ROW; ++i)
    {
      //Get first element of i row
      result=matrix[i][0];

      for (int j = 1; j < COL; ++j)
      {
        if(result<matrix[i][j])
        {
          //Get new max element
          result=matrix[i][j];

        }
      }
      cout<<"  "<<result<<endl;
    }
  }
};
int main() {
  MyMatrix obj;
  int matrix[][COL] ={ 
    {1, 0, 2, 3},  
    {4, 1, 1, 4},  
    {0, -3, 1, 0},  
    {2, 5, 1, 2}
  };  
  obj.row_max_element(matrix);


  return 0;
}

Output

  3
  4
  1
  5
/*
  Java Program
  Find maximum element of each row in a matrix
*/
public class MyMatrix {


  //print max element in each row
  public void row_max_element(int [][]matrix)
  {
    int result=0;
    //Get the size of array
    int row = matrix.length;
    int col = matrix[0].length;
    for (int i = 0; i < row; ++i)
    {
      //Get first element of i row
      result=matrix[i][0];

      for (int j = 1; j < col; ++j)
      {
        if(result<matrix[i][j])
        {
          //Get new max element
          result=matrix[i][j];

        }
      }
      System.out.print("  "+result+"\n");
    }
  }
  public static void main(String[] args) {
    MyMatrix obj = new MyMatrix();
    //Define matrix 
    int [][]matrix ={ 
      {1, 0, 2, 3},  
      {4, 1, 1, 4},  
      {0, -3, 1, 0},  
      {2, 5, 1, 2}
    };  
    obj.row_max_element(matrix);



  }
}

Output

  3
  4
  1
  5
/*
  C# Program
  Find maximum element of each row in a matrix
*/
using System;
public class MyMatrix {
	//print max element in each row
	public void row_max_element(int[,] matrix) {
		int result = 0;
		//Get the size of array
		int row = matrix.GetLength(0);
		int col = matrix.GetLength(1);
		for (int i = 0; i < row; ++i) {
			//Get first element of i row
			result = matrix[i,0];
			for (int j = 1; j < col; ++j) {
				if (result < matrix[i,j]) {
					//Get new max element
					result = matrix[i,j];
				}
			}
			Console.Write(" " + result + "\n");
		}
	}
	public static void Main(String[] args) {
		MyMatrix obj = new MyMatrix();
		//Define matrix 
      	int[,] matrix = {
			{
				1,
				0,
				2,
				3
			},
			{
				4,
				1,
				1,
				4
			},
			{
				0,
				-3,
				1,
				0
			},
			{
				2,
				5,
				1,
				2
			}
		};
		obj.row_max_element(matrix);
	}
}

Output

 3
 4
 1
 5
<?php
/*
  Php Program
  Find maximum element of each row in a matrix
*/
class MyMatrix {
	//print max element in each row

	public 	function row_max_element($matrix) {
		$result = 0;
		//Get the size of array
		$row = count($matrix);
		$col = count($matrix[0]);
		for ($i = 0; $i < $row; ++$i) {
			//Get first element of i row
			$result = $matrix[$i][0];
			for ($j = 1; $j < $col; ++$j) {
				if ($result < $matrix[$i][$j]) {
					//Get new max element
					$result = $matrix[$i][$j];
				}
			}
			echo(" ". $result ."\n");
		}
	}
};

function main() {
	$obj = new MyMatrix();
	//Define matrix 
	$matrix = array(array(1, 0, 2, 3), 
                    array(4, 1, 1, 4), 
                    array(0, -3, 1, 0), 
                    array(2, 5, 1, 2));
	$obj->row_max_element($matrix);

}
main();

Output

 3
 4
 1
 5
/*
 Node Js Program
 Find maximum element of each row in a matrix
*/
class MyMatrix {
	//print max element in each row
	row_max_element(matrix) {
		var result = 0;
		//Get the size of array
		var row = matrix.length;
		var col = matrix[0].length;
		for (var i = 0; i < row; ++i) {
			//Get first element of i row
			result = matrix[i][0];
			for (var j = 1; j < col; ++j) {
				if (result < matrix[i][j]) {
					//Get new max element
					result = matrix[i][j];
				}
			}

			process.stdout.write(" " + result + "\n");
		}
	}
}

function main(args) {
	var obj = new MyMatrix();
	//Define matrix 
	var matrix = [
		[1, 0, 2, 3],
		[4, 1, 1, 4],
		[0, -3, 1, 0],
		[2, 5, 1, 2]
	];
	obj.row_max_element(matrix);
}
main();

Output

 3
 4
 1
 5
# Python 3 Program

# Find maximum element of each row in a matrix
class MyMatrix :
	# print max element in each row
	def row_max_element(self, matrix) :
		result = 0
		row = len(matrix)
		col = len(matrix[0])
		i = 0
		while (i < row) :
			# Get first element of i row
			result = matrix[i][0]
			j = 1
			while (j < col) :
				if (result < matrix[i][j]) :
					# Get new max element
					result = matrix[i][j]
				
				j += 1
			
			print(" ", result ,"\n", end = "")
			i += 1
		
	

def main() :
	obj = MyMatrix()
	matrix = [
		[1, 0, 2, 3],
		[4, 1, 1, 4],
		[0, -3, 1, 0],
		[2, 5, 1, 2]
	]
	obj.row_max_element(matrix)


if __name__ == "__main__":
	main()

Output

  3
  4
  1
  5
# Ruby Program 
# Find maximum element of each row in a matrix
class MyMatrix 
	# print max element in each row
	def row_max_element(matrix) 
		result = 0
		row = matrix.length
		col = matrix[0].length
		i = 0
		while (i < row) 
			# Get first element of i row
			result = matrix[i][0]
			j = 1
			while (j < col) 
				if (result < matrix[i][j]) 
					# Get new max element
					result = matrix[i][j]
				end
				j += 1
			end
			print(" ", result ,"\n")
			i += 1
		end
	end
end
def main() 
	obj = MyMatrix.new()
	matrix = [
		[1, 0, 2, 3],
		[4, 1, 1, 4],
		[0, -3, 1, 0],
		[2, 5, 1, 2]
	]
	obj.row_max_element(matrix)
end
main()

Output

 3
 4
 1
 5
/*
 Scala Program
 Find maximum element of each row in a matrix
*/
class MyMatrix {
	//print max element in each row
	def row_max_element(matrix: Array[Array[Int]]): Unit = {
		var result: Int = 0;
		val row: Int = matrix.length;
		val col: Int = matrix(0).length;
		var i: Int = 0;
		while (i < row) {
			//Get first element of i row
			result = matrix(i)(0);
			var j: Int = 1;
			while (j < col) {
				if (result < matrix(i)(j)) {
					//Get new max element
					result = matrix(i)(j);
				}
				j += 1;
			}
			print(" " + result + "\n");
			i += 1;
		}
	}
}
object Main {
	def main(args: Array[String]): Unit = {
		val obj: MyMatrix = new MyMatrix();
		val matrix: Array[Array[Int]] = Array(
			Array(1, 0, 2, 3),
			Array(4, 1, 1, 4),
			Array(0, -3, 1, 0),
			Array(2, 5, 1, 2));
		obj.row_max_element(matrix);
	}
}

Output

 3
 4
 1
 5
/*
  Swift 4 Program
  Find maximum element of each row in a matrix
*/
class MyMatrix {
	//print max element in each row
	func row_max_element(_ matrix: [
		[Int]
	]) {
		var result: Int = 0;
		let row: Int = matrix.count;
		let col: Int = matrix[0].count;
		var i: Int = 0;
		while (i < row) {
			//Get first element of i row
			result = matrix[i][0];
			var j: Int = 1;
			while (j < col) {
				if (result < matrix[i][j]) {
					//Get new max element
					result = matrix[i][j];
				}
				j += 1;
			}
			print(" ", result ,"\n", terminator: "");
			i += 1;
		}
	}
}
func main() {
	let obj: MyMatrix = MyMatrix();
	let matrix: [
		[Int]
	] = [
		[1, 0, 2, 3],
		[4, 1, 1, 4],
		[0, -3, 1, 0],
		[2, 5, 1, 2]
	];
	obj.row_max_element(matrix);
}
main();

Output

  3
  4
  1
  5

Time Complexity

The time complexity of this algorithm is O(ROW * COL), where ROW is the number of rows in the matrix and COL is the number of columns. This is because we iterate through each element of the matrix exactly once to find the maximum element in each row. The nested loops contribute to the complexity.

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