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.
Below is the implementation :
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<result.length;i++) { System.out.println(result[i]); } } // Driver code public static void main(String[] args) { int [][] arr = new int [][] { { 3 , 4 , 1 , 8 }, { 1 , 4 , 9 , 11 }, { 76 , 34 , 21 , 1 }, { 2 , 1 , 4 , 5 } }; // Calling the function maxelement( 4 , arr); } } |
8 11 76 5
Time Complexity: O(R*C) where R and C are numbers of rows and columns of a given matrix respectively.
Auxiliary Space: O(R), extra space for array result[]
Please refer complete article on Find maximum element of each row in a matrix for more details!
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