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Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN: 9780133594140
Author: James Kurose, Keith Ross
Publisher: PEARSON
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Question
KINDLY SEE ATTACH IMAGE FOR THE QUESTION. REARRANGE SHUFFLED

Transcribed Image Text:Maximum Element In Each Row
Given a problem and correct (shuffled) program, re-arrange the shuffled code segments in the correct order.
Consider a program that finds the maximum element in each row of the matrix.
Compilation Errors
Input Format:
The input consists of (m*n+2) integers. The first integer corresponds to m, the number of rows in the
matrix, and the second integer corresponds to n, the number of columns in the matrix. The remaining
for(j=0;j<c;j++)
a=(int **)malloc(r*sizeof(int ");
read(a,r,c);
{
int **a,r,c;
integers correspond to the elements in the matrix. The elements are read in row-wise order, the first row
fırst, then the second row, and so on. Assume that the maximum value of m and nis 10.
scanf("%d"*(a+i)+j):
{|
a=create(r.c):
{
{
for(j=0:j<c:j++)
max=0;
for(i=0;i<r;i++)
void read(int **a,int r,int c);
int ij.max=0;
Output Format:
display(a.r.c);
}
}
if("("(a+i)+j)>max)
return 0;
Refer to sample output for details.
int ** create(int r,int c):
for(i=0;i<r;i++)
printf("\n%d",max);
int **create(int rint c)
Sample Input 1:
int main()
for(i=0;i<r;i++)
max="("(a+i)+j);
{ |
{
3
void display(int **a,int r,int c)
int ij;
void read(int **a,int r.int c)
2
45
*(a+i)=(int ")malloc(c'sizeof(int *)):
int **ai:
return a;
void display(int**a,int r,int c):
69
scanf("%d%d",&r,&c);
#include<stdio.h>
#include<stdlib.h>
03
Sample Output 1:
3
Expert Solution
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