Answer:
JavaScript is the correct answer
Explanation:
HTML help creates the structural overview, CSS helps in the design and an application software helps in the accessibility, editing, typing of code, but JavaScript helps to make it interactive. Thank you.
Answer:
Option a. int max = aList.get(0); for (int count = 1; count < aList.size(); count++) { if (aList.get(count) > max) { max = aList.get(count); } }
is the correct code snippet.
Explanation:
Following is given the explanation for the code snippet to find largest value in an integer array list aList.
- From the array list aList, very first element having index 0 will be stored in the variable max (having data type int).
- By using for starting from count =1 to count = size of array (aList), we will compare each element of the array with first element of the array.
- If any of the checked element get greater from the first element, it gets replaced in the variable max and the count is increased by 1 so that the next element may be checked.
- When the loop will end, the variable max will have the greatest value from the array aList.
i hope it will help you!
Answer:
We first need to find out the number of variables, and if it is 3 or less, we can use excel for visualization or else we need to make use of the Microsoft Power BI or Tableau. And here, we have one variable as subscribers in various age groups, and the second variable is the number of hours spent in watching videos. And hence, we have only two variables. And thus, we can make use of the bar chart for visualization, and that is quite enough for visualizing this case. However, remember visualization is done for analyzing, and analyzing is made easy through visualization.
Explanation:
Please check the answer.
1. C) RAM
2. C) CPU
3. B) Accounts Payable
I believe these are the correct answers you are looking for.
Answer:
#include<stdio.h>
#include<conio.h>
int m=0,n=4;
int cal(int temp[10][10],int t[10][10])
{
int i,j,m=0;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
{
if(temp[i][j]!=t[i][j])
m++;
}
return m;
}
int check(int a[10][10],int t[10][10])
{
int i,j,f=1;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
if(a[i][j]!=t[i][j])
f=0;
return f;
}
void main()
{
int p,i,j,n=4,a[10][10],t[10][10],temp[10][10],r[10][10];
int m=0,x=0,y=0,d=1000,dmin=0,l=0;
clrscr();
printf("\nEnter the matrix to be solved,space with zero :\n");
for(i=0;i < n;i++)
for(j=0;j < n;j++)
scanf("%d",&a[i][j]);
printf("\nEnter the target matrix,space with zero :\n");
for(i=0;i < n;i++)
for(j=0;j < n;j++)
scanf("%d",&t[i][j]);
printf("\nEntered Matrix is :\n");
for(i=0;i < n;i++)
{
for(j=0;j < n;j++)
printf("%d\t",a[i][j]);
printf("\n");
}
printf("\nTarget Matrix is :\n");
for(i=0;i < n;i++)
{
for(j=0;j < n;j++)
printf("%d\t",t[i][j]);
printf("\n");
}
while(!(check(a,t)))
{
l++;
d=1000;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
{
if(a[i][j]==0)
{
x=i;
y=j;
}
}
//To move upwards
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(x!=0)
{
p=temp[x][y];
temp[x][y]=temp[x-1][y];
temp[x-1][y]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
//To move downwards
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(x!=n-1)
{
p=temp[x][y];
temp[x][y]=temp[x+1][y];
temp[x+1][y]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
//To move right side
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(y!=n-1)
{
p=temp[x][y];
temp[x][y]=temp[x][y+1];
temp[x][y+1]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
//To move left
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(y!=0)
{
p=temp[x][y];
temp[x][y]=temp[x][y-1];
temp[x][y-1]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
printf("\nCalculated Intermediate Matrix Value :\n");
for(i=0;i < n;i++)
{
for(j=0;j < n;j++)
printf("%d\t",r[i][j]);
printf("\n");
}
for(i=0;i < n;i++)
for(j=0;j < n;j++)
{
a[i][j]=r[i][j];
temp[i][j]=0;
}
printf("Minimum cost : %d\n",d);
}
getch();
}
Explanation:
