You'll need a helper variable for this, so depending on your programming language, the solution becomes:
int helper;
helper = arr[i];
arr[i] = arr[j];
arr[j] = helper;
Answer:
Explanation:
#include<iostream>
#include<ctime>
#include<bits/stdc++.h>
using namespace std;
double calculate(double arr[], int l)
{
double avg=0.0;
int x;
for(x=0;x<l;x++)
{
avg+=arr[x];
}
avg/=l;
return avg;
}
int biggest(int arr[], int n)
{
int x,idx,big=-1;
for(x=0;x<n;x++)
{
if(arr[x]>big)
{
big=arr[x];
idx=x;
}
}
return idx;
}
int main()
{
vector<pair<int,double> >result;
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
int choice;
cin>>choice;
while(choice!=2)
{
int n,m;
cout<<"Enter N"<<endl;
cin>>n;
cout<<"Enter M"<<endl;
cin>>m;
int c=m;
double running_time[c];
while(c>0)
{
int arr[n];
int x;
for(x=0;x<n;x++)
{
arr[x] = rand();
}
clock_t start = clock();
int pos = biggest(arr,n);
clock_t t_end = clock();
c--;
running_time[c] = 1000.0*(t_end-start)/CLOCKS_PER_SEC;
}
double avg_running_time = calculate(running_time,m);
result.push_back(make_pair(n,avg_running_time));
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
cin>>choice;
}
for(int x=0;x<result.size();x++)
{
cout<<result[x].first<<" "<<result[x].second<<endl;
}
}
Answer:
Inverted page tables
Explanation:
Inverted page tables are indeed the ones used during broad address space, including such 64-bit addresses.
Answer:
Number of strings = (10, 2)×(8,3) = 2520
Explanation:
The number of possible combinations for taking two 0's is C(10, 2)
It remains 8 Positions
The number of possible combinations for taking three 1's is C(8,3)
So there remains 5 spots
Making guesses, as opposed to the other answers, this is the only one that forges new ideas and doing such will only lead to an illegitimate conclusion/study.
