Coursera's Algorithmic toolbox assignment solutions( Car Fueling, Maximum Advertisement Revenue ) :

Problem 3: Car Fueling

Solution: (in c++)

( please guys before moving to the solution try it yourself at least 3-4 times , if you really wanna become a good coder)

#include<bits/stdc++.h>

using namespace std;

int main()

{

long long d,m,n,i,nr=0,cr=0,lr; //d=distance between the cities , cr=currentrefill,

cin>>d>>m>>n; // m=max. distance by the car in a full tank

long long a[n+1]; //lr=lastrefill , nr=no. of refills

a[0]=0; // array a represents gas stations positions

for(i=1;i<=n;i++)

{

cin>>a[i];

}

a[n+1]=d;

while(cr<=n) //this algorithm is explained perfectly

{ //in the lecture

lr=cr;

while(cr<=n && a[cr+1]-a[lr]<=m)

{

cr=cr+1;

}

if(cr==lr)

{

cout<<"-1"<<endl;

break;

}

if(cr<=n)

{

nr=nr+1;

}

}

if(cr!=lr)

{

cout<<nr<<endl;

}

}

This code is simple. There's no need for any explanation.

Problem 4: Maximum Advertisement Revenue

Solution: (in c++)

( please guys before moving to the solution try it yourself at least 3-4 times , if you really wanna become a good coder)

#include<bits/stdc++.h>

using namespace std;

int main()

{

    long long n,i,j,k,res=0;

    cin>>n;

    long long a[n],b[n];

    for(i=0;i<n;i++)

{

cin>>a[i];

}

for(j=0;j<n;j++)

{

cin>>b[j];

}

sort(a,a+n); //see note below

sort(b,b+n);

for(k=0;k<n;k++)

{

res+=a[k]*b[k];

}

cout<<res;

}

Since we want to maximize the sum of the products , what we can do is first we will select the maximum element from array a and multiply it to the maximum element of array b and then we will select the second maximum element from array a and multiply it to the second maximum element of the array b and repeat this procedure and add each product to get the maximum result.

Guys , if you have any queries related to a question or need more explanation, comment down below!