Can you help me translate the following code into English steps by steps, and ex
ID: 3582786 • Letter: C
Question
Can you help me translate the following code into English steps by steps, and explain how it works?
#include <bits/stdc++.h>
using namespace std;
//implement this function using dynamic programming to get optimized result
bool isSumPossible(int arr[],int n,int s)
{
if(s<0||n<=0)
return false;
if(s==0)
return true;
//either this element will not produce result or produce the result
return ((isSumPossible(arr,n,s-arr[n-1])) || (isSumPossible(arr,n-1,s)));
}
int main()
{
int* arr;
int n;
cout<<"Enter the Size of Array ";
cin>>n;
arr=new int[n];
cout<<"Enter the value in Array ";
int prod=1;
for(int i=0;i<n;i++)
{
cin>>arr[i];
prod*=arr[i];
}
int i=0;
//instead of linear search ,do it with binary search for optimized
for(i=prod;i>=1;i--)
{
if(isSumPossible(arr,n,i)==false)
{
break;
}
}
cout<<"Maximum Number not Possible : "<<i<<endl;
return 0;
}
Explanation / Answer
// This is a program to check whether the sum of product of all elements till 1 can be represented with the elements in array
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
//implement this function using dynamic programming to get optimized result
bool isSumPossible(int arr[],int n,int s)
{
// if s is less than 0, which means prod can't be represented with nth element
// which returns true and will check for n-1th element then on
if(s<0||n<=0)
return false;
if(s==0)
return true;
// if the sum is possible with nth element, returning true from here
// if it is not possible checking with n-1th element
// doing it recursively
return ((isSumPossible(arr,n,s-arr[n-1])) || (isSumPossible(arr,n-1,s)));
}
int main()
{
int* arr;
int n;
cout<<"Enter the Size of Array ";
cin>>n;
arr=new int[n];
cout<<"Enter the value in Array ";
int prod=1;
// taking user input
for(int i=0;i<n;i++)
{
cin>>arr[i];
prod*=arr[i]; // calculating the product of every element entered
}
int i=0;
// checking from prod till 1 whether every number can be sum of the numbers of given array
for(i=prod;i>=1;i--)
{
// if any number can't be a sum then breaking the for loop
if(isSumPossible(arr,n,i)==false)
{
break;
}
}
cout<<"Maximum Number not Possible : "<<i<<endl;
return 0;
}