backtrack value which were chosen - c++

Question

I have a c++ program which calculates maximum of a array provided no two consecutive elements of array can be taken.
For eg:
7 3 4 6 will result in a answer of 13 .Here we chose 7 and 6 for optimal maximum.
Here is my recursive program for it.
#include <iostream>
using namespace std;
int n;
int findMax(int x,int ar[])
{
if(x < n)
return max( ar[x]+findMax(x+2,ar), findMax(x+1,ar));
return 0;
}
int main(){
int ar[]={1,7,4,4,9,5,12};
n = sizeof(ar)/sizeof(ar[0]);
cout<<findMax(0,ar);
return 0;
}
However I am more interested in the indices of array which were chosen for this purpose by my program .How can I do that efficiently.
In the above program answer should be 1,4,6 as we chose 1st , 4th and 6th element of the array for the maximum.
Note: I am using 0 based indexing.
Thanks.

Answer 1

A recurrence relation R(k) for the maximum sum of the first k elements of the array (with no adjacent terms) is:
R(0) = 0, R(1) = max(0, a[0])
R(k) = max(a[k] + R(k-2), R(k-1))
This is almost the same recurrence you're using in your code, but in your code your function returns the maximum sum of elements k and later.
Anyway, you can build a table of these values in linear time using dynamic programming. In pseudocode:
R = new array of length n+1
R[0] = 0
R[1] = max(0, a[0])
for i = 2 .. n
R[i] = max(a[i-1] + R[i-2], R[i-1])
If you just want the maximum sum, you can return R[n]. But you can also reconstruct the indices easily. In pseudo-code:
indices(a, R):
result = new empty vector
i = n
while i > 0
if (i == 1 and a[0] > 0) or R[i] == a[i-1] + R[i-2]
result.push_back(i-1)
i -= 2
else
i -= 1
You'll have to reverse result to get the indices in increasing order.

Answer 2

This is definitely not the most effective solution but probably the one with least implementation effort:
#include <iostream>
#include <vector>
using namespace std;
int n;
pair<int, vector<int> > findMax(int x, int ar[])
{
if (x < n) {
pair<int, vector<int> > max1 = findMax(x + 2, ar);
const pair<int, vector<int> > max2 = findMax(x + 1, ar);
max1.first += ar[x];
max1.second.insert(max1.second.begin(), x);
return max1.first >= max2.first ? max1 : max2;
}
return make_pair(0, vector<int>());
}
ostream& operator<<(ostream &out, const vector<int> &vec)
{
const char *sep = "";
for (int value : vec) {
out << sep << value; sep = ", ";
}
return out;
}
int main()
{
int ar[]={1,7,4,4,9,5,12};
n = sizeof ar / sizeof *ar;
const pair<int, vector<int> > maxAr = findMax(0, ar);
cout << maxAr.first << '\n'
<< maxAr.second << '\n';
return 0;
}
Output:
28
1, 4, 6
Life demo on coliru
Thereby, the return value is extended with a std::vector<int> which holds the used indices beside of the current sum.
std::max() could be used if I would provide a suitable (overloadeded) operator<() for std::pair<int, std::vector<int> >. To not make things over-complicated, I just replaced std::max() by the resp. condition.

Answer 3

I think below code will satisfy your need.
#include<bits/stdc++.h>
using namespace std;
int n;
void findMax(int arr[], int in, pair< int, vector<int> > tempStore,
pair< int, vector<int> > &resStore) {
if(in >=n) {
if(resStore.first < tempStore.first) {
resStore.first = tempStore.first;
resStore.second = tempStore.second;
}
return;
}
findMax(arr, in+1, tempStore, resStore);
tempStore.first += arr[in];
tempStore.second.push_back(in);
findMax(arr, in+2, tempStore, resStore);
}
int main() {
int ar[]={1,7,4,4,9,5,12};
n = sizeof(ar)/sizeof(ar[0]);
pair< int, vector<int> > resStore, tempStore;
findMax(ar, 0,tempStore,resStore);
cout<<"Result Value: "<<resStore.first;
cout<<"\nResult Index:\n";
for(int i=0; i<resStore.second.size(); i++) {
cout<<resStore.second[i]<<" ";
}
return 0;
}