For finding min. cost , we need to find optimal height where there will be minima.

Now , there is a little twist because one can move the blocks also.

Check out my code for detailed explanation:

Approach: Binary Search

int A,R,M;
vector<int> height;
int getCost(int mid)
{
    int adds = 0; //number of blocks to be added
    int removes = 0; //number of blocks to be removed

    for(int i=0;i<n;i++)
    {
        if(height[i] > mid) removes += (abs(height[i]-mid));
        else if(height[i] < mid) adds += (abs(height[i]-mid));
    }

    //This is the cost where we have done only adds or only removes
    int cost = (adds > removes)?(adds-removes)*A:(removes-adds)*R;

    //Tricky part is that if (adds = removes) ,then we have to decide whether we need to move block or add/delete.
    cost += min(adds,removes)*min(A+R,M);

    return cost;
}

void solve()
{
    cin >> n;
    cin >> A >> R >> M;
    height.resize(n);
    for(int i=0;i<n;i++) cin>>height[i];

    int lo = 0,hi = *max_element(height.begin(),height.end());
    int tmp=hi;
    int ans ,ht;
    while(lo<=hi)
    {
        int mid = lo +(hi-lo)/2;
        int left = 0,right=0;

        //left and right or just checking whether left/right slope decreasing or not (-1 means dec.)
        if(mid>0 && getCost(mid-1) >= getCost(mid)) left = -1;
        if(mid<tmp && getCost(mid+1) >= getCost(mid)) right=-1;

        //If both slops are decreasing , we got minima!
        if(right == -1 && left==-1)
        {
            ans = getCost(mid);
            ht = mid;
            break;
        }

        //Else reducing search space
        if(right==-1) hi = mid;
        else if(left==-1) lo = mid;
    }

    //Cost as well as req. height
    cout << ans <<" " << ht<< endl;

}