Removal of billboards from given ones

Question

I came across this question

ADZEN is a very popular advertising firm in your city. In every road you can see their advertising billboards. Recentl

Answer 1

I coded it in c++ using DP in O(nlogk). Idea is to maintain a multiset with next k values for a given position. This multiset will typically have k values in mid processing. Each time you move an element and push new one. Art is how to maintain this list to have the profit[i] + answer[i+2]. More details on set:

/*
 * Observation 1: ith state depends on next k states i+2....i+2+k
 * We maximize across this states added on them "accumulative" sum
 *
 * Let Say we have list of numbers of state i+1, that is list of {profit + state solution}, How to get states if ith solution
 *
 * Say we have following data k = 3
 *
 * Indices:     0   1   2   3   4
 * Profits:     1   3   2   4   2
 * Solution:    ?   ?   5   3   1
 *
 * Answer for [1] = max(3+3, 5+1, 9+0) = 9
 *
 * Indices:     0   1   2   3   4
 * Profits:     1   3   2   4   2
 * Solution:    ?   9   5   3   1
 *
 * Let's find answer for [0], using set of [1].
 *
 * First, last entry should be removed. then we have (3+3, 5+1)
 *
 * Now we should add 1+5, but entries should be incremented with 1
 *      (1+5, 4+3, 6+1) -> then find max.
 *
 *  Could we do it in other way but instead of processing list. Yes, we simply add 1 to all elements
 *
 *  answer is same as: 1 + max(1-1+5, 3+3, 5+1)
 *
 */

---

ll dp()
{
multiset<ll, greater<ll> > set;

mem[n-1] = profit[n-1];

ll sumSoFar = 0;

lpd(i, n-2, 0)
{
    if(sz(set) == k)
        set.erase(set.find(added[i+k]));

    if(i+2 < n)
    {
        added[i] = mem[i+2] - sumSoFar;
        set.insert(added[i]);
        sumSoFar += profit[i];
    }

    if(n-i <= k)
        mem[i] = profit[i] + mem[i+1];
    else 
        mem[i] = max(mem[i+1], *set.begin()+sumSoFar);
}

return mem[0];
 }