Constrained Subsequence Sum in C++

Suppose we have an array called nums and an integer k, we have to find the maximum sum of a non-empty subsequence of that array such that for every two consecutive numbers in the subsequence, nums[i] and nums[j], where i < j, the condition j - i <= k is true.

As we know a subsequence of an array is obtained by deleting some number of elements from the array, leaving the remaining elements in their original order.

So, if the input is like [10,2,-9,5,19] and k = 2, then the output will be 36 as the subsequence is [10,2,5,19].

To solve this, we will follow these steps −

ret := -inf
Define an array dp and copy given array into it
Define one deque dq
insert v[0] at the begining of dq
n := size of v
ret := v[0]
for initialize i := 1, when i < n, update (increase i by 1), do −
- if i > k and first element of dq is same as dp[i - k - 1], then
  - delete front element from dq
- dp[i] := maximum of dp[i] and (if dq is empty, then dp[i] + 0, otherwise first element of dp + dq[i])
- while (not dq is empty and last element of dq < dp[i]), do −
  - delete last element from dq
- insert dp[i] at the end of dq
- ret := maximum of ret and dp[i]
return ret

Let us see the following implementation to get better understanding −

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 10;
class Solution {
   public:
   int constrainedSubsetSum(vector<int>& v, int k) {
      int ret = -inf;
      vector<int> dp(v.begin(), v.end());
      deque<int> dq;
      dq.push_front(v[0]);
      int n = v.size();
      ret = v[0];
      for (int i = 1; i < n; i++) {
         if (i > k && dq.front() == dp[i - k - 1])
         dq.pop_front();
         dp[i] = max(dp[i], dq.empty() ? dp[i] + 0 : dp[i] +
         dq.front());
         while (!dq.empty() && dq.back() < dp[i])
         dq.pop_back();
         dq.push_back(dp[i]);
         ret = max(ret, dp[i]);
      }
      return ret;
   }
};
main(){
   Solution ob;
   vector<int> v = {10,2,-9,5,19};
   cout << (ob.constrainedSubsetSum(v, 2));
}