The Problem In Discussion is : https://mirror.codeforces.com/contest/1839/problem/D
Observations:
- The solution will be decreasing.
- Say you place some zeroes then each zero removes some contiguous subarray, the 0 is present in this subarray and subarrays of two zeroes don't overlap.
- A subarray that zero removes, give cost equal to the number of elements in the subarray.
- Now it is equivalent that each zero is present in the starting of the subarray that it is removing.
- So this can be solved with dp.
- This problem is very identical to LIS.
- In fact answer for the last query is always n — size_of_lis of the array.

State of dp
dp[i][k][2]:
1. dp[i][k][0] -> minimum no of removed elements upto index i such that you have used k zeros and i'th element is not removed.
2. dp[i][k][1] -> minimum no of removed elements upto index i such that you have used k zeroes and i'th element is removed.