PROBLEM LINK:

Practice
Contest

Author: Vitaliy Herasymiv
Tester: Istvan Nagy
Editorialist: Amit Pandey

DIFFICULTY:

Easy.

PREREQUISITES:

BIT, segment tree.

PROBLEM:

Determine minimum amount of minutes required to paint a fence consisting of $N$ parts, in green color. Each part is initially painted either red or green. We can chose an index $X$, and flip colors in range $[X,\min(X+K,N-1)]$ in one minute.

QUICK EXPLANATION:

Consider the index where first red is occurring as $T$, flip the colors in range $[T,\min(N,T+K-1)]$. Keep repeating this step until each part is colored green.

EXPLANATION:

First lets see how to solve the problem, later we can see the proof of correctness.

Consider the index at which first red is occurring and call it $T$, flip the colors in range $[T+\min(N,T+K-1)]$. Keep repeating the given procedure unless whole array is green. After each iteration value of $T$ will be increasing, so the given algorithm will terminate in less than or equal to $N$ steps.

Each iteration may take $O(N)$ amount of time, hence the given algorithm will be taking $O(N^2)$ time. As the $N$ can be $10^5$ in the given problem, we need to optimize our solution.

We can solve the problem in $O(N\log N)$ time. Range flip can be seen as adding $1$ to the range and value of color can be obtain using sum at that particular index modulo $2$. If the $\text{sum} \%2$ is $0$, then color at that index will be same as the initial one or flipped otherwise. The range updates and queries can be done in $O(\log N)$ time using BIT or segment tree.

O(N) Solution:

It can be done in $O(N)$ as well. Adding $1$ to range $[L,R]$ can be seen as adding $1$ to index $L$ and adding $-1$ to index $(R+1)$. When we want to retrieve the value at specific index, we need to take the prefix sum upto that index. See sub problem 1 of this editorial for more details.

Proof:

The given procedure can be proved correct using the following lemmas.

1. Order of the updates doesn't matter.
2. Starting index of any 2 updates will be different, as 2 updates at same starting position will cancel each other.
3. If we sort the minimum update sequence and suppose first update happens at index $i$, then at each index in range $[0,i-1]$ color must be green and at index $i$, color must be red.

Solution:

Setter's solution can be found here
Tester's solution can be found here