2612. Minimum Reverse Operations

Description

You are given an integer n and an integer p representing an array arr of length n where all elements are set to 0's, except position p which is set to 1. You are also given an integer array banned containing restricted positions. Perform the following operation on arr:

Reverse a subarray with size k if the single 1 is not set to a position in banned.

Return an integer array answer with n results where the i^th result is the minimum number of operations needed to bring the single 1 to position i in arr, or -1 if it is impossible.

Example 1:

Input: n = 4, p = 0, banned = [1,2], k = 4

Output: [0,-1,-1,1]

Explanation:

Initially 1 is placed at position 0 so the number of operations we need for position 0 is 0.
We can never place 1 on the banned positions, so the answer for positions 1 and 2 is -1.
Perform the operation of size 4 to reverse the whole array.
After a single operation 1 is at position 3 so the answer for position 3 is 1.

Example 2:

Input: n = 5, p = 0, banned = [2,4], k = 3

Output: [0,-1,-1,-1,-1]

Explanation:

Initially 1 is placed at position 0 so the number of operations we need for position 0 is 0.
We cannot perform the operation on the subarray positions [0, 2] because position 2 is in banned.
Because 1 cannot be set at position 2, it is impossible to set 1 at other positions in more operations.

Example 3:

Input: n = 4, p = 2, banned = [0,1,3], k = 1

Output: [-1,-1,0,-1]

Explanation:

Perform operations of size 1 and 1 never changes its position.

Constraints:

1 <= n <= 10⁵
0 <= p <= n - 1
0 <= banned.length <= n - 1
0 <= banned[i] <= n - 1
1 <= k <= n
banned[i] != p
all values in banned are unique

Solutions

Solution 1: Ordered Set + BFS

We notice that for any index $i$ in the subarray interval $[l,..r]$, the flipped index $j = l + r - i$.

If the subarray moves one position to the right, then $j = l + 1 + r + 1 - i = l + r - i + 2$, that is, $j$ will increase by $2$.

Similarly, if the subarray moves one position to the left, then $j = l - 1 + r - 1 - i = l + r - i - 2$, that is, $j$ will decrease by $2$.

Therefore, for a specific index $i$, all its flipped indices form an arithmetic progression with common difference $2$, that is, all the flipped indices have the same parity.

Next, we consider the range of values of the index $i$ after flipping $j$.

If the boundary is not considered, the range of values of $j$ is $[i - k + 1, i + k - 1]$.
If the subarray is on the left, then $[l, r] = [0, k - 1]$, so the flipped index $j$ of $i$ is $0 + k - 1 - i$, that is, $j = k - i - 1$, so the left boundary $mi = max(i - k + 1, k - i - 1)$.
If the subarray is on the right, then $[l, r] = [n - k, n - 1]$, so the flipped index $j= n - k + n - 1 - i$ is $j = n \times 2 - k - i - 1$, so the right boundary of $j$ is $mx = min(i + k - 1, n \times 2 - k - i - 1)$.

We use two ordered sets to store all the odd indices and even indices to be searched, here we need to exclude the indices in the array $banned$ and the index $p$.

Then we use BFS to search, each time searching all the flipped indices $j$ of the current index $i$, that is, $j = mi, mi + 2, mi + 4, \dots, mx$, updating the answer of index $j$ and adding index $j$ to the search queue, and removing index $j$ from the corresponding ordered set.

When the search is over, the answer to all indices can be obtained.

The time complexity is $O(n \times \log n)$ and the space complexity is $O(n)$. Where $n$ is the given array length in the problem.

Python3JavaC++GoTypeScript

from sortedcontainers import SortedSet


class Solution:
    def minReverseOperations(
        self, n: int, p: int, banned: List[int], k: int
    ) -> List[int]:
        ans = [-1] * n
        ans[p] = 0
        ts = [SortedSet() for _ in range(2)]
        for i in range(n):
            ts[i % 2].add(i)
        ts[p % 2].remove(p)
        for i in banned:
            ts[i % 2].remove(i)
        ts[0].add(n)
        ts[1].add(n)
        q = deque([p])
        while q:
            i = q.popleft()
            mi = max(i - k + 1, k - i - 1)
            mx = min(i + k - 1, n * 2 - k - i - 1)
            s = ts[mi % 2]
            j = s.bisect_left(mi)
            while s[j] <= mx:
                q.append(s[j])
                ans[s[j]] = ans[i] + 1
                s.remove(s[j])
                j = s.bisect_left(mi)
        return ans

class Solution {
    public int[] minReverseOperations(int n, int p, int[] banned, int k) {
        int[] ans = new int[n];
        TreeSet<Integer>[] ts = new TreeSet[] {new TreeSet<>(), new TreeSet<>()};
        for (int i = 0; i < n; ++i) {
            ts[i % 2].add(i);
            ans[i] = i == p ? 0 : -1;
        }
        ts[p % 2].remove(p);
        for (int i : banned) {
            ts[i % 2].remove(i);
        }
        ts[0].add(n);
        ts[1].add(n);
        Deque<Integer> q = new ArrayDeque<>();
        q.offer(p);
        while (!q.isEmpty()) {
            int i = q.poll();
            int mi = Math.max(i - k + 1, k - i - 1);
            int mx = Math.min(i + k - 1, n * 2 - k - i - 1);
            var s = ts[mi % 2];
            for (int j = s.ceiling(mi); j <= mx; j = s.ceiling(mi)) {
                q.offer(j);
                ans[j] = ans[i] + 1;
                s.remove(j);
            }
        }
        return ans;
    }
}