The bug jumps according to the following rules:

• It can jump exactly a positions forward (to the right).
• It can jump exactly b positions backward (to the left).
• It cannot jump backward twice in a row.
• It cannot jump to any forbidden positions.

The bug may jump forward beyond its home, but it cannot jump to positions numbered with negative integers.

Given an array of integers forbidden, where forbidden[i] means that the bug cannot jump to the position forbidden[i], and integers a, b, and x, return the minimum number of jumps needed for the bug to reach its home. If there is no possible sequence of jumps that lands the bug on position x, return -1.

Example 1:

Input: forbidden = [14,4,18,1,15], a = 3, b = 15, x = 9
Output: 3
Explanation: 3 jumps forward (0 -> 3 -> 6 -> 9) will get the bug home.

Example 2:

Input: forbidden = [8,3,16,6,12,20], a = 15, b = 13, x = 11
Output: -1

Example 3:

Input: forbidden = [1,6,2,14,5,17,4], a = 16, b = 9, x = 7
Output: 2
Explanation: One jump forward (0 -> 16) then one jump backward (16 -> 7) will get the bug home.

Constraints:

• 1 <= forbidden.length <= 1000
• 1 <= a, b, forbidden[i] <= 2000
• 0 <= x <= 2000
• All the elements in forbidden are distinct.
• Position x is not forbidden.

Solution: BFS

Normal BFS with two tricks:
1. For each position, we need to track whether it’s reached via a forward jump or backward jump
2. How far should we go? If we don’t limit, it can go forever which leads to TLE/MLE. We can limit the distance to 2*max_jump, e.g. 4000, that’s maximum distance we can jump back to home in one shot.

Time complexity: O(max_distance * 2)
Space complexity: O(max_distance * 2)

// Author: Huahua
class Solution {
public:
  int minimumJumps(vector<int>& forbidden, int a, int b, int x) {
    constexpr int kMaxPosition = 4000;
    if (x == 0) return 0;
    queue<pair<int, bool>> q{{{0, true}}};
    unordered_set<int> seen1, seen2;
    for (int f : forbidden) seen1.insert(f), seen2.insert(f);
    seen1.insert(0);
    int steps = 0;
    while (!q.empty()) {      
      int size = q.size();
      while (size--) {
        auto [cur, forward] = q.front(); q.pop();        
        if (cur == x) return steps;
        if (cur > kMaxPosition) continue; // no way to go back
        if (seen1.insert(cur + a).second)      
          q.emplace(cur + a, true);
        if (cur - b >= 0 && forward && seen2.insert(cur - b).second)
          q.emplace(cur - b, false);
      }
      ++steps;
    }
    return -1;
  }
};

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