In one step you can jump from index i to index:

• i + 1 where: i + 1 < arr.length.
• i - 1 where: i - 1 >= 0.
• j where: arr[i] == arr[j] and i != j.

Return the minimum number of steps to reach the last index of the array.

Notice that you can not jump outside of the array at any time.

Example 1:

Input: arr = [100,-23,-23,404,100,23,23,23,3,404]
Output: 3
Explanation: You need three jumps from index 0 --> 4 --> 3 --> 9. Note that index 9 is the last index of the array.

Example 2:

Input: arr = [7]
Output: 0
Explanation: Start index is the last index. You don't need to jump.

Example 3:

Input: arr = [7,6,9,6,9,6,9,7]
Output: 1
Explanation: You can jump directly from index 0 to index 7 which is last index of the array.

Example 4:

Input: arr = [6,1,9]
Output: 2

Example 5:

Input: arr = [11,22,7,7,7,7,7,7,7,22,13]
Output: 3

Constraints:

• 1 <= arr.length <= 5 * 10^4
• -10^8 <= arr[i] <= 10^8

Solution: HashTable + BFS

Use a hashtable to store the indices of each unique number.

each index i has neighbors (i-1, i + 1, hashtable[arr[i]])

Use BFS to find the shortest path in this unweighted graph.

Key optimization, clear hashtable[arr[i]] after the first use, since all nodes are already on queue, no longer needed.

Time complexity: O(n)
Space complexity: O(n)

// Author: Huahua, 124 ms, 30.5 MB
class Solution {
public:
  int minJumps(vector<int>& arr) {
    const int n = arr.size();
    unordered_map<int, vector<int>> m;
    for (int i = 0; i < n; ++i)
      m[arr[i]].push_back(i);
    vector<int> seen(n);
    queue<int> q({0});    
    seen[0] = 1;
    int steps = 0;
    while (!q.empty()) {
      int size = q.size();
      while (size--) {
        int i = q.front(); q.pop();        
        if (i == n - 1) return steps;
        if (i - 1 >= 0 && !seen[i - 1]++) q.push(i - 1);
        if (i + 1 < n && !seen[i + 1]++) q.push(i + 1);
        auto it = m.find(arr[i]);
        if (it == m.end()) continue;
        for (int nxt : it->second)
          if (!seen[nxt]++) q.push(nxt);
        m.erase(it); // no longer needed.
      }
      ++steps;
    }
    return -1;
  }
};

Related Problems

• https://zxi.mytechroad.com/blog/greedy/leetcode-55-jump-game/
• https://zxi.mytechroad.com/blog/greedy/leetcode-45-jump-game-ii/
• https://zxi.mytechroad.com/blog/searching/leetcode-1306-jump-game-iii/
• https://zxi.mytechroad.com/blog/dynamic-programming/leetcode-1344-jump-game-v/

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