1345.Jump Game IV
===

###### tags: `Hard`,`Array`,`Hash Table`,`BFS`

[1345. Jump Game IV](https://leetcode.com/problems/jump-game-iv/)

### 題目描述

Given an array of integers `arr`, you are initially positioned at the first index of the array.

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.

### 範例

**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 do not 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.

**Constraints**:

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

### 解答
#### C++

class Solution {
public:
    int minJumps(vector<int>& arr) {
        int n = arr.size();
        unordered_map<int, vector<int>> m;
        for (int i = 0; i < n; i++) m[arr[i]].push_back(i);

        vector<bool>visited(n, false);
        queue<int>q;
        q.push(0);
        for (int step = 0; !q.empty(); step++) {
            for (int t = q.size(); t > 0; t--) {
                int i = q.front();
                q.pop();
                if (i == n - 1) return step;
                if (i + 1 < n && !visited[i + 1]) {
                    visited[i + 1] = true;
                    q.push(i + 1);
                }
                if (i - 1 >= 0 && !visited[i - 1]) {
                    visited[i - 1] = true;
                    q.push(i - 1);
                }
                for (int j : m[arr[i]]) {
                    if (!visited[j]) {
                        visited[j] = true;
                        q.push(j);
                    }
                }
                m[arr[i]].clear();
            }
        }
        return 0;
    }
};

> [name=Yen-Chi Chen][time=Mon, Mar 6, 2023]

#### Python

class Solution:
    def minJumps(self, arr: List[int]) -> int:
        n = len(arr)
        d = defaultdict(list)
        for i in range(n):
            d[arr[i]].append(i)
        
        curr = [0]
        visited = {0}
        step = 0
        while curr:
            children = []
            for i in curr:
                if i == n - 1: return step
                for j in d[arr[i]]:
                    if j not in visited:
                        visited.add(j)
                        children.append(j)
                d[arr[i]].clear()
                for j in [i - 1, i + 1]:
                    if 0 <= j < n and j not in visited:
                        visited.add(j)
                        children.append(j)
            curr = children
            step += 1
        return -1

> [name=Yen-Chi Chen][time=Mon, Mar 6, 2023]

### Reference

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