1345.Jump Game IV

tags: Hard,Array,Hash Table,BFS

1345. 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 * 10⁴
• -10⁸ <= arr[i] <= 10⁸

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;
    }
};

Yen-Chi ChenMon, 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

Yen-Chi ChenMon, Mar 6, 2023

Reference

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