1345

1345. Jump Game IV (Hard)

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.

 

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

Companies: Amazon, Google, TikTok

Related Topics:
Array, Hash Table, Breadth-First Search

Similar Questions:

• Jump Game VII (Medium)
• Jump Game VIII (Medium)
• Maximum Number of Jumps to Reach the Last Index (Medium)

Solution 1. BFS

Since we are looking for the shortest distance, BFS should be our first option.

// OJ: https://leetcode.com/problems/jump-game-iv/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int minJumps(vector<int>& A) {
        unordered_map<int, vector<int>> m;
        int N = A.size(), step = 0;
        for (int i = 0; i < N; ++i) m[A[i]].push_back(i);
        vector<bool> seen(N);
        seen[0] = true;
        queue<int> q{{0}};
        while (q.size()) {
            int cnt = q.size();
            while (cnt--) {
                int u = q.front();
                q.pop();
                if (u == N - 1) return step;
                if (u - 1 >= 0 && !seen[u - 1]) {
                    q.push(u - 1);
                    seen[u - 1] = true;
                }
                if (u + 1 < N && !seen[u + 1]) {
                    q.push(u + 1);
                    seen[u + 1] = true;
                }
                if (m.count(A[u])) {
                    for (int v : m[A[u]]) {
                        if (seen[v]) continue;
                        seen[v] = true;
                        q.push(v);
                    }
                    m.erase(A[u]); // Note that we should remove this set of nodes since there is no point revisiting them later.
                }
            }
            ++step;
        }
        return -1;
    }
};