1345. Jump Game IV

Problem Statement

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

Click to open Hints

• Build a graph of n nodes where nodes are the indices of the array and edges for node i are nodes i+1, i-1, j where arr[i] == arr[j].
• Start bfs from node 0 and keep distance. The answer is the distance when you reach node n-1.

Solution:

Rust

use std::collections::{HashMap, VecDeque};

impl Solution {
    pub fn min_jumps(arr: Vec<i32>) -> i32 {
        let mut map = HashMap::new();

        for (i, &v) in arr.iter().enumerate() {
            map.entry(v).or_insert(vec![]).push(i);
        }

        let mut q = VecDeque::new();
        q.push_back((0, 0_i32));

        let mut visited = vec![false; arr.len()];
        visited[0] = true;

        while let Some((i, mut step)) = q.pop_front() {
            if i == arr.len() - 1 {
                return step;
            }

            step += 1;

            if let Some(v) = map.remove(&arr[i]) {
                for j in v {
                    if !visited[j] {
                        visited[j] = true;
                        q.push_back((j, step));
                    }
                }
            }

            if i + 1 < arr.len() && !visited[i + 1] {
                visited[i + 1] = true;
                q.push_back((i + 1, step));
            }

            if i >= 1 && !visited[i - 1] {
                visited[i - 1] = true;
                q.push_back((i - 1, step));
            }
        }

        -1
    }
}

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