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1345. Jump Game IV

Description

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

Solutions

Solution 1

Thinking

A jump may go to a neighbor or to any equal value; we want the fewest jumps to the last index. \(n \le 5 \times 10^4\), so rescanning an equal-value list is too slow. Layered BFS enqueues \(i\pm 1\) and the equal indices, then \(\textit{pop}\)s that value's list so each such edge is used once.

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class Solution:
    def minJumps(self, arr: List[int]) -> int:
        g = defaultdict(list)
        for i, x in enumerate(arr):
            g[x].append(i)
        q = deque([0])
        vis = {0}
        ans = 0
        while 1:
            for _ in range(len(q)):
                i = q.popleft()
                if i == len(arr) - 1:
                    return ans
                for j in (i + 1, i - 1, *g.pop(arr[i], [])):
                    if 0 <= j < len(arr) and j not in vis:
                        q.append(j)
                        vis.add(j)
            ans += 1

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class Solution {
    public int minJumps(int[] arr) {
        Map<Integer, List<Integer>> g = new HashMap<>();
        int n = arr.length;
        for (int i = 0; i < n; i++) {
            g.computeIfAbsent(arr[i], k -> new ArrayList<>()).add(i);
        }
        boolean[] vis = new boolean[n];
        Deque<Integer> q = new ArrayDeque<>();
        q.offer(0);
        vis[0] = true;
        for (int ans = 0;; ++ans) {
            for (int k = q.size(); k > 0; --k) {
                int i = q.poll();
                if (i == n - 1) {
                    return ans;
                }
                for (int j : g.get(arr[i])) {
                    if (!vis[j]) {
                        vis[j] = true;
                        q.offer(j);
                    }
                }
                g.get(arr[i]).clear();
                for (int j : new int[] {i - 1, i + 1}) {
                    if (0 <= j && j < n && !vis[j]) {
                        vis[j] = true;
                        q.offer(j);
                    }
                }
            }
        }
    }
}

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class Solution {
public:
    int minJumps(vector<int>& arr) {
        unordered_map<int, vector<int>> g;
        int n = arr.size();
        for (int i = 0; i < n; ++i) {
            g[arr[i]].push_back(i);
        }
        vector<bool> vis(n);
        queue<int> q{{0}};
        vis[0] = true;
        for (int ans = 0;; ++ans) {
            for (int k = q.size(); k; --k) {
                int i = q.front();
                q.pop();
                if (i == n - 1) {
                    return ans;
                }
                for (int j : g[arr[i]]) {
                    if (!vis[j]) {
                        vis[j] = true;
                        q.push(j);
                    }
                }
                g[arr[i]].clear();
                for (int j : {i - 1, i + 1}) {
                    if (0 <= j && j < n && !vis[j]) {
                        vis[j] = true;
                        q.push(j);
                    }
                }
            }
        }
    }
};

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func minJumps(arr []int) int {
    g := map[int][]int{}
    for i, x := range arr {
        g[x] = append(g[x], i)
    }
    n := len(arr)
    q := []int{0}
    vis := make([]bool, n)
    vis[0] = true
    for ans := 0; ; ans++ {
        for k := len(q); k > 0; k-- {
            i := q[0]
            q = q[1:]
            if i == n-1 {
                return ans
            }
            for _, j := range g[arr[i]] {
                if !vis[j] {
                    vis[j] = true
                    q = append(q, j)
                }
            }
            g[arr[i]] = nil
            for _, j := range []int{i - 1, i + 1} {
                if 0 <= j && j < n && !vis[j] {
                    vis[j] = true
                    q = append(q, j)
                }
            }
        }
    }
}

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function minJumps(arr: number[]): number {
    const g: Map<number, number[]> = new Map();
    const n = arr.length;
    for (let i = 0; i < n; ++i) {
        if (!g.has(arr[i])) {
            g.set(arr[i], []);
        }
        g.get(arr[i])!.push(i);
    }
    let q: number[] = [0];
    const vis: boolean[] = Array(n).fill(false);
    vis[0] = true;
    for (let ans = 0; ; ++ans) {
        const nq: number[] = [];
        for (const i of q) {
            if (i === n - 1) {
                return ans;
            }
            for (const j of g.get(arr[i])!) {
                if (!vis[j]) {
                    vis[j] = true;
                    nq.push(j);
                }
            }
            g.get(arr[i])!.length = 0;
            for (const j of [i - 1, i + 1]) {
                if (j >= 0 && j < n && !vis[j]) {
                    vis[j] = true;
                    nq.push(j);
                }
            }
        }
        q = nq;
    }
}

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