Question

Formatted question description: https://leetcode.ca/all/1345.html

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

Algorithm

Reference: https://leetcode.com/problems/jump-game-iv/solution/

store nodes with the same value together in a graph dictionary

First loop over arr to obtain each element and the indices at which the element is in arr. Then loop over arr again to obtain the next indices of each index.

Use dynamic programming. Starting from index 0, each time find the next possible indices that can be jumped to, and update the indices’ minimum jumps. Repeat the process until the last index is reached, and return the number of jumps to reach the last index.

To reduce runtime, consider some special cases first. If arr.length == 1, then the first index is also the last index, so return 0. If the elements at the first index and the last index are the same, then jump once to the last index, so return 1. If the elements at the first index and the second last index are the same, then jump once to the second last index and jump another time to the last index, so return 2.

Code

class Solution {
    public int minJumps(int[] arr) {
        int n = arr.length;
        if (n <= 1) {
            return 0;
        }

        // store nodes with the same value together in a graph dictionary
        Map<Integer, List<Integer>> graph = new HashMap<>();
        for (int i = 0; i < n; i++) {
            graph.computeIfAbsent(arr[i], v -> new LinkedList<>()).add(i);
        }

        List<Integer> curs = new LinkedList<>(); // store current layer
        curs.add(0);
        Set<Integer> visited = new HashSet<>();
        int step = 0;

        // when current layer exists
        while (!curs.isEmpty()) {
            List<Integer> nex = new LinkedList<>();

            // iterate the layer
            for (int node : curs) {
                // check if index 'node' reached end index
                if (node == n - 1) {
                    return step;
                }

                // 1. check same value
                for (int child : graph.get(arr[node])) {
                    if (!visited.contains(child)) {
                        visited.add(child);
                        nex.add(child);
                    }
                }

                // clear the list to prevent redundant search
                graph.get(arr[node]).clear();

                // 2. check left/right neighbors
                if (node + 1 < n && !visited.contains(node + 1)) {
                    visited.add(node + 1);
                    nex.add(node + 1);
                }
                if (node - 1 >= 0 && !visited.contains(node - 1)) {
                    visited.add(node - 1);
                    nex.add(node - 1);
                }
            }

            curs = nex;
            step++;
        }

        return -1;
    }
}

############

class Solution {
    public int minJumps(int[] arr) {
        Map<Integer, List<Integer>> idx = new HashMap<>();
        int n = arr.length;
        for (int i = 0; i < n; ++i) {
            idx.computeIfAbsent(arr[i], k -> new ArrayList<>()).add(i);
        }
        Deque<int[]> q = new LinkedList<>();
        Set<Integer> vis = new HashSet<>();
        vis.add(0);
        q.offer(new int[] {0, 0});
        while (!q.isEmpty()) {
            int[] e = q.pollFirst();
            int i = e[0], step = e[1];
            if (i == n - 1) {
                return step;
            }
            int v = arr[i];
            ++step;
            for (int j : idx.getOrDefault(v, new ArrayList<>())) {
                if (!vis.contains(j)) {
                    vis.add(j);
                    q.offer(new int[] {j, step});
                }
            }
            idx.remove(v);
            if (i + 1 < n && !vis.contains(i + 1)) {
                vis.add(i + 1);
                q.offer(new int[] {i + 1, step});
            }
            if (i - 1 >= 0 && !vis.contains(i - 1)) {
                vis.add(i - 1);
                q.offer(new int[] {i - 1, step});
            }
        }
        return -1;
    }
}

// OJ: https://leetcode.com/problems/jump-game-iv/
// 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]);
                }
            }
            ++step;
        }
        return -1;
    }
};

'''

list.remove(v): for list, remove by value, the first occurrence of v


del my_dict[v]: for dict, delete by key 'v'

    my_dict = {'a': 1, 'b': 2, 'c': 3}
    del my_dict['b']
    print(my_dict)  # {'a': 1, 'c': 3}


my_dict.pop('b')

    my_dict = {'a': 1, 'b': 2, 'c': 3}
    val = my_dict.pop('b')
    print(my_dict)  # {'a': 1, 'c': 3}
    print(val)  # 2


my_dict.popitem()

    my_dict = {'a': 1, 'b': 2, 'c': 3}
    item = my_dict.popitem()
    print(my_dict)  # {'a': 1, 'b': 2}
    print(item)  # ('c', 3)


filter() and lambda
    my_dict = {'a': 1, 'b': 2, 'c': 3}
    my_dict = dict(filter(lambda item: item[0] != 'b', my_dict.items()))
    print(my_dict)  # {'a': 1, 'c': 3}

'''

from collections import deque

class Solution:
    def minJumps(self, arr: List[int]) -> int:
        idx = defaultdict(list)
        for i, v in enumerate(arr):
            idx[v].append(i)
        q = deque([(0, 0)]) # (index, step)
        vis = {0}
        while q:
            i, step = q.popleft()
            if i == len(arr) - 1:
                return step
            v = arr[i]
            step += 1
            for j in idx[v]:
                if j not in vis:
                    vis.add(j)
                    q.append((j, step))
            # without this del, it will be huge/duplicated loop
            # over time limit for input [7,7,7,7,7,....7,7]
            del idx[v] # avoid dedup
            if i + 1 < len(arr) and (i + 1) not in vis:
                vis.add(i + 1)
                q.append((i + 1, step))
            if i - 1 >= 0 and (i - 1) not in vis:
                vis.add(i - 1)
                q.append((i - 1, step))

############

class Solution:
    def minJumps(self, arr: List[int]) -> int:
        idx = defaultdict(list)
        for i, v in enumerate(arr):
            idx[v].append(i)
        q = deque([(0, 0)])
        vis = {0}
        while q:
            i, step = q.popleft()
            if i == len(arr) - 1:
                return step
            v = arr[i]
            step += 1
            for j in idx[v]:
                if j not in vis:
                    vis.add(j)
                    q.append((j, step))
            del idx[v]
            if i + 1 < len(arr) and (i + 1) not in vis:
                vis.add(i + 1)
                q.append((i + 1, step))
            if i - 1 >= 0 and (i - 1) not in vis:
                vis.add(i - 1)
                q.append((i - 1, step))

func minJumps(arr []int) int {
	idx := map[int][]int{}
	for i, v := range arr {
		idx[v] = append(idx[v], i)
	}
	vis := map[int]bool{0: true}
	type pair struct{ idx, step int }
	q := []pair{ {0, 0} }
	for len(q) > 0 {
		e := q[0]
		q = q[1:]
		i, step := e.idx, e.step
		if i == len(arr)-1 {
			return step
		}
		step++
		for _, j := range idx[arr[i]] {
			if !vis[j] {
				vis[j] = true
				q = append(q, pair{j, step})
			}
		}
		delete(idx, arr[i])
		if i+1 < len(arr) && !vis[i+1] {
			vis[i+1] = true
			q = append(q, pair{i + 1, step})
		}
		if i-1 >= 0 && !vis[i-1] {
			vis[i-1] = true
			q = append(q, pair{i - 1, step})
		}
	}
	return -1
}

1345 - Jump Game IV

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