Question

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:

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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:

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Input: arr = [7]
Output: 0
Explanation: Start index is the last index. You do not need to jump.

Example 3:

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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

Source: https://leetcode.com/problems/jump-game-iv/

Solution

The idea of using queue size is to achieve layer-order traversal, like LeetCode 102. Binary Tree Level Order Traversal. So we can record the length of current search paths easily.

Repeated visits to nodes with the same value are not helpful in getting the min number of steps to the end. Thus, we can prune the BFS tree. Note that marking those nodes as visited is not enough. We need to clear corresponding key-value pair in the valueIndexMap. Because as long as these nodes are pushed to the queue, they will increase time cost.

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public int minJumps(int[] arr) {
    int len = arr.length;
    // value -> list of indexes that have the same value arr[i]
    Map<Integer, List<Integer>> valueIndexMap = new HashMap<>();
    for (int i = 0; i < len; i++) {
        if (valueIndexMap.containsKey(arr[i])) {
            List<Integer> list = valueIndexMap.get(arr[i]);
            list.add(i);
        } else {
            List<Integer> list = new ArrayList<>();
            list.add(i);
            valueIndexMap.put(arr[i], list);
        }
    }

    int pathLen = 0;
    boolean[] visited = new boolean[len]; // initialized with false
    Queue<Integer> queue = new LinkedList<>(); // stored indexes
    queue.add(0);
    while (!queue.isEmpty()) {
        int size = queue.size();
        // traverse in layer order
        for (int i = 0; i < size; i++) {
            int currIndex = queue.poll();
            if (visited[currIndex]) {
                continue;
            }
            if (currIndex == len - 1) {
                // the first return path is the shortest path
                return pathLen;
            }
            visited[currIndex] = true;
            if (currIndex - 1 >= 0) {
                queue.add(currIndex - 1);
            }
            if (currIndex + 1 < len) {
                queue.add(currIndex + 1);
            }
            for (int index : valueIndexMap.get(arr[currIndex])) {
                if (index != currIndex) {
                    queue.add(index);
                }
            }
            // there is no meaning to teleport to the same value on current and further iterations
            // only produce paths with identical or larger length
            // so that to waive repeated inner loop
            valueIndexMap.get(arr[currIndex]).clear();
        }
        pathLen++;
    }
    return -1;
}

Question

Given an array of non-negative integers arr, you are initially positioned at start index of the array. When you are at index i, you can jump to i + arr[i] or i - arr[i], check if you can reach to any index with value 0.

Notice that you can not jump outside of the array at any time.

Question

There is an undirected graph with n nodes, where each node is numbered between 0 and n - 1. You are given a 2D array graph, where graph[u] is an array of nodes that node u is adjacent to. More formally, for each v in graph[u], there is an undirected edge between node u and node v. The graph has the following properties:

• There are no self-edges (graph[u] does not contain u).
• There are no parallel edges (graph[u] does not contain duplicate values).
• If v is in graph[u], then u is in graph[v] (the graph is undirected).
• The graph may not be connected, meaning there may be two nodes u and v such that there is no path between them.

A graph is bipartite if the nodes can be partitioned into two independent sets A and B such that every edge in the graph connects a node in set A and a node in set B.

Return true if and only if it is bipartite.

Question

Given the root of a binary tree, return the level order traversal of its nodes’ values. (i.e., from left to right, level by level).