Problem Description

The problem involves an array of integers arr and a character that starts at the first index of the array. This character can jump to a new index in one of three ways. They can jump to the next index (i + 1), jump back to the previous index (i - 1), or jump to any index j that contains the same value as the current index (arr[i] == arr[j] and i != j). The goal is to determine the minimum number of steps required to reach the last index of the array. It is important to note that jumps can only occur within the bounds of the array, meaning that the character cannot jump to an index outside of the array.

Flowchart Walkthrough

Let's analyze LeetCode problem 1345. Jump Game IV using the algorithm flowchart. Here's a step-by-step walkthrough to deduce the appropriate algorithm:

Is it a graph?

• Yes: Although not explicitly structured as a graph, the connections dictated by the indices and values (i.e., jumps to indices of the same value, forward and backward by one) allow it to be represented as a graph.

Is it a tree?

• No: The graph isn't necessarily hierarchical, as it includes multiple pathways via index values to potentially re-visit nodes.

Is the problem related to directed acyclic graphs (DAGs)?

• No: This problem involves a general graph where cycles may exist due to the ability to jump backward and forward to elements with the same value.

Is the problem related to shortest paths?

• Yes: The objective is to find the minimum number of jumps to get from the start to the end of the array, which can be framed as finding the shortest path in a graph.

Is the graph weighted?

• No: Every jump, irrespective of its length (either index difference or value-based connectivity), is treated equally with a basic increment in count.

Conclusion: Following the flowchart, the emphasis on shortest paths in an unweighted setting suggests the application of the Breadth-First Search (BFS) pattern. BFS is ideal as it effectively explores each level or layer (i.e., connections or jumps that can be made in one move), ensuring the shortest path is found in an unweighted graph.

Intuition

To arrive at the solution, we use breadth-first search (BFS) as the main strategy since the goal is to find the minimum number of steps. BFS is suitable for this kind of problem because it explores the nearest neighbors first and only goes deeper once there are no more neighbors to explore. Given this approach, we start from the first index and explore all possible jumps. To avoid revisiting indices and getting stuck in loops, we use a visited set (vis) to keep track of which indices have already been explored. Also, to avoid repeatedly performing the same jumps, we use a dictionary (idx) that maps each value to its indices in the array.

As we progress, we store in a queue (q) each index we can jump to, along with the number of steps taken to reach there. Once the jump to the last index is made, we return the number of steps that led to it as the result. To make the BFS efficient, after checking all indices that have the same value as the current index, we delete the entry from the idx dictionary to prevent redundant jumps in the future. This speeds up the BFS process significantly, as it reduces the number of potential jumps that are not helpful in reaching the end of the array quickly.

Learn more about Breadth-First Search patterns.

Solution Approach

To solve this problem, the code uses the Breadth-First Search (BFS) algorithm, which is a standard approach for finding the shortest path in an unweighted graph or the fewest number of steps to reach a certain point.

Here's a walkthrough of the implementation:

1. Index Dictionary (idx): A defaultdict from Python's collections module is used to create a dictionary where each value in the array is a key, and the corresponding value is a list of all indices in the array with that value.

   idx = defaultdict(list)
   for i, v in enumerate(arr):
       idx[v].append(i)

2. Queue (q): A queue is implemented with deque from the collections module, and it's initialized with a tuple containing the starting index (0) and the initial step count (0).

   q = deque([(0, 0)])

3. Visited Set (vis): A set is used to keep track of the indices that have been visited. This prevents the algorithm from processing an index more than once and ensures not to get caught in an infinite loop.

   vis = {0}

4. BFS Loop: The algorithm processes the queue q until it's empty, doing the following:

   • It dequeues an element, which is a tuple of the current index (i) and the number of steps taken to reach this index (step).
   • If the dequeued index is the last index of the array, the number of steps is immediately returned.
   • All indices j in the array where arr[i] == arr[j] are added to the queue, with the number of steps incremented by 1, and marked as visited.
   • After processing all jumps to the same value, that value is deleted from idx to prevent future unnecessary jumps to those indices.
   • The algorithm then looks to add the neighbor indices i + 1 and i - 1 to the queue (if they haven't been visited yet).

   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))

By employing BFS, which explores all possible paths breadthwise and visits each node exactly once, the solution finds the minimum number of steps needed to reach the last index of the array.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach for the array arr = [100, 23, 23, 100, 23, 100].

1. Initialize the Index Dictionary (idx): We create a dictionary mapping each value in the array to a list of all indices in the array with that value.

   • idx[100] = [0, 3, 5]
   • idx[23] = [1, 2, 4]

2. Initialize the Queue (q): We start with the first index and 0 steps as we haven't yet taken any steps.

   • q = deque([(0, 0)])

3. Initialize the Visited Set (vis): To begin, we've only visited the start index 0.

   • vis = {0}

4. Begin the BFS Loop:

   • Dequeue (0, 0). Current index i = 0, steps taken step = 0.

     • We found the value 100 at index 0. Indices with value 100 are [0, 3, 5].
     • Queue up index 3 and 5 with step+1 since 0 is already visited.
     • Update queue: q = deque([(3, 1), (5, 1)]) and vis = {0, 3, 5}.
     • Remove 100 from the idx dictionary to prevent future jumps to these indices: del idx[100].

   • Dequeue (3, 1). Current index i = 3, steps taken step = 1.

     • Value at index 3 is 100, but we've already deleted it from idx.
     • We check the neighbors, and 4 has not been visited, so we queue it up.
     • Update queue: q = deque([(5, 1), (4, 2)]) and vis = {0, 3, 4, 5}.

   • Dequeue (5, 1). Current index i = 5, steps taken step = 1.

     • We've reached the last index of the array, so we return the number of steps: 1.

In this example, by using BFS and our optimizations, we've found that we can reach the end of the array arr in just 1 step through the value jumps. We can jump directly from the first index to the last index since they share the same value of 100.

Solution Implementation

Time and Space Complexity

The time complexity of the provided code is O(N + E) where N is the number of elements in arr and E is the total number of edges in the graph constructed where each value in arr has edges to indices with the same value. The while loop runs for each index in arr once, and each index is added to the queue at most once. The inner loops contribute to the edges. Deletion of keys in the idx dictionary ensures that each value's list is iterated over at most once, hence, contributing to E.

The space complexity is O(N) as we need to store the indices in the idx dictionary, which in the worst case will store all indices, and the vis set, which also could potentially store all indices if they are visited. The queue could also hold up to N-1 elements in the worst case scenario.

Learn more about how to find time and space complexity quickly using problem constraints.