Problem Statement
In this challenge, your task is to determine the minimum number of moves required to traverse from the first index to the last index of a given integer array, arr. You start at the first element of the array and you can jump under certain conditions:
- To the adjacent index in either direction (left or right).
- Directly to any index that contains the same value as the current index.
The objective is to find out the least number of jumps required to reach the very last element of the array. It's crucial to plan jumps intelligently as you can leap over multiple indices in one move if a value repeats itself in the array, potentially reducing the total number of steps required drastically.
Examples
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
Approach and Intuition
The problem at hand essentially deals with finding the shortest path in a special kind of graph where each element of the array represents a node and the possible jumps are the edges. Here's a breakdown of the approach:
- Visualize each element of the array
arras a node in a graph. - Construct edges based on the possible moves:
- From node
itoi+1, ifi+1is within array bounds. - From node
itoi-1, ifi-1is within array bounds. - From node
ito any nodej(wherej != i) ifarr[i] == arr[j].
- From node
- The task now is to find the shortest path from the first node (index
0) to the last node (indexarr.length - 1). - A Breadth-First Search (BFS) is suitable for this scenario:
- Start with the first index and explore all reachable indices using the aforementioned jump rules.
- For each vertex, record the minimum number of steps taken to reach it.
- Use a queue to manage the exploration and a set or boolean array to keep track of already visited nodes to avoid cyclic paths and redundant work.
- For values that occur multiple times in the array, utilize a hashmap to quickly access all indices of the same value, allowing for rapid jumps across distant parts of the array.
Example insights based on given examples:
- In certain configurations where the array has only one element or where direct jumps can be made to the last index (because of repeated values), the number of steps required can be minimized effectively to zero or one respectively.
- In more complex arrays with multiple repeated values or longer lengths, strategic use of jumps to indices with the same values can cut down numerous incremental steps.
This method should be efficient, given the constraints, although the need to keep track of visited nodes and the dynamic construction of connections (especially where values are repeated frequently) can affect the speed of execution. Thus, the BFS algorithm, supplemented with appropriate data structures for tracking and retrieving connections, stands out as a potent approach to solve the problem.
Solutions
- Java
- Python
class Solution {
public static int minimumJumps(int[] array) {
int length = array.length;
if (length <= 1) {
return 0;
}
Map<Integer, List<Integer>> indexMap = new HashMap<>();
for (int i = 0; i < length; i++) {
indexMap.computeIfAbsent(array[i], v -> new LinkedList<>()).add(i);
}
HashSet<Integer> currentSet = new HashSet<>(); // store layers from start
currentSet.add(0);
Set<Integer> visitedIndices = new HashSet<>();
visitedIndices.add(0);
visitedIndices.add(length - 1);
int jumps = 0;
HashSet<Integer> targetSet = new HashSet<>(); // store layers from end
targetSet.add(length - 1);
// when current layer exists
while (!currentSet.isEmpty()) {
// search from the side with fewer nodes
if (currentSet.size() > targetSet.size()) {
HashSet<Integer> temp = currentSet;
currentSet = targetSet;
targetSet = temp;
}
HashSet<Integer> nextSet = new HashSet<>();
// iterate the layer
for (int current : currentSet) {
// check same value
for (int connected : indexMap.get(array[current])) {
if (targetSet.contains(connected)) {
return jumps + 1;
}
if (!visitedIndices.contains(connected)) {
visitedIndices.add(connected);
nextSet.add(connected);
}
}
// clear the list to prevent redundant search
indexMap.get(array[current]).clear();
// check neighbors
if (targetSet.contains(current + 1) || targetSet.contains(current - 1)) {
return jumps + 1;
}
if (current + 1 < length && !visitedIndices.contains(current + 1)) {
visitedIndices.add(current + 1);
nextSet.add(current + 1);
}
if (current - 1 >= 0 && !visitedIndices.contains(current - 1)) {
visitedIndices.add(current - 1);
nextSet.add(current - 1);
}
}
currentSet = nextSet;
jumps++;
}
return -1;
}
}
The provided Java code defines a solution for the "Jump Game IV" problem. This problem requires determining the minimum number of jumps needed to move from the start to the end of an array where each element indicates jump to any index with the same value, the next or the previous index.
Follow these primary steps used in the code to achieve the solution:
- Check if the array length is less than or equal to 1, in which case, return 0 since no jumps are required.
- Create a mapping of array values to their indices using a HashMap. This facilitates quick access to all indices with a particular value.
- Initialize two sets:
currentSetfor the start layer which initially contains the index 0, andtargetSetfor the end layer which contains the last index. - Use a HashSet
visitedIndicesto track visited indices, adding the first and last indices to it. - Iterate, calculating possible jumps until the
currentSetis empty. Determine if any index incurrentSetcan directly connect to an index intargetSetwith:- Using indices array elements are identical which can help jump to any other index with the same value.
- Considering immediate neighbors (i.e., indices
current+1andcurrent-1).
- In each iteration, swap
currentSetandtargetSetifcurrentSetis larger thantargetSetfor efficient processing. - Update the
currentSetwith new indices to be processed (neighbors and same value jumps) that haven't been visited. - Increment the jump counter after processing each layer.
- If a connection is found between the
currentSetandtargetSet, return the jump count incremented by one. - If no connection can be made, return -1 indicating it's not possible to jump to the end.
The code effectively uses Breadth-First Search (BFS) principles, where each layer (set of nodes) explored represents a potential jump. Using two sets for BFS from both ends assists in speeding up the search, essentially reducing the search space by always expanding the smaller set.
class Solution:
def shortestPath(self, arr) -> int:
size = len(arr)
if size <= 1:
return 0
connections = {}
for index in range(size):
if arr[index] in connections:
connections[arr[index]].append(index)
else:
connections[arr[index]] = [index]
from_start = set([0]) # initial layer from start
visited_nodes = {0, size-1}
distance = 0
from_end = set([size-1]) # initial layer from end
while from_start:
if len(from_start) > len(from_end):
from_start, from_end = from_end, from_start
next_layer = set()
for current in from_start:
for adj in connections[arr[current]]:
if adj in from_end:
return distance + 1
if adj not in visited_nodes:
visited_nodes.add(adj)
next_layer.add(adj)
connections[arr[current]].clear()
for adj in [current-1, current+1]:
if adj in from_end:
return distance + 1
if 0 <= adj < size and adj not in visited_nodes:
visited_nodes.add(adj)
next_layer.add(adj)
from_start = next_layer
distance += 1
return -1
The provided Python solution tackles the problem of calculating the shortest path in the "Jump Game IV" challenge using a bidirectional BFS approach. Here's a breakdown of the key aspects and operations performed in the solution:
Initialization: The function accepts an array
arr, and sets up basic variables likesizefor the length of the array, a dictionaryconnectionsto map elements to their indices, and setsfrom_startandfrom_endto initiate the BFS from both ends of the array.Connections Setup: Iterates through the array to populate the
connectionsdictionary, keying array values to lists of their indices. This helps in quickly looking up possible "jumps" from any given position.Bidirectional BFS:
- The algorithm alternates layers of the graph starting from both ends (
from_startandfrom_end). - For each node in the current layer, it explores all adjacent nodes (either by array value or +/-1 positions).
- If a node from the opposite end is found during exploration, it returns the current distance incremented by one (indicating the ends have met).
- The BFS leverages a
visited_nodesset to avoid revisiting nodes, and the search from either end might swap in each iteration to enhance performance by always expanding the smaller fringe.
- The algorithm alternates layers of the graph starting from both ends (
Termination: If the BFS layers meet, the shortest distance is returned. If the search exhausts all possibilities without meeting, it returns -1, indicating no possible path.
Overall, the approach effectively uses a double-ended BFS to minimize the search space and efficiently find the shortest path between two points in the graph represented by the game's positions and allowed jumps.
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