How do you solve Jump Game IV on LeetCode?

Jump Game IV (LeetCode #1345) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using BFS ensures that we find the shortest path to the last index efficiently.

What is the brute force solution for Jump Game IV?

The brute force approach for Jump Game IV is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible jump from each index to find the minimum steps to reach the last index. It explores all paths recursively, which can be inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Jump Game IV?

Jump Game IV uses the following concepts: Array, Hash Table, Breadth-First Search. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Jump Game IV?

Always think about edge cases, such as arrays of length 1. Be clear about the data structures you choose; they can significantly affect performance. Practice explaining your thought process as you code; it helps solidify your understanding.

What are common mistakes when solving Jump Game IV?

Not using a visited set, which can lead to infinite loops. Failing to clear the graph after processing to avoid revisiting indices.

#1345

Jump Game IV

Hard

ArrayHash TableBreadth-First SearchHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses Breadth-First Search (BFS) to explore the shortest path to the last index. By leveraging a graph representation of the indices and their connections based on values, we can efficiently find the minimum steps.

⚙️

Algorithm

4 steps

1Step 1: Create a graph where each value in the array points to all indices that have the same value.
2Step 2: Use BFS starting from index 0, keeping track of visited indices to avoid cycles.
3Step 3: For each index, explore the next index (i + 1), previous index (i - 1), and all indices with the same value as arr[i].
4Step 4: Return the number of steps when reaching the last index.

solution.py27 lines

1# Full working Python code
2from collections import defaultdict, deque
3
4def minJumps(arr):
5    n = len(arr)
6    if n <= 1:
7        return 0
8    visited = set()
9    graph = defaultdict(list)
10    for i in range(n):
11        graph[arr[i]].append(i)
12
13    queue = deque([(0, 0)])  # (index, steps)
14    visited.add(0)
15
16    while queue:
17        index, steps = queue.popleft()
18        if index == n - 1:
19            return steps
20        # Check next and previous indices
21        for next_index in [index + 1, index - 1] + graph[arr[index]]:
22            if 0 <= next_index < n and next_index not in visited:
23                visited.add(next_index)
24                queue.append((next_index, steps + 1))
25        graph[arr[index]].clear()  # Clear to prevent future jumps
26
27    return -1

ℹ

Complexity note: The time complexity is O(n) because each index is processed once, and the space complexity is O(n) due to the storage of the graph and visited set.

1Using BFS ensures that we find the shortest path to the last index efficiently.
2Clearing the graph after processing an index prevents unnecessary future jumps.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.