#1345

Hard

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Jump Game IV

Jump Game IV requires minimizing steps to reach the last index using jumps across equal values and neighbors efficiently.

array hash table breadth first search

Problem Statement

Given an integer array arr, start at index 0. You can jump to index i+1, i-1, or any index j where arr[i] == arr[j].

Determine the minimum number of jumps required to reach the last index of the array. Design an approach that efficiently handles repeated values to avoid unnecessary traversals.

Examples

Example 1

Input: arr = [100,-23,-23,404,100,23,23,23,3,404]

Output: 3

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

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

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

Solution Approach

Map Values to Indices

Build a hash table mapping each unique value in arr to a list of all its indices. This allows O(1) lookup for jumps to indices with the same value, crucial for BFS efficiency.

Breadth-First Search Traversal

Use BFS starting from index 0, enqueuing reachable neighbors (i+1, i-1, and indices from the value map). Track visited indices to prevent cycles. Increment steps at each BFS level to count jumps.

Prune Visited Value Groups

After processing all jumps for a value, clear its list in the hash table. This avoids redundant exploration of indices with the same value, which is a common bottleneck in large arrays with duplicates.

Complexity Analysis

Metric	Value
Time	\mathcal{O}(N)
Space	\mathcal{O}(N)

Time complexity is O(N) since each index is visited at most once in BFS. Space complexity is O(N) to store the hash table mapping values to indices and the BFS queue.

What Interviewers Usually Probe

Candidate immediately considers BFS with value-based jumps.
Candidate correctly prunes visited value groups to prevent repeated traversals.
Candidate discusses time and space trade-offs related to large arrays with duplicates.

Common Pitfalls or Variants

Common pitfalls

Failing to clear processed value lists leading to repeated jumps.
Ignoring boundary conditions for i-1 and i+1 jumps at edges of the array.
Attempting DFS instead of BFS, causing inefficient exploration and wrong minimal steps.

Follow-up variants

Allow jumps to multiples of the current value instead of equality, modifying the value map approach.
Restrict jumps to only equal values, removing i+1 and i-1, emphasizing hash lookup efficiency.
Count all possible minimal jump sequences instead of only one, extending BFS to track multiple paths.

FAQ

What is the main pattern in Jump Game IV?

The core pattern is array scanning plus hash lookup to jump across indices with equal values efficiently.

How do I handle large arrays with many duplicates?

Use a hash map to track indices for each value and clear lists after visiting to avoid redundant BFS expansions.

Can BFS guarantee minimal jumps in Jump Game IV?

Yes, BFS explores all reachable indices level by level, ensuring the first time you reach the last index is minimal steps.

Why not use DFS for this problem?

DFS may explore deeper paths first and miss shorter sequences, leading to non-optimal jump counts.

What are typical mistakes to avoid in Jump Game IV?

Common errors include revisiting indices of the same value, mishandling array boundaries, or not pruning processed value lists.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def minJumps(self, arr: List[int]) -> int:
        g = defaultdict(list)
        for i, x in enumerate(arr):
            g[x].append(i)
        q = deque([0])
        vis = {0}
        ans = 0
        while 1:
            for _ in range(len(q)):
                i = q.popleft()
                if i == len(arr) - 1:
                    return ans
                for j in (i + 1, i - 1, *g.pop(arr[i], [])):
                    if 0 <= j < len(arr) and j not in vis:
                        q.append(j)
                        vis.add(j)
            ans += 1

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