Jump Game IV

Jump Game IV

1. What is this problem about?

The Jump Game IV interview question is a high-level graph search problem. You are given an array of integers. From any index i, you can jump to:

1. i + 1
2. i - 1
3. Any index j such that arr[i] == arr[j] and i != j. Your goal is to find the minimum number of jumps to get from index 0 to the last index of the array.

2. Why is this asked in interviews?

Companies like Meta, Amazon, and Google use the Jump Game IV coding problem to test your ability to optimize Breadth-First Search (BFS). The "jump to same value" rule can create a dense graph where a single value (like 7) appears thousands of times. A naive BFS would be $O(N^2)$. It tests if you can identify and prune redundant edges to achieve $O(N)$ time.

3. Algorithmic pattern used

This problem uses BFS with Map-based Grouping and Edge Pruning.

1. Pre-processing: Create a Hash Map valToIndices where the key is the number and the value is a list of all indices where that number appears.
2. BFS: Start a standard BFS from index 0.
3. Pruning (The Critical Step): After you've explored all "same-value" jumps for a specific value $X$, clear the list for $X$ in the map. This ensures you never traverse those same-value edges again, reducing the complexity from $O(N^2)$ to $O(N)$.

4. Example explanation

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

1. Level 0: {0}
2. Level 1: 0 -> 1 (neighbor) and 0 -> 4 (same value 100). Queue: {1, 4}. Clear map entry for 100.
3. Level 2: From 4, jump to 3 (neighbor) or 9 (neighbor/same value 404). Queue: {1, 3, 9}.
4. Index 9 is the target. Result: 3 jumps.

5. Common mistakes candidates make

• $O(N^2)$ BFS: Failing to clear the map after the first time a value's indices are added to the queue.
• Directional Bias: Forgetting that you can jump backwards (i - 1).
• Memory Limit: Not being careful with the "visited" set, leading to massive memory usage in very large arrays.

6. Interview preparation tip

Master the BFS Edge Pruning technique. In many graph problems where "all nodes with property X are connected," you can treat property X as a single "virtual node" or simply process its edges once and then "delete" them. This is a common Hash Table interview pattern optimization.