Given an array of integers arr, you are initially positioned at the first index of the array.
In one step you can jump from index i to index:
i + 1 where: i + 1 < arr.length.i - 1 where: i - 1 >= 0.j where: arr[i] == arr[j] and i != j.Return the minimum number of steps to reach the last index of the array.
Notice that you can not jump outside of the array at any time.
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] <= 108In this approach, we use a Breadth-First Search (BFS) to find the minimum number of jumps required. We start from the first index and explore all possible indices we can jump to. We maintain a queue that helps us process nodes level by level, where each level represents a jump.
We use a map to track all indices with the same value to make efficient jumps to similar value indices. Once an index is processed, we clear it from our map entries to avoid unnecessary recalculation.
This Python solution uses BFS for level-order traversal. Each level in BFS corresponds to a jump. We utilize a deque for efficient pop from the queue and a set to track visited indices to avoid revisiting and redundant processing of indices. The defaultdict helps in storing and accessing indices sharing the same array values efficiently.
Java
Time Complexity: O(n), where n is the length of the array. The worst-case scenario processes every element once.
Space Complexity: O(n), due to additional storage for graph mapping and visited set.
While the first BFS approach involved clearing the graph entry after processing, this approach incorporates an even lazier clearing technique. Instead of clearing as soon as we're done processing, we mark for future removal unless revisiting an edge is necessary. This maintains a balance between redundant work and excessive memory usage.
The C++ solution leverages an unordered_map for organizing jumps and a queue to perform BFS similar to the Java approach. The difference lies in simplifying memory operations with lazy eviction of processed nodes to maintain simpler code and potentially decrease memory overhead.
C#
Time Complexity: O(n), because we explore each node a constant number of times.
Space Complexity: O(n), accommodating the graph structure and tracking nodes via additional containers.
| Approach | Complexity |
|---|---|
| Approach 1: Breadth-First Search (BFS) | Time Complexity: O(n), where n is the length of the array. The worst-case scenario processes every element once. Space Complexity: O(n), due to additional storage for graph mapping and visited set. |
| Approach 2: Optimized BFS with Lazy Clearing | Time Complexity: O(n), because we explore each node a constant number of times. Space Complexity: O(n), accommodating the graph structure and tracking nodes via additional containers. |
Jump Game IV | Live Coding with Explanation | Leetcode - 1345 • Algorithms Made Easy • 6,861 views views
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