```
public class JumpGameIV {
public int jump(int[] nums) {
if (nums == null || nums.length == 0) return 0;
int jumps = 0;
int currentEnd = 0;
int farthest = 0;
for (int i = 0; i < nums.length - 1; i++) {
// Update the farthest point that can be reached
farthest = Math.max(farthest, i + nums[i]);
// If we have reached the end of the current jump range
if (i == currentEnd) {
jumps++;
currentEnd = farthest;
}
}
return jumps;
}
}
```

### Explanation of Key Parts
- **Initialization**: The variables `jumps`, `currentEnd`, and `farthest` are initialized to keep track of the number of jumps, the end of the current jump range, and the farthest point that can be reached, respectively.
- **Updating Farthest Point**: For each index, the farthest point is updated using `Math.max(farthest, i + nums[i])`.
- **Incrementing Jumps**: When the current index reaches `currentEnd`, the jump count is incremented, and `currentEnd` is updated to `farthest`.
## Complexity Analysis
- **Time Complexity**: O(n), where n is the length of the array. The array is traversed once.
- **Space Complexity**: O(1), as no additional space is used except for a few variables.
### Comparison
The optimized solution significantly improves over the naive approach by reducing the time complexity from exponential to linear.
## Edge Cases
- **Single Element Array**: The output should be 0 as no jumps are needed.
- **Unreachable Last Index**: If the last index is unreachable, the algorithm should handle it gracefully.
### Example
- **Input**: [0]
**Output**: 0
- **Input**: [1, 2, 3]
**Output**: 2
## Testing
To test the solution comprehensively:
- **Simple Cases**: Arrays with a few elements.
- **Complex Cases**: Larger arrays with varying jump lengths.
- **Edge Cases**: Single element arrays, arrays with unreachable last index.
### Test Cases
java
public static void main(String[] args) {
JumpGameIV solution = new JumpGameIV();
// Test case 1
int[] nums1 = {2, 3, 1, 1, 4};
System.out.println(solution.jump(nums1)); // Output: 2
// Test case 2
int[] nums2 = {2, 3, 0, 1, 4};
System.out.println(solution.jump(nums2)); // Output: 2
// Edge case
int[] nums3 = {0};
System.out.println(solution.jump(nums3)); // Output: 0
}
## Thinking and Problem-Solving Tips
- **Break Down the Problem**: Understand the problem requirements and constraints.
- **Start with a Naive Solution**: Implement a simple solution to understand the problem better.
- **Optimize**: Look for patterns and optimize the solution step-by-step.
- **Practice**: Solve similar problems to improve problem-solving skills.
## Conclusion
Understanding and solving the Jump Game IV problem helps in mastering dynamic programming and greedy algorithms. Practice and explore further to enhance your problem-solving skills.
## Additional Resources
- [LeetCode Problems](https://leetcode.com/problemset/all/)
- [GeeksforGeeks Tutorials](https://www.geeksforgeeks.org/)
- [Dynamic Programming Patterns](https://www.interviewbit.com/courses/programming/topics/dynamic-programming/)
By following this comprehensive guide, you can effectively solve the Jump Game IV problem and improve your algorithmic skills. Happy coding!