Given an array of non-negative integers, you are initially positioned at the first index of the array.
Each element in the array represents your maximum jump length at that position.
Your goal is to reach the last index in the minimum number of jumps.
Example:
Input: [2,3,1,1,4] Output: 2 Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.
Your algorithm should run in O(n^2) time and use O(n) extra space.
The core challenge of this problem is to determine the minimum number of jumps needed to reach the end of the array. Each element in the array specifies the maximum number of steps you can jump forward from that position. The goal is to find the optimal path with the fewest jumps.
This problem is significant in scenarios where you need to find the shortest path or minimum steps in a constrained environment, such as game development, robotics, and network routing.
Potential pitfalls include misunderstanding the jump lengths and not considering all possible paths, which can lead to incorrect solutions.
To solve this problem, we can start with a naive approach and then optimize it:
The naive approach involves exploring all possible paths and keeping track of the minimum jumps required. This can be done using a recursive or iterative method, but it is not optimal due to its high time complexity.
We can optimize the solution using a greedy algorithm. The idea is to keep track of the farthest point that can be reached and the end of the current jump range. We increment the jump count each time we reach the end of the current range.
Here is a step-by-step breakdown of the optimized algorithm:
def jump(nums):
# Initialize variables
jumps = 0
current_end = 0
farthest = 0
# Iterate through the array
for i in range(len(nums) - 1):
# Update the farthest point that can be reached
farthest = max(farthest, i + nums[i])
# If we reach the end of the current jump range
if i == current_end:
# Increment the jump count
jumps += 1
# Update the end of the current jump range
current_end = farthest
return jumps
# Example usage
print(jump([2, 3, 1, 1, 4])) # Output: 2
The time complexity of this approach is O(n) because we iterate through the array once. The space complexity is O(1) as we use a constant amount of extra space.
Consider the following edge cases:
To test the solution comprehensively, consider the following test cases:
assert jump([2, 3, 1, 1, 4]) == 2
assert jump([1, 1, 1, 1, 1]) == 4
assert jump([0]) == 0
assert jump([10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 0]) == 1
When approaching such problems, consider the following tips:
In this blog post, we discussed the Jump Game problem, explored a naive approach, and then optimized it using a greedy algorithm. We provided a detailed explanation of the algorithm, code implementation, complexity analysis, and testing. Understanding and solving such problems is crucial for improving problem-solving skills and preparing for technical interviews.
For further reading and practice, consider the following resources: