Jump Game II Leetcode Problem 45 [Python Solution]

Welcome back to another coding session!

Today, we are going to tackle the Jump Game II problem, a classic problem in the world of coding.

This LeetCode problem, specifically number 45, falls under the category of "Greedy" and has been known to be a favorite of interviewers at companies like Google.

In this blog post, we'll explore the problem statement, devise a Python solution, and understand the reasoning behind our approach.

We'll also discuss the time and space complexity of our solution, ensuring it's beginner-friendly and informative.

Problem Overview

Let's dive into the problem statement:

Question:
You are given a 0-indexed array of integers nums of length n.

You are initially positioned at nums[0].

Each element nums[i] represents the maximum length of a forward jump from index i.

In other words, if you are at nums[i], you can jump to any nums[i + j] where:

• 0 <= j <= nums[i]
• i + j < n

Return the minimum number of jumps to reach nums[n - 1].

The test cases are generated in such a way that you can reach nums[n - 1].

Let's consider an example to illustrate the problem:

Example 1:
Input: nums = [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.

Example 2:
Input: nums = [2,3,0,1,4]
Output: 2

Constraints:
– 1 <= nums.length <= 104
– 0 <= nums[i] <= 1000
– It's guaranteed that you can reach nums[n - 1].

Now that we understand the problem, let's take a closer look at the constraints and the approach to solving it.

Understanding Constraints

Before we dive into the solution, it's essential to understand the constraints of the problem.

These constraints play a vital role in determining the efficiency of our approach.

1. The length of the nums array can be as large as 10,000. This means that our solution needs to be efficient, preferably with a time complexity of O(n).
2. The values in the nums array are non-negative and can go up to 1000. We must consider all possible jump lengths.
3. The problem statement guarantees that you can reach the last index.

This assures us that we won't encounter unsolvable instances.

With these constraints in mind, we can proceed to devise a Python solution.

Jump Game II LeetCode Problem Solution

1. Bruteforce Approach

Let's start by exploring a brute-force approach to solving this problem.

While this might not be the most efficient solution, it's crucial to understand it as it forms the foundation for our optimized approach.

The brute-force approach involves exploring all possible paths, counting the number of jumps, and selecting the minimum.

In code, it might look something like this:

def jump(self, nums: List[int]) -> int:
    def jumpFromPosition(pos):
        if pos == len(nums) - 1:
            return 0
        maxJump = nums[pos]
        minJumps = float('inf')
        for step in range(1, maxJump + 1):
            if pos + step < len(nums):
                nextJump = jumpFromPosition(pos + step)
                if nextJump != float('inf'):
                    minJumps = min(minJumps, 1 + nextJump)
        return minJumps
    
    return jumpFromPosition(0)

In this brute-force approach, we define a recursive function jumpFromPosition(pos) that explores all possible jumps from a given position.

It keeps track of the minimum number of jumps required to reach the end of the array.

We iterate through all possible jump lengths from 1 to maxJump, where maxJump is the maximum jump length from the current position.

If we reach the end of the array, we return 0 jumps.

Otherwise, we return the minimum number of jumps by considering all possible next positions.

While this approach is correct, it's highly inefficient and not suitable for large input arrays.

The time complexity of this approach can be exponential, making it impractical for our problem's constraints.

2. An Efficient Approach with Greedy Algorithm

Now, let's explore a more efficient approach using a greedy algorithm.

The key insight here is to find the farthest position we can reach with the minimum number of jumps.

We'll use two pointers, left and right, to keep track of our current window and a farthest variable to track the farthest position we can reach within that window.

Here's the Python code for this efficient approach:

def jump(self, nums: List[int]) -> int:
    l, r = 0, 0  # Initialize left and right pointers.

res = 0  # Initialize the result (number of jumps).

while r < (len(nums) - 1):
        maxJump = 0  # Initialize the maximum jump within the current window.

for i in range(l, r + 1):
            maxJump = max(maxJump, i + nums[i])

        l = r + 1  # Update the left pointer to the rightmost position within the current window.

r = maxJump  # Update the right pointer to the farthest position we can reach within the current window.

res += 1  # Increment the result (number of jumps).

return res

In this efficient approach, we start with two pointers, l and r, both initialized to 0. The res variable keeps track of the number of jumps.

We iterate through the array while the r pointer is less than the last index of the array.

Within each iteration, we find the maximum jump within the current window by iterating through the positions from l to r.

We update the l and r pointers and increment the result based on the farthest position we can reach within the current window.

This approach efficiently calculates the minimum number of jumps needed to reach the end of the array.

It's essentially a simplified breadth-first search on a one-dimensional array, which is why it's so effective.

Time and Space Complexity

Now, let's discuss the time and space complexity of our efficient solution.

Time Complexity:
The time complexity of our approach is O(n), where n is the length of the nums array.

This is because we traverse the array once, and within each iteration, we perform constant time operations.

Space Complexity:
The space complexity is O(1), which means that we use a constant amount of extra space regardless of the size of the input array.

We only use a few variables to keep track of our pointers and results.

Reasoning Behind Our Approach

The reasoning behind our efficient approach lies in identifying the minimum number of jumps needed to reach the end of the array while considering the farthest position we can reach with

the minimum number of jumps.

By using a greedy algorithm and maintaining a window defined by the left and right pointers, we ensure that we explore the most efficient path while avoiding unnecessary jumps.

This approach is highly optimized and suitable for the problem's constraints.

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Conclusion

In this blog post, we tackled the Jump Game II problem, a LeetCode problem that falls under the "Greedy" category.

We started by understanding the problem statement, explored a brute-force approach, and then introduced an efficient approach using a greedy algorithm.

Our efficient approach, with a time complexity of O(n), ensures that we find the minimum number of jumps needed to reach the end of the array.

It's a great example of how a simple and well-thought-out algorithm can lead to an optimal solution.

I hope this explanation and Python solution have been helpful to you.

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If you have any questions, suggestions, or insights, please feel free to comment and share your thoughts.

Happy coding, and see you in the next session!

Link to the original problem on LeetCode.