Hello, Welcome to my July Challenge of Problem Solving and finding solutions to Leetcode's Data Structures and Algorithm questions. Here, I choose a question daily and try to solve it, and finally pen down my process of flow in solving the question.

I hope you find it helpful. Today I will be solving Leetcode's question 55 - Jump Game

Problem Statement

You are given an integer array nums. You are initially positioned at the array's first index, and each element in the array represents your maximum jump length at that position.

Return true if you can reach the last index, or false otherwise.

Example 1:

Input: nums = [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.

Lets Dive Into Solving it

The above question is rated as Medium. There are two approaches to solving the problem, either through Dynamic Programming or Greedy Algorithm.

Greedy Algorithm

• The greedy algorithm approaches the problem by making the locally optimal choice at each step without considering the overall structure of the problem.

• In the case of the "Jump Game," the greedy algorithm iterates through the array in reverse order, updating the goal variable whenever it finds a position that can reach the current goal.

• The greedy algorithm has a time complexity of O(n), where n is the length of the input array. It only requires a single pass through the array.

• The space complexity of the greedy algorithm is O(1) since it only uses a constant amount of additional space.

Dynamic Programming Algorithm:

• The dynamic programming algorithm approaches the problem by breaking it down into smaller subproblems and storing the solutions to those subproblems to avoid redundant calculations.

• In the case of the "Jump Game," the dynamic programming algorithm can be implemented using a bottom-up approach. We can define a dynamic programming array to store the information about whether each position is reachable from the start.

The greedy algorithm provides an efficient and simple solution to the "Jump Game" problem, with a time complexity of O(n) and constant space complexity. On the other hand, the dynamic programming algorithm can solve the problem as well, but it has a higher time complexity of O(n^2) and requires additional space to store the dynamic programming array. The choice between the two approaches depends on the specific requirements of the problem and the trade-off between time and space complexity. In most cases, the greedy algorithm is preferred for its simplicity and efficiency.

What the Question Means - (Analogy)

The question is asking whether it is possible to jump from the first position to the last position in an array of numbers. Each number in the array represents the maximum distance you can jump from that position.

Imagine you are at the bottom of a staircase, and each step represents an element in the given array. The value of each step represents the maximum number of stairs you can climb in a single jump.

The goal is to reach the top of the staircase, which corresponds to the last element in the array. The question is asking if it is possible to reach the top of the staircase using the given set of maximum jump lengths.

To solve this, you need to determine if it is possible to climb the stairs, starting from the first step and using the maximum number of stairs you can jump at each step. The question wants you to determine if you can successfully reach the top of the staircase. If it is possible, you should return "true." Otherwise, if it's not possible to reach the top step, you should return "false."

The Code

class Solution:
    def canJump(self, nums: List[int]) -> bool:
        goal = len(nums) - 1

        for i in range(len(nums) -1, -1, -1):
            if i + nums[i] >= goal:
                goal = i

        return True if goal == 0 else False

How the Code Works

• The code defines a class named Solution with a method called canJump. This method takes in a parameter called "nums" which in essence is a list of intergers and returns a boolean indicating if its possible to reach the last index or not

• "goal" is initialized with the value of "len(nums)-1" which represents the last index.

• The 'for" loop iterates over the array but in reverse order starting from "len(nums)-1" and moving towards the last index. For every iteration, the code checks if the current position "i" plus the value of "nums[i]" is greater or equal to "goal". By iterating through the array in reverse order, we start from the last index and work backward, updating 'goal' whenever we find a position that we can reach with the current 'goal'

• Finally, we return True if the "goal" is equal to 0 indicating that we can reach the last index starting from the first index. If goal is not 0, it means we couldn't reach the last index, so the code returns False