0055 - Jump Game (Medium)

Problem Link

https://leetcode.com/problems/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.

Example 2:

Input: nums = [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximum jump length is 0, which makes it impossible to reach the last index.

Constraints:

• 1 <= nums.length <= 10^4
• 0 <= nums[i] <= 10^5

Approach 1: Greedy

We can solve this backwards by tracking the furthest position we can jump, and moving it backwards if we can reach that position from the current position.

By initializing a $finish$ at $nums.length$ and iterating backwards through the numbers and setting our $finish$ to $i$ if the current number plus $i$ is greater than or equal to our finish, we can find the answer if we can eventually get our $finish$ to reach 0.

Time Complexity: $O(n)$ where $n$ is the length of $nums$. We only need to iterate through each number in the $nums$ array once.

Space Compelxity: $O(1)$. We only need constant space to track our $finish$ line.

Python

Written by @ColeB2

class Solution:
    def canJump(self, nums: List[int]) -> bool:
        # initialize a finish line
        finish = len(nums) - 1
        # loop through numbers backwards
        for i in range(len(nums) - 1, -1, -1):
            # if we can jump past our finish line from current position:
            if nums[i] + i >= finish:
                # move our finish line to current position, as we know
                # if we ever reach this position, we can reach the finish
                finish = i
        # if our finish reaches 0, we know we can successfully reach
        # the finish line, if not, this will return False.
        return finish == 0