55. Jump Game

Medium LeetCode

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

How to solve the problem

• Brute Force O(n^2)

class Solution:
    def canJump(self, nums: List[int]) -> bool:
        reachable = {0}
        if len(nums) == 1:
            return True

        for i in range(len(nums)):
            if i in reachable:
                for j in range(1, nums[i] + 1):
                    if i + j >= len(nums) - 1:
                        return True
                    reachable.add(i + j)
        return False

• Greedy O(n)

class Solution:
    def canJump(self, nums: List[int]) -> bool:
        max_reach = 0
        for i in range(len(nums)):
            if max_reach < i:
                return False
            max_reach = max(max_reach, nums[i] + i)
        return True

Complexity

• Brute Force: Time complexity: O(n^2), Space complexity: O(n) - Where n is the length of the array. We check all possible jumps from each position.
• Greedy: Time complexity: O(n), Space complexity: O(1) - We traverse the array once and use constant extra space.