How do you solve Jump Game on LeetCode?

Jump Game (LeetCode #55) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The key observation is that if you can reach an index, you can potentially reach further indices based on the jump value at that index.

What is the brute force solution for Jump Game?

The brute force approach for Jump Game is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible jump from each index to see if we can reach the last index. This means recursively exploring all jump possibilities, which can be very slow for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Jump Game?

Jump Game uses the following concepts: Array, Dynamic Programming, Greedy. The recommended patterns to study are: Greedy, Dynamic Programming.

What are the interview tips for Jump Game?

Always clarify the problem requirements and constraints before diving into coding. Think aloud while solving the problem to demonstrate your thought process. Consider edge cases and test your solution with them before finalizing.

What are common mistakes when solving Jump Game?

Failing to check if the current index is reachable before updating maxReach. Not considering edge cases, such as when the array has only one element.

#55

Jump Game

Medium

ArrayDynamic ProgrammingGreedyGreedyDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a greedy strategy to track the furthest index we can reach at any point. If we can reach the last index or beyond, we return true.

⚙️

Algorithm

5 steps

1Step 1: Initialize a variable 'maxReach' to 0, which keeps track of the furthest index we can reach.
2Step 2: Iterate through the array, and for each index, check if it's reachable (i.e., index <= maxReach).
3Step 3: Update 'maxReach' to the maximum of its current value and the index plus the jump length at that index.
4Step 4: If at any point 'maxReach' is greater than or equal to the last index, return true.
5Step 5: If the loop ends and we haven't reached the last index, return false.

solution.py9 lines

1def canJump(nums):
2    maxReach = 0
3    for i in range(len(nums)):
4        if i > maxReach:
5            return False
6        maxReach = max(maxReach, i + nums[i])
7        if maxReach >= len(nums) - 1:
8            return True
9    return False

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once, and the space complexity is O(1) since we use a constant amount of extra space.

1The key observation is that if you can reach an index, you can potentially reach further indices based on the jump value at that index.
2Greedy algorithms often yield optimal solutions for problems involving making a series of choices that lead to a final goal.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.