How do you solve Reach End of Array With Max Score on LeetCode?

Reach End of Array With Max Score (LeetCode #3282) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: From each index, the optimal jump is to the nearest index with a higher value.

What is the time complexity of Reach End of Array With Max Score?

The optimal solution for Reach End of Array With Max Score runs in O(n) time and uses O(n) space. This complexity is due to the single pass through the array with a stack that may store up to n elements in the worst case.

What is the brute force solution for Reach End of Array With Max Score?

The brute force approach for Reach End of Array With Max Score is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach explores all possible jumps from each index to calculate the maximum score. It systematically checks every possible jump to find the best path to the end. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Reach End of Array With Max Score?

Reach End of Array With Max Score uses the following concepts: Array, Greedy. The recommended patterns to study are: Greedy, Stack.

What are the interview tips for Reach End of Array With Max Score?

Clearly explain your thought process and how you derive the optimal solution. Be prepared to discuss the trade-offs between brute force and optimal approaches. Practice writing clean and efficient code under time constraints.

What are common mistakes when solving Reach End of Array With Max Score?

Failing to consider all possible jumps and missing out on higher scores. Not using a stack or greedy approach, leading to unnecessary computations.

#3282

Reach End of Array With Max Score

Medium

ArrayGreedyGreedyStack

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a greedy approach to jump to the next index that gives the highest score based on the hint provided. This reduces unnecessary calculations and focuses on the most promising jumps.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to keep track of the maximum score.
2Step 2: Use a loop to iterate through the array, maintaining a stack to store indices of potential jumps.
3Step 3: For each index, pop from the stack while the current number is greater than the number at the index stored in the stack, calculating scores as you go.
4Step 4: Push the current index onto the stack and continue until reaching the last index.

solution.py15 lines

1# Full working Python code
2
3def maxScore(nums):
4    n = len(nums)
5    max_score = 0
6    stack = []
7    for i in range(n):
8        while stack and nums[stack[-1]] < nums[i]:
9            j = stack.pop()
10            max_score = max(max_score, (i - j) * nums[j])
11        stack.append(i)
12    return max_score
13
14# Example usage:
15print(maxScore([1, 3, 1, 5]))  # Output: 7

ℹ

Complexity note: This complexity is due to the single pass through the array with a stack that may store up to n elements in the worst case.

1From each index, the optimal jump is to the nearest index with a higher value.
2Using a stack helps efficiently track potential jump indices and calculate scores.

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