How do you solve Get the Maximum Score on LeetCode?

Get the Maximum Score (LeetCode #1537) 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: Common elements serve as transition points between the two arrays.

What is the brute force solution for Get the Maximum Score?

The brute force approach for Get the Maximum Score is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible paths by traversing both arrays and checking for common elements. This method is straightforward but inefficient, as it checks every combination of paths. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Get the Maximum Score?

Get the Maximum Score uses the following concepts: Array, Two Pointers, Dynamic Programming, Greedy. The recommended patterns to study are: Two Pointers, Dynamic Programming.

What are the interview tips for Get the Maximum Score?

Always consider edge cases, such as when there are no common elements. Think about the properties of the input (e.g., sorted arrays) to optimize your approach. Explain your thought process clearly during the interview to showcase your understanding.

What are common mistakes when solving Get the Maximum Score?

Failing to account for the modulo operation when returning the result. Not handling the case where one array is fully traversed before the other.

#1537

Get the Maximum Score

Hard

ArrayTwo PointersDynamic ProgrammingGreedyTwo PointersDynamic 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 two-pointer technique to traverse both arrays simultaneously, leveraging the fact that they are sorted. This allows us to efficiently calculate the maximum score by only considering segments between common elements.

⚙️

Algorithm

5 steps

1Step 1: Initialize two pointers for nums1 and nums2, and variables to keep track of the current sum and maximum score.
2Step 2: Traverse both arrays using the pointers, adding values to the current sum until a common element is found.
3Step 3: When a common element is encountered, compare the current sum with the maximum score and update accordingly.
4Step 4: Move the pointer of the array that has the smaller current element to continue traversing.
5Step 5: After reaching the end of both arrays, return the maximum score.

solution.py30 lines

1# Full working Python code
2
3def maxScore(nums1, nums2):
4    i, j = 0, 0
5    sum1, sum2 = 0, 0
6    max_score = 0
7    MOD = 10**9 + 7
8
9    while i < len(nums1) and j < len(nums2):
10        if nums1[i] < nums2[j]:
11            sum1 += nums1[i]
12            i += 1
13        elif nums1[i] > nums2[j]:
14            sum2 += nums2[j]
15            j += 1
16        else:
17            max_score = max(max_score, sum1 + sum2 + nums1[i])
18            sum1 = sum2 = 0
19            i += 1
20            j += 1
21
22    while i < len(nums1):
23        sum1 += nums1[i]
24        i += 1
25    while j < len(nums2):
26        sum2 += nums2[j]
27        j += 1
28
29    max_score = max(max_score, sum1 + sum2)
30    return max_score % MOD

ℹ

Complexity note: The time complexity is O(n) because we traverse each array only once using two pointers, leading to a linear number of operations.

1Common elements serve as transition points between the two arrays.
2The problem can be efficiently solved using a two-pointer technique due to the sorted nature of the arrays.

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