How do you solve Maximum Score Using Exactly K Pairs on LeetCode?

Maximum Score Using Exactly K Pairs (LeetCode #3836) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m * k) time complexity and O(k) space complexity. Key insight: Choosing larger values from both arrays maximizes score.

What is the time complexity of Maximum Score Using Exactly K Pairs?

The optimal solution for Maximum Score Using Exactly K Pairs runs in O(n * m * k) time and uses O(k) space. The time complexity arises from the nested loops iterating through nums1, nums2, and k pairs.

What is the brute force solution for Maximum Score Using Exactly K Pairs?

The brute force approach for Maximum Score Using Exactly K Pairs is Brute Force, with O(n²) time complexity and O(n²) space complexity. Generate all possible pairs of indices from nums1 and nums2, calculate their scores, and select the top k scores. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n * m * k).

What algorithm or data structure is used to solve Maximum Score Using Exactly K Pairs?

Maximum Score Using Exactly K Pairs uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Maximum Score Using Exactly K Pairs?

Explain your thought process clearly. Discuss edge cases and constraints. Practice similar problems to build confidence.

What are common mistakes when solving Maximum Score Using Exactly K Pairs?

Forgetting to maintain index order. Not considering negative values in arrays.

#3836

Maximum Score Using Exactly K Pairs

Hard

ArrayDynamic ProgrammingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m * k)
Space	O(n²)	O(k)

💡

Intuition

Time O(n * m * k)Space O(k)

Use dynamic programming to maintain the maximum score achievable with k pairs by iterating through the arrays and selecting optimal pairs based on previous results.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP array where dp[k] keeps track of the maximum score achievable with k pairs.
2Step 2: Iterate through nums1 and nums2, updating the DP array based on the current element's contribution to the score.
3Step 3: Return dp[k] as the maximum score using exactly k pairs.

solution.py8 lines

1def maxScore(nums1, nums2, k):
2    n, m = len(nums1), len(nums2)
3    dp = [0] * (k + 1)
4    for i in range(n):
5        for j in range(m):
6            for x in range(k, 0, -1):
7                dp[x] = max(dp[x], dp[x - 1] + nums1[i] * nums2[j])
8    return dp[k]

ℹ

Complexity note: The time complexity arises from the nested loops iterating through nums1, nums2, and k pairs.

1Choosing larger values from both arrays maximizes score.
2Maintaining order in indices is crucial.

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