How do you solve Max Dot Product of Two Subsequences on LeetCode?

Max Dot Product of Two Subsequences (LeetCode #1458) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n * m) space complexity. Key insight: Subsequences can be tricky; focus on lengths and positions.

What is the time complexity of Max Dot Product of Two Subsequences?

The optimal solution for Max Dot Product of Two Subsequences runs in O(n * m) time and uses O(n * m) space. The complexity is O(n * m) because we fill a DP table of size n x m, where n and m are the lengths of the two input arrays.

What is the brute force solution for Max Dot Product of Two Subsequences?

The brute force approach for Max Dot Product of Two Subsequences is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsequences of both arrays and calculating their dot products. This is straightforward but inefficient due to the exponential number of subsequences. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Max Dot Product of Two Subsequences?

Max Dot Product of Two Subsequences uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Array.

What are the interview tips for Max Dot Product of Two Subsequences?

Explain your thought process clearly. Discuss edge cases and how your solution handles them. Be prepared to optimize your solution if asked.

What are common mistakes when solving Max Dot Product of Two Subsequences?

Forgetting to handle negative numbers properly. Not initializing the DP table correctly.

#1458

Max Dot Product of Two Subsequences

Hard

ArrayDynamic ProgrammingDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

The optimal solution uses dynamic programming to build a table that captures the maximum dot product for subsequences of different lengths. This avoids the need to generate all subsequences explicitly, leading to a more efficient solution.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP table where DP[i][j] represents the maximum dot product of subsequences ending at nums1[i-1] and nums2[j-1].
2Step 2: Iterate through each element of nums1 and nums2, updating the DP table based on previous values and the current elements.
3Step 3: Return the maximum value found in the DP table.

solution.py7 lines

1def maxDotProduct(nums1, nums2):
2    n, m = len(nums1), len(nums2)
3    dp = [[float('-inf')] * (m + 1) for _ in range(n + 1)]
4    for i in range(1, n + 1):
5        for j in range(1, m + 1):
6            dp[i][j] = max(dp[i-1][j-1] + nums1[i-1] * nums2[j-1], nums1[i-1] * nums2[j-1], dp[i][j])
7    return max(max(row) for row in dp)

ℹ

Complexity note: The complexity is O(n * m) because we fill a DP table of size n x m, where n and m are the lengths of the two input arrays.

1Subsequences can be tricky; focus on lengths and positions.
2Dynamic programming can significantly reduce complexity.

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