How do you solve Distinct Subsequences on LeetCode?

Distinct Subsequences (LeetCode #115) 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: Dynamic programming is essential for optimizing problems involving subsequences.

What is the brute force solution for Distinct Subsequences?

The brute force approach for Distinct Subsequences is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach generates all possible subsequences of string `s` and counts how many of them match string `t`. This method is simple but inefficient because it explores every combination, leading to a high time complexity. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Distinct Subsequences?

Distinct Subsequences uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Two Pointers, Backtracking.

What are the interview tips for Distinct Subsequences?

Explain your thought process clearly when discussing dynamic programming. Practice constructing DP tables with smaller examples before tackling larger problems. Always consider edge cases, such as when one string is empty.

What are common mistakes when solving Distinct Subsequences?

Not initializing the DP table correctly. Confusing the indices when accessing characters in strings.

#115

Distinct Subsequences

Hard

StringDynamic ProgrammingDynamic ProgrammingTwo PointersBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

The optimal solution uses dynamic programming to build a table that counts distinct subsequences efficiently. This approach avoids redundant calculations by storing results of subproblems, leading to a significant reduction in time complexity.

⚙️

Algorithm

4 steps

1Step 1: Create a 2D array `dp` where `dp[i][j]` represents the number of distinct subsequences of `s[0..i-1]` that equals `t[0..j-1]`.
2Step 2: Initialize the first row and column of the `dp` array.
3Step 3: Fill the `dp` array based on whether characters match or not.
4Step 4: Return `dp[s.length()][t.length()]`.

solution.py13 lines

1# Full working Python code
2
3def numDistinct(s, t):
4    dp = [[0] * (len(t) + 1) for _ in range(len(s) + 1)]
5    for i in range(len(s) + 1):
6        dp[i][0] = 1
7    for i in range(1, len(s) + 1):
8        for j in range(1, len(t) + 1):
9            if s[i - 1] == t[j - 1]:
10                dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]
11            else:
12                dp[i][j] = dp[i - 1][j]
13    return dp[len(s)][len(t)]

ℹ

Complexity note: The time complexity is O(n * m) because we fill a 2D array where `n` is the length of `s` and `m` is the length of `t`. Each cell is computed based on previous cells, leading to efficient calculations.

1Dynamic programming is essential for optimizing problems involving subsequences.
2Understanding how to build and utilize a DP table is crucial.

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