How do you solve Number of Unique Good Subsequences on LeetCode?

Number of Unique Good Subsequences (LeetCode #1987) 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: Good subsequences must not have leading zeros unless they are '0'.

What is the brute force solution for Number of Unique Good Subsequences?

The brute force approach for Number of Unique Good Subsequences is Brute Force, with O(n²) time complexity and O(2^n) space complexity. The brute force approach generates all possible subsequences of the binary string and checks each one for uniqueness and validity as a good subsequence. This is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Unique Good Subsequences?

Number of Unique Good Subsequences uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Bit Manipulation.

What are the interview tips for Number of Unique Good Subsequences?

Always clarify the definition of 'good subsequence' before starting. Consider edge cases such as strings with all zeros or all ones. Think about how to optimize brute force solutions using dynamic programming.

What are common mistakes when solving Number of Unique Good Subsequences?

Forgetting to handle leading zeros correctly. Not using a set or map to track unique subsequences.

#1987

Number of Unique Good Subsequences

Hard

StringDynamic ProgrammingDynamic ProgrammingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(2^n)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses dynamic programming to count unique good subsequences efficiently by leveraging previously computed results. This avoids the need to generate all subsequences explicitly.

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Algorithm

3 steps

1Step 1: Initialize a DP array where dp[i] represents the number of unique good subsequences ending at index i.
2Step 2: Iterate through the binary string, updating the DP array based on whether the current character is '0' or '1'.
3Step 3: Use a last occurrence map to handle duplicates and ensure uniqueness.

solution.py14 lines

1def uniqueGoodSubseq(binary):
2    MOD = 10**9 + 7
3    n = len(binary)
4    dp = [0] * (n + 1)
5    last = {'0': -1, '1': -1}
6    dp[0] = 1
7    for i in range(1, n + 1):
8        dp[i] = (2 * dp[i - 1]) % MOD
9        if binary[i - 1] == '0':
10            dp[i] = (dp[i] - (dp[last['0']] if last['0'] != -1 else 0)) % MOD
11        else:
12            dp[i] = (dp[i] - (dp[last['1']] if last['1'] != -1 else 0)) % MOD
13        last[binary[i - 1]] = i - 1
14    return (dp[n] - 1) % MOD

ℹ

Complexity note: The time complexity is O(n) because we iterate through the string once, updating our DP array. The space complexity is O(n) for storing the DP values.

1Good subsequences must not have leading zeros unless they are '0'.
2Dynamic programming can efficiently count subsequences by building on previous results.

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