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Leetcode #1987: Number of Unique Good Subsequences

In this guide, we solve Leetcode #1987 Number of Unique Good Subsequences in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

You are given a binary string binary. A subsequence of binary is considered good if it is not empty and has no leading zeros (with the exception of "0").

Quick Facts

  • Difficulty: Hard
  • Premium: No
  • Tags: String, Dynamic Programming

Intuition

The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.

A carefully chosen DP state captures exactly what we need to build the final answer.

Approach

Define the DP state and recurrence, then compute states in the correct order.

Optionally compress space once the recurrence is clear.

Steps:

  • Choose a DP state definition.
  • Write the recurrence and base cases.
  • Compute states in the correct order.

Example

Input: binary = "001" Output: 2 Explanation: The good subsequences of binary are ["0", "0", "1"]. The unique good subsequences are "0" and "1".

Python Solution

class Solution: def numberOfUniqueGoodSubsequences(self, binary: str) -> int: f = g = 0 ans = 0 mod = 10**9 + 7 for c in binary: if c == "0": g = (g + f) % mod ans = 1 else: f = (f + g + 1) % mod ans = (ans + f + g) % mod return ans

Complexity

The time complexity is O(n)O(n)O(n), where nnn is the length of the string. The space complexity is O(1)O(1)O(1).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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