LeetCode #1987 — HARD

Number of Unique Good Subsequences

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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").

Find the number of unique good subsequences of binary.

For example, if binary = "001", then all the good subsequences are ["0", "0", "1"], so the unique good subsequences are "0" and "1". Note that subsequences "00", "01", and "001" are not good because they have leading zeros.

Return the number of unique good subsequences of binary. Since the answer may be very large, return it modulo 10⁹ + 7.

A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

Example 1:

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

Example 2:

Input: binary = "11"
Output: 2
Explanation: The good subsequences of binary are ["1", "1", "11"].
The unique good subsequences are "1" and "11".

Example 3:

Input: binary = "101"
Output: 5
Explanation: The good subsequences of binary are ["1", "0", "1", "10", "11", "101"]. 
The unique good subsequences are "0", "1", "10", "11", and "101".

Constraints:

1 <= binary.length <= 10⁵
binary consists of only '0's and '1's.

Step 01

Brute Force Baseline

Problem summary: 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"). Find the number of unique good subsequences of binary. For example, if binary = "001", then all the good subsequences are ["0", "0", "1"], so the unique good subsequences are "0" and "1". Note that subsequences "00", "01", and "001" are not good because they have leading zeros. Return the number of unique good subsequences of binary. Since the answer may be very large, return it modulo 109 + 7. A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming

Example 1

"001"

Example 2

"11"

Example 3

"101"

Core Insight

What unlocks the optimal approach

The number of unique good subsequences is equal to the number of unique decimal values there are for all possible subsequences.
Find the answer at each index based on the previous indexes' answers.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1987: Number of Unique Good Subsequences
class Solution {
    public int numberOfUniqueGoodSubsequences(String binary) {
        final int mod = (int) 1e9 + 7;
        int f = 0, g = 0;
        int ans = 0;
        for (int i = 0; i < binary.length(); ++i) {
            if (binary.charAt(i) == '0') {
                g = (g + f) % mod;
                ans = 1;
            } else {
                f = (f + g + 1) % mod;
            }
        }
        ans = (ans + f + g) % mod;
        return ans;
    }
}

// Accepted solution for LeetCode #1987: Number of Unique Good Subsequences
func numberOfUniqueGoodSubsequences(binary string) (ans int) {
	const mod int = 1e9 + 7
	f, g := 0, 0
	for _, c := range binary {
		if c == '0' {
			g = (g + f) % mod
			ans = 1
		} else {
			f = (f + g + 1) % mod
		}
	}
	ans = (ans + f + g) % mod
	return
}

# Accepted solution for LeetCode #1987: Number of Unique Good Subsequences
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

// Accepted solution for LeetCode #1987: Number of Unique Good Subsequences
/**
 * [1987] Number of Unique Good Subsequences
 *
 * 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").
 * Find the number of unique good subsequences of binary.
 *
 * 	For example, if binary = "001", then all the good subsequences are ["0", "0", "1"], so the unique good subsequences are "0" and "1". Note that subsequences "00", "01", and "001" are not good because they have leading zeros.
 *
 * Return the number of unique good subsequences of binary. Since the answer may be very large, return it modulo 10^9 + 7.
 * A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.
 *  
 * Example 1:
 *
 * Input: binary = "001"
 * Output: 2
 * Explanation: The good subsequences of binary are ["0", "0", "1"].
 * The unique good subsequences are "0" and "1".
 *
 * Example 2:
 *
 * Input: binary = "11"
 * Output: 2
 * Explanation: The good subsequences of binary are ["1", "1", "11"].
 * The unique good subsequences are "1" and "11".
 * Example 3:
 *
 * Input: binary = "101"
 * Output: 5
 * Explanation: The good subsequences of binary are ["1", "0", "1", "10", "11", "101"].
 * The unique good subsequences are "0", "1", "10", "11", and "101".
 *
 *  
 * Constraints:
 *
 * 	1 <= binary.length <= 10^5
 * 	binary consists of only '0's and '1's.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/number-of-unique-good-subsequences/
// discuss: https://leetcode.com/problems/number-of-unique-good-subsequences/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn number_of_unique_good_subsequences(binary: String) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_1987_example_1() {
        let binary = "001".to_string();

        let result = 2;

        assert_eq!(Solution::number_of_unique_good_subsequences(binary), result);
    }

    #[test]
    #[ignore]
    fn test_1987_example_2() {
        let binary = "11".to_string();

        let result = 2;

        assert_eq!(Solution::number_of_unique_good_subsequences(binary), result);
    }

    #[test]
    #[ignore]
    fn test_1987_example_3() {
        let binary = "101".to_string();

        let result = 5;

        assert_eq!(Solution::number_of_unique_good_subsequences(binary), result);
    }
}

// Accepted solution for LeetCode #1987: Number of Unique Good Subsequences
function numberOfUniqueGoodSubsequences(binary: string): number {
    let [f, g] = [0, 0];
    let ans = 0;
    const mod = 1e9 + 7;
    for (const c of binary) {
        if (c === '0') {
            g = (g + f) % mod;
            ans = 1;
        } else {
            f = (f + g + 1) % mod;
        }
    }
    ans = (ans + f + g) % mod;
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.