LeetCode #3351 — HARD

Sum of Good Subsequences

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

The Problem

Problem Statement

You are given an integer array nums. A good subsequence is defined as a subsequence of nums where the absolute difference between any two consecutive elements in the subsequence is exactly 1.

Return the sum of all possible good subsequences of nums.

Since the answer may be very large, return it modulo 10⁹ + 7.

Note that a subsequence of size 1 is considered good by definition.

Example 1:

Input: nums = [1,2,1]

Output: 14

Explanation:

Good subsequences are: [1], [2], [1], [1,2], [2,1], [1,2,1].
The sum of elements in these subsequences is 14.

Example 2:

Input: nums = [3,4,5]

Output: 40

Explanation:

Good subsequences are: [3], [4], [5], [3,4], [4,5], [3,4,5].
The sum of elements in these subsequences is 40.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. A good subsequence is defined as a subsequence of nums where the absolute difference between any two consecutive elements in the subsequence is exactly 1. Return the sum of all possible good subsequences of nums. Since the answer may be very large, return it modulo 109 + 7. Note that a subsequence of size 1 is considered good by definition.

Baseline thinking

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

Pattern signal: Array · Hash Map · Dynamic Programming

Example 1

[1,2,1]

Example 2

[3,4,5]

Step 02

Core Insight

What unlocks the optimal approach

Consider counting how many times each element occurs in all possible good subsequences. This can help you derive the final answer more easily.
Use dynamic programming to track both the count and the sum of subsequences where the last element is <code>nums[i]</code>.

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 #3351: Sum of Good Subsequences
class Solution {
    public int sumOfGoodSubsequences(int[] nums) {
        final int mod = (int) 1e9 + 7;
        int mx = 0;
        for (int x : nums) {
            mx = Math.max(mx, x);
        }
        long[] f = new long[mx + 1];
        long[] g = new long[mx + 1];
        for (int x : nums) {
            f[x] += x;
            g[x] += 1;
            if (x > 0) {
                f[x] = (f[x] + f[x - 1] + g[x - 1] * x % mod) % mod;
                g[x] = (g[x] + g[x - 1]) % mod;
            }
            if (x + 1 <= mx) {
                f[x] = (f[x] + f[x + 1] + g[x + 1] * x % mod) % mod;
                g[x] = (g[x] + g[x + 1]) % mod;
            }
        }
        long ans = 0;
        for (long x : f) {
            ans = (ans + x) % mod;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #3351: Sum of Good Subsequences
func sumOfGoodSubsequences(nums []int) (ans int) {
	mod := int(1e9 + 7)
	mx := slices.Max(nums)

	f := make([]int, mx+1)
	g := make([]int, mx+1)

	for _, x := range nums {
		f[x] += x
		g[x] += 1

		if x > 0 {
			f[x] = (f[x] + f[x-1] + g[x-1]*x%mod) % mod
			g[x] = (g[x] + g[x-1]) % mod
		}

		if x+1 <= mx {
			f[x] = (f[x] + f[x+1] + g[x+1]*x%mod) % mod
			g[x] = (g[x] + g[x+1]) % mod
		}
	}

	for _, x := range f {
		ans = (ans + x) % mod
	}
	return
}

# Accepted solution for LeetCode #3351: Sum of Good Subsequences
class Solution:
    def sumOfGoodSubsequences(self, nums: List[int]) -> int:
        mod = 10**9 + 7
        f = defaultdict(int)
        g = defaultdict(int)
        for x in nums:
            f[x] += x
            g[x] += 1
            f[x] += f[x - 1] + g[x - 1] * x
            g[x] += g[x - 1]
            f[x] += f[x + 1] + g[x + 1] * x
            g[x] += g[x + 1]
        return sum(f.values()) % mod

// Accepted solution for LeetCode #3351: Sum of Good Subsequences
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3351: Sum of Good Subsequences
// class Solution {
//     public int sumOfGoodSubsequences(int[] nums) {
//         final int mod = (int) 1e9 + 7;
//         int mx = 0;
//         for (int x : nums) {
//             mx = Math.max(mx, x);
//         }
//         long[] f = new long[mx + 1];
//         long[] g = new long[mx + 1];
//         for (int x : nums) {
//             f[x] += x;
//             g[x] += 1;
//             if (x > 0) {
//                 f[x] = (f[x] + f[x - 1] + g[x - 1] * x % mod) % mod;
//                 g[x] = (g[x] + g[x - 1]) % mod;
//             }
//             if (x + 1 <= mx) {
//                 f[x] = (f[x] + f[x + 1] + g[x + 1] * x % mod) % mod;
//                 g[x] = (g[x] + g[x + 1]) % mod;
//             }
//         }
//         long ans = 0;
//         for (long x : f) {
//             ans = (ans + x) % mod;
//         }
//         return (int) ans;
//     }
// }

// Accepted solution for LeetCode #3351: Sum of Good Subsequences
function sumOfGoodSubsequences(nums: number[]): number {
    const mod = 10 ** 9 + 7;
    const mx = Math.max(...nums);
    const f: number[] = Array(mx + 1).fill(0);
    const g: number[] = Array(mx + 1).fill(0);
    for (const x of nums) {
        f[x] += x;
        g[x] += 1;
        if (x > 0) {
            f[x] = (f[x] + f[x - 1] + ((g[x - 1] * x) % mod)) % mod;
            g[x] = (g[x] + g[x - 1]) % mod;
        }
        if (x + 1 <= mx) {
            f[x] = (f[x] + f[x + 1] + ((g[x + 1] * x) % mod)) % mod;
            g[x] = (g[x] + g[x + 1]) % mod;
        }
    }
    return f.reduce((acc, cur) => (acc + cur) % mod, 0);
}

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.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.