LeetCode #2597 — MEDIUM

The Number of Beautiful Subsets

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given an array nums of positive integers and a positive integer k.

A subset of nums is beautiful if it does not contain two integers with an absolute difference equal to k.

Return the number of non-empty beautiful subsets of the array nums.

A subset of nums is an array that can be obtained by deleting some (possibly none) elements from nums. Two subsets are different if and only if the chosen indices to delete are different.

Example 1:

Input: nums = [2,4,6], k = 2
Output: 4
Explanation: The beautiful subsets of the array nums are: [2], [4], [6], [2, 6].
It can be proved that there are only 4 beautiful subsets in the array [2,4,6].

Example 2:

Input: nums = [1], k = 1
Output: 1
Explanation: The beautiful subset of the array nums is [1].
It can be proved that there is only 1 beautiful subset in the array [1].

Constraints:

1 <= nums.length <= 18
1 <= nums[i], k <= 1000

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: You are given an array nums of positive integers and a positive integer k. A subset of nums is beautiful if it does not contain two integers with an absolute difference equal to k. Return the number of non-empty beautiful subsets of the array nums. A subset of nums is an array that can be obtained by deleting some (possibly none) elements from nums. Two subsets are different if and only if the chosen indices to delete are different.

Baseline thinking

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

Pattern signal: Array · Hash Map · Math · Dynamic Programming · Backtracking

Example 1

[2,4,6]
2

Example 2

[1]
1

Related Problems

Construct the Lexicographically Largest Valid Sequence (construct-the-lexicographically-largest-valid-sequence)

Step 02

Core Insight

What unlocks the optimal approach

Sort the array nums and create another array cnt of size nums[i].
Use backtracking to generate all the beautiful subsets. If cnt[nums[i] - k] is positive, then it is impossible to add nums[i] in the subset, and we just move to the next index. Otherwise, it is also possible to add nums[i] in the subset, in this case, increase cnt[nums[i]], and move to the next index.
Bonus: Can you solve the problem in O(n log n)?

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

Upper-end input sizes

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 #2597: The Number of Beautiful Subsets
class Solution {
    private int[] nums;
    private int[] cnt = new int[1010];
    private int ans = -1;
    private int k;

    public int beautifulSubsets(int[] nums, int k) {
        this.k = k;
        this.nums = nums;
        dfs(0);
        return ans;
    }

    private void dfs(int i) {
        if (i >= nums.length) {
            ++ans;
            return;
        }
        dfs(i + 1);
        boolean ok1 = nums[i] + k >= cnt.length || cnt[nums[i] + k] == 0;
        boolean ok2 = nums[i] - k < 0 || cnt[nums[i] - k] == 0;
        if (ok1 && ok2) {
            ++cnt[nums[i]];
            dfs(i + 1);
            --cnt[nums[i]];
        }
    }
}

// Accepted solution for LeetCode #2597: The Number of Beautiful Subsets
func beautifulSubsets(nums []int, k int) int {
	ans := -1
	n := len(nums)
	cnt := [1010]int{}
	var dfs func(int)
	dfs = func(i int) {
		if i >= n {
			ans++
			return
		}
		dfs(i + 1)
		ok1 := nums[i]+k >= len(cnt) || cnt[nums[i]+k] == 0
		ok2 := nums[i]-k < 0 || cnt[nums[i]-k] == 0
		if ok1 && ok2 {
			cnt[nums[i]]++
			dfs(i + 1)
			cnt[nums[i]]--
		}
	}
	dfs(0)
	return ans
}

# Accepted solution for LeetCode #2597: The Number of Beautiful Subsets
class Solution:
    def beautifulSubsets(self, nums: List[int], k: int) -> int:
        def dfs(i: int) -> None:
            nonlocal ans
            if i >= len(nums):
                ans += 1
                return
            dfs(i + 1)
            if cnt[nums[i] + k] == 0 and cnt[nums[i] - k] == 0:
                cnt[nums[i]] += 1
                dfs(i + 1)
                cnt[nums[i]] -= 1

        ans = -1
        cnt = Counter()
        dfs(0)
        return ans

// Accepted solution for LeetCode #2597: The Number of Beautiful Subsets
/**
 * [2597] The Number of Beautiful Subsets
 */
pub struct Solution {}

// submission codes start here
use std::collections::{BTreeMap, HashMap};

impl Solution {
    pub fn beautiful_subsets(nums: Vec<i32>, k: i32) -> i32 {
        let mut map = HashMap::new();

        for i in nums {
            let value = i % k;
            let entry = map.entry(value).or_insert(BTreeMap::new());
            let group_entry = entry.entry(i).or_insert(0);
            *group_entry += 1;
        }

        let mut result = 1;

        for group in map.values() {
            let n = group.len();
            // (x, y) x表示不选择第i个数
            // y表示选择第i个数
            let mut dp = vec![(0, 0); n];
            // 懒得用迭代器移来移去
            // 直接复制到数组开干
            let array: Vec<(i32, i32)> = group.iter().map(|(x, y)| (*x, *y)).collect();

            for (i, &(key, count)) in array.iter().enumerate() {
                if i == 0 {
                    dp[0].0 = 1;
                    dp[0].1 = (1 << count) - 1;
                } else {
                    dp[i].0 = dp[i - 1].0 + dp[i - 1].1;

                    if key - array[i - 1].0 == k {
                        dp[i].1 = dp[i - 1].0 * ((1 << count) - 1);
                    } else {
                        dp[i].1 = (dp[i - 1].0 + dp[i - 1].1) * ((1 << count) - 1);
                    }
                }
            }

            result *= dp[n - 1].0 + dp[n - 1].1;
        }

        result - 1
    }
}

// submission codes end

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

    #[test]
    fn test_2597() {
        assert_eq!(4, Solution::beautiful_subsets(vec![2, 4, 6], 2));
        assert_eq!(1, Solution::beautiful_subsets(vec![1], 1));
    }
}

// Accepted solution for LeetCode #2597: The Number of Beautiful Subsets
function beautifulSubsets(nums: number[], k: number): number {
    let ans: number = -1;
    const cnt: number[] = new Array(1010).fill(0);
    const n: number = nums.length;
    const dfs = (i: number) => {
        if (i >= n) {
            ++ans;
            return;
        }
        dfs(i + 1);
        const ok1: boolean = nums[i] + k >= 1010 || cnt[nums[i] + k] === 0;
        const ok2: boolean = nums[i] - k < 0 || cnt[nums[i] - k] === 0;
        if (ok1 && ok2) {
            ++cnt[nums[i]];
            dfs(i + 1);
            --cnt[nums[i]];
        }
    };
    dfs(0);
    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(2^n)

Space

O(n)

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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.