LeetCode #2902 — HARD

Count of Sub-Multisets With Bounded Sum

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

Solve on LeetCode
The Problem

Problem Statement

You are given a 0-indexed array nums of non-negative integers, and two integers l and r.

Return the count of sub-multisets within nums where the sum of elements in each subset falls within the inclusive range of [l, r].

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

A sub-multiset is an unordered collection of elements of the array in which a given value x can occur 0, 1, ..., occ[x] times, where occ[x] is the number of occurrences of x in the array.

Note that:

  • Two sub-multisets are the same if sorting both sub-multisets results in identical multisets.
  • The sum of an empty multiset is 0.

Example 1:

Input: nums = [1,2,2,3], l = 6, r = 6
Output: 1
Explanation: The only subset of nums that has a sum of 6 is {1, 2, 3}.

Example 2:

Input: nums = [2,1,4,2,7], l = 1, r = 5
Output: 7
Explanation: The subsets of nums that have a sum within the range [1, 5] are {1}, {2}, {4}, {2, 2}, {1, 2}, {1, 4}, and {1, 2, 2}.

Example 3:

Input: nums = [1,2,1,3,5,2], l = 3, r = 5
Output: 9
Explanation: The subsets of nums that have a sum within the range [3, 5] are {3}, {5}, {1, 2}, {1, 3}, {2, 2}, {2, 3}, {1, 1, 2}, {1, 1, 3}, and {1, 2, 2}.

Constraints:

  • 1 <= nums.length <= 2 * 104
  • 0 <= nums[i] <= 2 * 104
  • Sum of nums does not exceed 2 * 104.
  • 0 <= l <= r <= 2 * 104
Patterns Used
📊Dynamic Programming→🪟Sliding Window→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array nums of non-negative integers, and two integers l and r. Return the count of sub-multisets within nums where the sum of elements in each subset falls within the inclusive range of [l, r]. Since the answer may be large, return it modulo 109 + 7. A sub-multiset is an unordered collection of elements of the array in which a given value x can occur 0, 1, ..., occ[x] times, where occ[x] is the number of occurrences of x in the array. Note that: Two sub-multisets are the same if sorting both sub-multisets results in identical multisets. The sum of an empty multiset is 0.

Baseline thinking

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

Pattern signal: Array · Hash Map · Dynamic Programming · Sliding Window

Example 1

[1,2,2,3]
6
6

Example 2

[2,1,4,2,7]
1
5

Example 3

[1,2,1,3,5,2]
3
5

Related Problems

  • Coin Change (coin-change)
  • Coin Change II (coin-change-ii)
Step 02

Core Insight

What unlocks the optimal approach

  • Since the sum of <code>nums</code>is at most <code>20000</code>, the number of distinct elements of nums is <code>200</code>.
  • Let <code>dp[x]</code> be the number of submultisets of <code>nums</code> with sum <code>x</code>.
  • The answer to the problem is <code>dp[l] + dp[l+1] + … + dp[r]</code>.
  • Use coin change dp to transition between states.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #2902: Count of Sub-Multisets With Bounded Sum
class Solution {
    static final int MOD = 1_000_000_007;
    public int countSubMultisets(List<Integer> nums, int l, int r) {
        Map<Integer, Integer> count = new HashMap<>();
        int total = 0;
        for (int num : nums) {
            total += num;
            if (num <= r) {
                count.merge(num, 1, Integer::sum);
            }
        }
        if (total < l) {
            return 0;
        }
        r = Math.min(r, total);
        int[] dp = new int[r + 1];
        dp[0] = count.getOrDefault(0, 0) + 1;
        count.remove(Integer.valueOf(0));
        int sum = 0;
        for (Map.Entry<Integer, Integer> e : count.entrySet()) {
            int num = e.getKey();
            int c = e.getValue();
            sum = Math.min(sum + c * num, r);
            // prefix part
            // dp[i] = dp[i] + dp[i - num] + ... + dp[i - c*num] + dp[i-(c+1)*num] + ... + dp[i %
            // num]
            for (int i = num; i <= sum; i++) {
                dp[i] = (dp[i] + dp[i - num]) % MOD;
            }
            int temp = (c + 1) * num;
            // correction part
            // subtract dp[i - (freq + 1) * num] to the end part.
            // leves dp[i] = dp[i] + dp[i-num] +...+ dp[i - c*num];
            for (int i = sum; i >= temp; i--) {
                dp[i] = (dp[i] - dp[i - temp] + MOD) % MOD;
            }
        }
        int ans = 0;
        for (int i = l; i <= r; i++) {
            ans += dp[i];
            ans %= MOD;
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

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

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.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.

Related Problems
#1477Find Two Non-overlapping Sub-arrays Each With Target Summedium→#3505Minimum Operations to Make Elements Within K Subarrays Equalhard→#139Word Breakmedium→#140Word Break IIhard→#219Contains Duplicate IIeasy→