What is the key approach for solving the Count of Sub-Multisets With Bounded Sum problem?

The key approach is using dynamic programming with hash maps to track possible subset sums and applying sliding window techniques to count sums within the specified range.

How do you handle large inputs in this problem?

Efficient use of dynamic programming and hash maps, along with optimizations such as frequency counting, helps manage large inputs within time limits.

Why is a hash map useful in this problem?

A hash map stores subset sums efficiently and allows quick lookups to count how many subsets sum to a specific value, reducing redundant calculations.

How does the sliding window technique apply here?

The sliding window technique is used to efficiently count how many subset sums fall within the given range `[l, r]` without repeatedly iterating over all subsets.

What is the time complexity of this problem?

The time complexity is approximately `O(n * sum)`, where `n` is the array length and `sum` is the maximum possible sum, due to the dynamic programming approach.

#2902

Hard

auto_awesome数组·哈希·扫描

LeetCode 题解工作台

和带限制的子多重集合的数目

给你一个下标从 0 开始的非负整数数组 nums 和两个整数 l 和 r 。请你返回 nums 中子多重集合的和在闭区间 [l, r] 之间的子多重集合的数目。由于答案可能很大，请你将答案对 10 9 + 7 取余后返回。子多重集合指的是从数组中选出一些元素构成的无序集合，每个元素 …

数组哈希表动态规划滑动窗口

题目描述

给你一个下标从 0 开始的非负整数数组 nums 和两个整数 l 和 r 。

请你返回 nums 中子多重集合的和在闭区间 [l, r] 之间的 子多重集合的数目 。

由于答案可能很大，请你将答案对 10⁹+ 7 取余后返回。

子多重集合 指的是从数组中选出一些元素构成的无序集合，每个元素 x 出现的次数可以是 0, 1, ..., occ[x] 次，其中 occ[x] 是元素 x 在数组中的出现次数。

注意：

如果两个子多重集合中的元素排序后一模一样，那么它们两个是相同的 子多重集合 。
空集合的和是 0 。

示例 1：

输入：nums = [1,2,2,3], l = 6, r = 6
输出：1
解释：唯一和为 6 的子集合是 {1, 2, 3} 。

示例 2：

输入：nums = [2,1,4,2,7], l = 1, r = 5
输出：7
解释：和在闭区间 [1, 5] 之间的子多重集合为 {1} ，{2} ，{4} ，{2, 2} ，{1, 2} ，{1, 4} 和 {1, 2, 2} 。

示例 3：

输入：nums = [1,2,1,3,5,2], l = 3, r = 5
输出：9
解释：和在闭区间 [3, 5] 之间的子多重集合为 {3} ，{5} ，{1, 2} ，{1, 3} ，{2, 2} ，{2, 3} ，{1, 1, 2} ，{1, 1, 3} 和 {1, 2, 2} 。

提示：

1 <= nums.length <= 2 * 10⁴
0 <= nums[i] <= 2 * 10⁴
nums 的和不超过 2 * 10⁴ 。
0 <= l <= r <= 2 * 10⁴

lightbulb

解题思路

方法一

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class Solution:
    def countSubMultisets(self, nums: List[int], l: int, r: int) -> int:
        kMod = 1_000_000_007
        # dp[i] := # of submultisets of nums with sum i
        dp = [1] + [0] * r
        count = collections.Counter(nums)
        zeros = count.pop(0, 0)

        for num, freq in count.items():
            # stride[i] := dp[i] + dp[i - num] + dp[i - 2 * num] + ...
            stride = dp.copy()
            for i in range(num, r + 1):
                stride[i] += stride[i - num]
            for i in range(r, 0, -1):
                if i >= num * (freq + 1):
                    # dp[i] + dp[i - num] + dp[i - freq * num]
                    dp[i] = stride[i] - stride[i - num * (freq + 1)]
                else:
                    dp[i] = stride[i]

        return (zeros + 1) * sum(dp[l : r + 1]) % kMod

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Look for understanding of dynamic programming and hash map usage in counting subsets.
question_mark
Test for optimization techniques when handling large arrays and sums, such as using frequency counts.
question_mark
Assess the candidate's ability to handle modular arithmetic efficiently in large numbers.

warning

常见陷阱

外企场景

error
Failing to properly count subsets with repeated elements can lead to incorrect results.
error
Inefficient range counting methods may lead to time limit exceeded errors.
error
Overlooking the modulo constraint may result in integer overflow or incorrect final answers.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Modify the problem to count only subsets where all elements are distinct.
arrow_right_alt
Change the sum range to be exclusive, not inclusive.
arrow_right_alt
Introduce constraints on the number of elements in each subset and track those as well.

help

常见问题

外企场景

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