Is Subsequence Sum After Capping Elements easy or hard?

The problem is rated Medium because the individual ideas—capping values, subsequences, sorting, and DP—are straightforward, but combining them efficiently requires recognizing patterns that avoid exponential enumeration.

Subsequence Sum After Capping Elements Python/Java solution

Most Python or Java solutions first transform the array with value = min(value, cap), then sort the array and apply a DP or two pointer technique to compute valid subsequence sums. The overall complexity is typically O(n log n) time with linear auxiliary space.

How to solve Subsequence Sum After Capping Elements in O(n)?

Pure O(n) solutions are possible after preprocessing if the array is already sorted. Once values are capped and ordered, prefix sums and a two pointer sweep can evaluate valid subsequences in a single linear pass.

What is the best approach for Subsequence Sum After Capping Elements?

The most practical solution sorts the capped array and applies a two pointers or prefix-sum based strategy to evaluate subsequence combinations efficiently. This avoids enumerating all subsequences and typically runs in O(n log n) time due to the sorting step with O(n) extra space.

Is Subsequence Sum After Capping Elements asked at Google/Amazon/Meta?

Problems combining subsequences, capping constraints, and optimized scanning patterns are common in interviews at companies like Google, Amazon, and Meta. They test familiarity with dynamic programming, sorting strategies, and pointer-based optimizations.

What data structure is used in Subsequence Sum After Capping Elements?

Common implementations rely on arrays with prefix sums, dynamic programming tables for subset sums, and sorting utilities. Two pointers are often used after sorting to efficiently evaluate subsequence combinations.

What is the time complexity of Subsequence Sum After Capping Elements?

The optimal implementation usually runs in O(n log n) time because the array is sorted before applying a linear scan or two pointer technique. Brute force enumeration takes O(2^n) time, which is impractical for large inputs.

Description Hints 4 Video 23:00

3685. Subsequence Sum After Capping Elements

Medium25.2% acceptance1 company

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

An array capped by value x is obtained by replacing every element nums[i] with min(nums[i], x).

For each integer x from 1 to n, determine whether it is possible to choose a subsequence from the array capped by x such that the sum of the chosen elements is exactly k.

Return a 0-indexed boolean array answer of size n, where answer[i] is true if it is possible when using x = i + 1, and false otherwise.

Example 1:

Input: nums = [4,3,2,4], k = 5

Output: [false,false,true,true]

Explanation:

For x = 1, the capped array is [1, 1, 1, 1]. Possible sums are 1, 2, 3, 4, so it is impossible to form a sum of 5.
For x = 2, the capped array is [2, 2, 2, 2]. Possible sums are 2, 4, 6, 8, so it is impossible to form a sum of 5.
For x = 3, the capped array is [3, 3, 2, 3]. A subsequence [2, 3] sums to 5, so it is possible.
For x = 4, the capped array is [4, 3, 2, 4]. A subsequence [3, 2] sums to 5, so it is possible.

Example 2:

Input: nums = [1,2,3,4,5], k = 3

Output: [true,true,true,true,true]

Explanation:

For every value of x, it is always possible to select a subsequence from the capped array that sums exactly to 3.

Constraints:

1 <= n == nums.length <= 4000
1 <= nums[i] <= n
1 <= k <= 4000

Asked at 1 company

Google

Topics

Hints open one at a time

Hint 1

Sort the array nums in descending order.

Hint 2

Build a knapsack DP table dp[idx][sum] where dp[idx][sum] == true if and only if there exists a subsequence of nums[idx] through the end that sums exactly to sum.

Hint 3

Observe that capping all values above x to x in a sorted array is equivalent to taking the first t elements (those originally > x) and treating each as x. Precompute which multiples of x up to t * x are selectable.

Hint 4

For each cap value x from 1 to n, let t be the number of elements in nums originally greater than x. Then iterate over all sums s with dp[t][s] == true and all counts m from 0 to t; if there exists an m such that s + m * x == k, set answer[x-1] to true. Otherwise, it remains false.

LanguageLoading editor…

1class Solution:
2    def subsequenceSumAfterCapping(self, nums: List[int], k: int) -> List[bool]:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[4,3,2,4]
5

Input

[1,2,3,4,5]
3

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