3685. Subsequence Sum After Capping Elements

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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

Hints

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.