Watch 10 video solutions for Subsequence Sum After Capping Elements, a medium level problem involving Array, Two Pointers, Dynamic Programming. This walkthrough by Sanyam IIT Guwahati has 2,232 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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:
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.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.x = 3, the capped array is [3, 3, 2, 3]. A subsequence [2, 3] sums to 5, so it is possible.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 <= 40001 <= nums[i] <= n1 <= k <= 4000Problem Overview: You are given an array where elements larger than a specified cap are reduced to that cap. After applying this transformation, the task is to evaluate subsequences and compute valid sums based on the capped values. The challenge comes from handling many subsequences efficiently without enumerating all possibilities.
Approach 1: Brute Force Subsequence Enumeration (O(2^n) time, O(n) space)
The most direct strategy generates every possible subsequence using recursion or bitmasking. For each subsequence, apply the capping rule (value = min(value, cap)) and compute the resulting sum. While simple, this requires checking all 2^n subsequences, which quickly becomes infeasible for moderate input sizes. This approach is useful only for validating small test cases or understanding the effect of the capping operation on subsequence sums.
Approach 2: Dynamic Programming on Subsequence Sums (O(n * S) time, O(S) space)
After transforming each element to its capped value, the problem becomes a classic subsequence-sum style dynamic programming task. Maintain a DP structure where dp[s] represents whether or how many subsequences can produce sum s. Iterate through the array and update states in reverse to avoid reusing elements in the same step. This technique leverages patterns from dynamic programming and avoids explicit subsequence generation. It works well when the maximum possible sum S is reasonably bounded.
Approach 3: Sorting + Two Pointers Optimization (O(n log n) time, O(n) space)
An efficient solution sorts the capped values first, enabling structured exploration of valid subsequences. Sorting helps group identical capped values and allows pointer-based scans when evaluating combinations that satisfy sum constraints. Using prefix sums and a two pointers sweep, you can efficiently determine how many elements can participate in subsequences without exceeding target conditions. The preprocessing step uses sorting, after which pointer movement avoids nested enumeration. This reduces the effective complexity while keeping memory usage linear.
Recommended for interviews: Interviewers expect you to move beyond brute force quickly. Demonstrating the exponential enumeration first shows understanding of the subsequence space. The optimized approach—sorting combined with two pointers or DP—shows algorithmic maturity by exploiting order and cumulative sums. Most production solutions use the optimized method because it scales to large arrays while maintaining predictable complexity.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Subsequence Enumeration | O(2^n) | O(n) | Small arrays or conceptual understanding of subsequences |
| Dynamic Programming on Sums | O(n * S) | O(S) | When the maximum possible sum is limited and DP states remain manageable |
| Sorting + Two Pointers Optimization | O(n log n) | O(n) | General case with large arrays where enumeration is infeasible |