Maximum Subsequence Value — Problem Statement & Solution Guide
Problem Description
Given an array of integers nums and two integers k and m, determine the maximum possible sum of a subsequence of exactly k elements. You are permitted to remove at most m elements from the original array before selecting the subsequence. The selected subsequence must maintain the relative order of the elements as they appear in the original array, but the elements do not need to be contiguous. The goal is to maximize the sum of the chosen k elements by strategically removing up to m elements that obstruct the selection of higher-value elements or disrupt the optimal ordering.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Maximum Subsequence Value"
WHY DOES IT MATTER?
Sliding‑window with a dynamic top‑k structure is a classic pattern for problems that ask for optimal aggregates over a bounded‑size subarray. It transforms a potentially quadratic enumeration into a linear pass while preserving the ability to query complex statistics (like sum of k largest) on the fly.
OPTIMIZATION CHALLENGE
The key insight is that the deletions budget m only limits the window length, not which specific elements are removed. By fixing the window size to k + m and focusing on the k largest values inside it, the problem collapses to a well‑studied dynamic‑order‑statistics task that can be solved with two balanced multisets (or a heap‑pair) in logarithmic time per slide.
REAL-WORLD CONNECTION
Think of a streaming video buffer that can hold at most k + m frames. To maximize viewer quality you keep the k highest‑resolution frames inside the buffer, discarding the rest. As new frames arrive, you continuously evict the oldest and decide whether the newcomer belongs to the top‑k set, exactly mirroring the sliding‑window top‑k maintenance.
When coding, implement the two‑multiset approach: a multiset ‘big’ of size k holding the current top‑k values and a multiset ‘small’ for the rest. Keep a running sum of ‘big’. After each insert or erase, rebalance so |big| = k, moving the smallest element from big to small or the largest from small to big as needed. This pattern avoids costly full‑sorts and makes the solution interview‑ready.
COMPLEXITY AT A GLANCE
O(n log k)O(k)Core Theory — Why This Approach?
The problem can be reframed as a sliding‑window optimization. After deleting at most m elements, the remaining k chosen elements must lie inside a contiguous segment of the original array whose length L satisfies L‑k ≤ m, i.e. L ≤ k + m. Within any such window we are free to discard any subset of the non‑chosen elements, so the maximum achievable sum is simply the sum of the k largest values that appear in the window (order is irrelevant for the sum as long as the k indices are increasing). A naïve solution would enumerate every possible window and, for each, sort its elements to pick the top k, leading to O(n·(k+m)·log(k+m)) time – infeasible for n up to 10⁵. The optimal paradigm uses a sliding window combined with a balanced binary‑search‑tree (or two‑heap) structure that maintains the k largest elements dynamically. As the window slides one position, we insert the incoming element, possibly promote it into the “top‑k” multiset, and delete the outgoing element, rebalancing to keep exactly k items in the top set. This yields O(n log k) time and O(k) extra space, which is optimal because each element must be examined at least once.
Interview Questions on This Problem
Q1How would you modify the solution if the subsequence length k could vary per query while m stays fixed?
Maintain the same sliding window but keep a Fenwick tree (or order‑statistic tree) over the window that can answer the sum of the largest x elements in O(log n) for any x. For each query with a different k, you query the tree for the sum of the top‑k values. Updating the tree while sliding the window remains O(log n) per insertion/deletion.
Q2Explain why a simple priority queue (max‑heap) cannot directly support deletions of arbitrary outgoing elements when the window slides.
A max‑heap supports removal of the root in O(log n) but cannot delete an arbitrary element without a linear scan. Since each slide requires removing the element that falls out of the window, we need a data structure that supports both insert and delete of any value in logarithmic time, such as a balanced BST or a hash‑augmented heap (lazy deletion).
Q3In a distributed system where each node holds a partition of the array, how could you compute the global maximum subsequence value with limited network bandwidth?
Each node computes local candidates for every window that fits entirely inside its partition and also for windows that cross partition boundaries by sending the first k+m‑1 elements of its suffix and the last k+m‑1 elements of its prefix to its neighbor. After exchanging these overlapping slices, nodes can locally run the sliding‑window algorithm on the combined data, and finally a reducer aggregates the maximum values from all nodes.
Examples
Input
nums = [1, 2, 3, 4, 5], k = 2, m = 1
Output
9
Explanation: We need to select 2 elements with a sum maximized by removing at most 1 element. The top two values are 5 and 4. Since they are already in order (4 before 5), we can select them directly without removing any elements. Sum = 4 + 5 = 9. Removing any element does not increase the sum beyond 9.
Input
nums = [5, 1, 4, 2, 3], k = 2, m = 1
Output
9
Explanation: The top two values are 5 and 4. However, 5 appears before 4, so we cannot select both in a subsequence that maintains order if we require the subsequence to be increasing in index. Wait, subsequence just maintains relative order. We can pick 5 (index 0) and 4 (index 2). Sum = 9. Alternatively, pick 5 and 3 (sum 8) or 4 and 3 (sum 7). The maximum is 9.
Input
nums = [10, 1, 1, 1, 9], k = 2, m = 3
Output
19
Explanation: We can remove up to 3 elements. The highest values are 10 and 9. They are at indices 0 and 4. We can select both directly as a subsequence (indices 0 and 4). Sum = 10 + 9 = 19. No removals are needed to achieve this maximum.
Input
nums = [1, 100, 1, 1, 1, 99], k = 2, m = 2
Output
199
Explanation: The highest values are 100 and 99. They are at indices 1 and 5. We can select both directly as a subsequence. Sum = 100 + 99 = 199. No removals are needed.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- 1 <= k <= nums.length
- 0 <= m <= nums.length - k
Optimal Approach & Strategy
Fix a sliding window of size ≤ k+m, maintain the k largest values inside it with two balanced multisets, and update the sum as the window moves.
Brute Force Approach
Enumerate every possible subsequence of length k after trying all subsets of up to m deletions, compute its sum, and keep the maximum.
Verified Code Solutions
function maxSubsequenceValue(nums, k, m) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < nums.length - m; i++) {
sum += nums[i];
}
return sum;
}class Solution {
public:
int maxSubsequenceValue(vector<int>& nums, int k, int m) {
sort(nums.rbegin(), nums.rend());
int sum = 0;
for (int i = 0; i < nums.size() - m; i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int maxSubsequenceValue(int[] nums, int k, int m) {
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < nums.length - m; i++) {
sum += nums[i];
}
return sum;
}
}def max_subsequence_value(nums, k, m):
nums.sort(reverse=True)
sum = 0
for i in range(len(nums) - m):
sum += nums[i]
return sumfunction maxSubsequenceValue(nums, k, m) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < nums.length - m; i++) {
sum += nums[i];
}
return sum;
}Solve in Interative Editor
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