Segmented Subsequence Sum — Problem Statement & Solution Guide

HeapEasyMin-Heap Extraction

TimeO(n log k)

SpaceO(k)

Problem Description

Given an array of integers, compute the segmented subsequence sum by applying the target algorithm rules. The target algorithm involves maintaining a max heap of the current maximum subsequence sum. For each element in the array, if it's greater than the root of the max heap, we pop the root and add the current element to the max heap. Finally, we return the sum of all elements in the max heap.

DSA Pattern Breakdown

"Segmented Subsequence Sum"

easy

WHY DOES IT MATTER?

Top‑k selection via a heap is a fundamental pattern for streaming and online problems where you cannot afford to store the entire solution space. It enables constant‑time access to the current extremum while keeping updates cheap, which is essential for real‑time analytics, recommendation engines, and any scenario that demands the best‑of‑many results on the fly.

OPTIMIZATION CHALLENGE

The breakthrough is recognising that you only need to keep the k highest sums, not every possible sum. By using a max‑heap (or min‑heap depending on orientation) you get O(log k) insertion and removal, turning a quadratic enumeration into a linear‑ithmic scan.

REAL-WORLD CONNECTION

Think of a news aggregator that must always display the k most‑read articles. As new reads arrive, the system compares the article's count with the smallest count in the current top‑k list; if it beats it, the article replaces the least popular one. The heap acts like the editorial board that constantly curates the headline list without scanning the entire archive.

During an interview, implement the heap logic first with a language‑provided priority queue, then focus on edge cases (negative numbers, k > n). Keep the code modular: a helper to push‑pop based on the comparison, and a final aggregation loop. This shows clean design and awareness of library utilities.

COMPLEXITY AT A GLANCE

⏱ Time:O(n log k)

💾 Space:O(k)

Core Theory — Why This Approach?

The segmented subsequence sum problem asks us to keep track of the most valuable contiguous (or non‑contiguous) pieces of an array as we scan it once. A naïve solution would recompute every possible subsequence, leading to O(n²) or worse time, which quickly becomes infeasible for n in the millions. The optimal paradigm leverages a priority queue (max‑heap) to maintain only the best‑scoring segments seen so far. By storing the current top‑k segment sums in a heap, each new element can be evaluated in O(log k) time: if the element alone or when combined with an existing segment yields a sum larger than the smallest entry in the heap, we replace that entry, guaranteeing that the heap always contains the k largest segment sums.

This approach works because the heap provides constant‑time access to the extremal value (the root) and logarithmic updates, allowing us to discard sub‑optimal candidates without enumerating them. The algorithm therefore reduces the combinatorial explosion of subsequence enumeration to a linear pass with logarithmic overhead, achieving O(n log k) time and O(k) auxiliary space. The key insight is that we never need to remember every possible sum—only the best k—so the heap acts as a compact summary of the search space.

In practice, this technique generalises to many “top‑k” selection problems across streams, sliding windows, and online decision making. It demonstrates the power of combining greedy selection with a suitable data structure, turning an exponential‑time brute force into a scalable solution suitable for production‑grade workloads.

Interview Questions on This Problem

Q1How would you modify the heap‑based solution if the required number of top segments (k) is not known beforehand and you must return the sum of all segments that are strictly greater than a dynamic threshold?

Maintain a max‑heap of all segment sums encountered and a running total. For each new element, compute its contribution, push it onto the heap, and while the heap’s root is ≤ threshold, pop it and subtract its value from the total. This keeps the total equal to the sum of all segments above the threshold in O(log m) per operation, where m is the current heap size.

Q2Explain why a simple sort‑and‑pick‑top‑k approach after generating all possible segment sums is unsuitable for large inputs, and how the heap approach overcomes this limitation.

Generating all segment sums is O(n²) in time and space, making sorting impossible for large n. The heap approach avoids materialising the full list; it incrementally keeps only the k best sums, guaranteeing O(n log k) time and O(k) space, which scales linearly with the input size.

Q3In a distributed system where each node processes a slice of the array, how can you combine local heap results to obtain the global segmented subsequence sum?

Each node runs the same heap algorithm on its slice, producing its local top‑k heap. A central coordinator merges these heaps by inserting all local entries into a new max‑heap of size k, discarding the smallest entries as needed. This merge step is O(p k log k) where p is the number of nodes, preserving the overall O(n log k) bound.

Examples

Example 1

Input

[1, 2, 3, 4, 5]

Output

Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we first initialize the max heap with the first element 1. Then, we add 2 to the max heap because 2 is greater than 1. Next, we add 3 to the max heap because 3 is greater than 2. Then, we add 4 to the max heap because 4 is greater than 3. Finally, we add 5 to the max heap because 5 is greater than 4. The max heap now contains all elements from the array, so the sum is actually the sum of all elements in the array, which is 15.

Example 2

Input

[5, 4, 3, 2, 1]

Output

Explanation: Step-by-step: with input [5, 4, 3, 2, 1], we first initialize the max heap with the first element 5. Then, we add 4 to the max heap because 4 is less than 5. Next, we add 3 to the max heap because 3 is less than 4. Then, we add 2 to the max heap because 2 is less than 3. Finally, we add 1 to the max heap because 1 is less than 2. The max heap now contains all elements from the array, so the sum is actually the sum of all elements in the array, which is 15.

Constraints

1 <= N <= 10^5
-10^9 <= arr[i] <= 10^9
Time Complexity expected: O(N) or O(N log N)
Space Complexity expected: O(1) or O(N)

Optimal Approach & Strategy

Maintain a max‑heap of size k while iterating once through the array, inserting new segment sums and discarding the smallest when the heap exceeds k. This runs in O(n log k) time and O(k) space.

Brute Force Approach

Generate every possible subsequence, compute its sum, sort all sums, and add the top k values. This requires O(n²) time and O(n²) space.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n log k)

function solution(nums) {
   const maxHeap = new MaxHeap();
   let sum = 0;
   for (let num of nums) {
       maxHeap.insert(num);
       if (num > maxHeap.peek()) {
           sum += maxHeap.extractMax();
       }
   }
   return sum;
}

class MaxHeap {
public:
   void insert(int num) {
       heap.push_back(num);
       heapifyUp(heap.size() - 1);
   }
   int peek() {
       return heap[0];
   }
   int extractMax() {
       if (heap.size() == 0) {
           return 0;
       }
       int maxVal = heap[0];
       heap[0] = heap.back();
       heap.pop_back();
       heapifyDown(0);
       return maxVal;
   }
   void heapifyUp(int index) {
       int parent = (index - 1) / 2;
       if (index <= 0) {
           return;
       }
       if (heap[parent] < heap[index]) {
           std::swap(heap[parent], heap[index]);
           heapifyUp(parent);
       }
   }
   void heapifyDown(int index) {
       int left = 2 * index + 1;
       int right = 2 * index + 2;
       int largest = index;
       if (left < heap.size() && heap[left] > heap[largest]) {
           largest = left;
       }
       if (right < heap.size() && heap[right] > heap[largest]) {
           largest = right;
       }
       if (largest != index) {
           std::swap(heap[largest], heap[index]);
           heapifyDown(largest);
       }
   }
   int solution(int nums[], int size) {
       std::priority_queue<int> maxHeap;
       int sum = 0;
       for (int i = 0; i < size; i++) {
           maxHeap.push(nums[i]);
           if (nums[i] > maxHeap.top()) {
               sum += maxHeap.top();
               maxHeap.pop();
           }
       }
       return sum;
   }
};

import java.util.PriorityQueue;
public class Solution {
   public int solution(int[] nums) {
       PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a, b) -> b - a);
       int sum = 0;
       for (int num : nums) {
           maxHeap.add(num);
           if (num > maxHeap.peek()) {
               sum += maxHeap.poll();
           }
       }
       return sum;
   }
}

class MaxHeap:
   def __init__(self):
       self.heap = []
   def insert(self, num):
       self.heap.append(num)
       self.heapifyUp(len(self.heap) - 1)
   def peek(self):
       return self.heap[0]
   def extractMax(self):
       if len(self.heap) == 0:
           return 0
       maxVal = self.heap[0]
       self.heap[0] = self.heap.pop()
       self.heapifyDown(0)
       return maxVal
   def heapifyUp(self, index):
       parent = Math.floor((index - 1) / 2)
       if index <= 0:
           return
       if self.heap[parent] < self.heap[index]:
           self.heap[parent], self.heap[index] = self.heap[index], self.heap[parent]
           self.heapifyUp(parent)
   def heapifyDown(self, index):
       left = 2 * index + 1
       right = 2 * index + 2
       largest = index
       if left < len(self.heap) and self.heap[left] > self.heap[largest]:
           largest = left
       if right < len(self.heap) and self.heap[right] > self.heap[largest]:
           largest = right
       if largest != index:
           self.heap[largest], self.heap[index] = self.heap[index], self.heap[largest]
           self.heapifyDown(largest)
   def solution(self, nums):
       maxHeap = MaxHeap()
       sum = 0
       for num in nums:
           maxHeap.insert(num)
           if num > maxHeap.peek():
               sum += maxHeap.extractMax()
       return sum

function solution(nums) {
   const maxHeap = new MaxHeap();
   let sum = 0;
   for (let num of nums) {
       maxHeap.insert(num);
       if (num > maxHeap.peek()) {
           sum += maxHeap.extractMax();
       }
   }
   return sum;
}

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Segmented Subsequence Sum Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Heap

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Segmented Subsequence Sum — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor