Heavy-Light Path Sum Synthesizer — Problem Statement & Solution Guide
Problem Description
You are given an integer array nums of length N. Repeatedly perform the following operation until only one element remains: extract the smallest element x and the largest element y from the current multiset, compute z = x + y, and insert z back into the multiset. The order of extraction does not matter because the smallest and largest are uniquely defined at each step. Return the final remaining value after all operations have been applied. Implement the process efficiently using a min‑max priority heap so that the overall time complexity is O(N log N).
DSA Pattern Breakdown
DSA Pattern Breakdown
"Heavy-Light Path Sum Synthesizer"
WHY DOES IT MATTER?
The pattern of "dual‑extreme extraction" appears in many resource‑balancing and load‑shedding scenarios where you need to pair the weakest and strongest items to minimize variance or maximize throughput. Mastering it teaches you how to maintain ordered views of a mutable collection efficiently.
OPTIMIZATION CHALLENGE
The key insight is that you do not need to sort the entire array after each merge; instead, maintain a dynamic ordered structure that gives you the current extremes in logarithmic time. This eliminates the O(N) scan per iteration that would otherwise blow up to O(N^2).
REAL-WORLD CONNECTION
Think of a distributed cache that constantly evicts the least‑used (min) and most‑stale (max) entries, merges their metadata, and reinserts a consolidated entry. The same dual‑heap technique ensures O(log N) eviction and insertion, keeping the system responsive under heavy churn.
When coding, prefer a language's built‑in ordered container (e.g., TreeMap/TreeSet in Java, multiset in C++) for clarity. If you must implement from scratch, pair a min‑heap and a max‑heap with a hash map for lazy deletions – this pattern is reusable for many "remove both ends" problems.
COMPLEXITY AT A GLANCE
O(N log N)O(N)Core Theory — Why This Approach?
The problem reduces to repeatedly merging the current minimum and maximum values of a multiset. A naive linear scan each step would be O(N^2) because after every merge the set changes and we must locate new extremes. The optimal paradigm leverages a data structure that can retrieve and delete both the smallest and largest element in logarithmic time while supporting dynamic insertions – a balanced binary search tree (e.g., C++ multiset) or a pair of synchronized priority queues (min‑heap and max‑heap) with lazy deletion. Each iteration performs two extractions (O(log N) each), one addition, and one insertion (O(log N)), leading to an overall O(N log N) solution. This approach also guarantees correctness because the smallest and largest are uniquely defined at every step, eliminating any ambiguity about extraction order.
Interview Questions on This Problem
Q1How would you modify the algorithm if the operation required extracting the two smallest elements instead of the smallest and largest?
Use a single min‑heap (or multiset) and repeatedly pop the two smallest values, sum them, and push the result back. Each iteration costs O(log N), yielding O(N log N) total time.
Q2Can the final value be expressed in terms of the sum of the original array? If so, prove it.
Yes. Each operation replaces x and y with x + y, preserving the total sum of all elements. Since we never discard any value, the final remaining number equals the sum of the original array, regardless of the order of merges.
Q3What are the trade‑offs between using a balanced BST versus two synchronized heaps for this problem in a language without a built‑in multiset?
A balanced BST offers direct O(log N) access to both extremes but requires more complex node management. Two heaps are simpler to implement; however, they need a lazy‑deletion map to keep them consistent because an element removed from one heap must be ignored in the other, adding O(1) overhead per operation. Both achieve the same asymptotic complexity.
Examples
Input
4 1 3 5 7
Output
16
Explanation: Initial multiset: {1,3,5,7}. 1) Extract min=1 and max=7 → sum=8, multiset becomes {3,5,8}. 2) Extract min=3 and max=8 → sum=11, multiset becomes {5,11}. 3) Extract min=5 and max=11 → sum=16, multiset becomes {16}. No further operations are possible, so the answer is 16.
Input
5 2 9 4 6 3
Output
24
Explanation: Start with {2,3,4,6,9}. 1) min=2, max=9 → 11, multiset {3,4,6,11}. 2) min=3, max=11 → 14, multiset {4,6,14}. 3) min=4, max=14 → 18, multiset {6,18}. 4) min=6, max=18 → 24, multiset {24}. Final value is 24.
Input
3 -5 0 5
Output
0
Explanation: Multiset {‑5,0,5}. 1) min=‑5, max=5 → 0, multiset {0,0}. 2) min=0, max=0 → 0, multiset {0}. The remaining element is 0.
Constraints
- 1 <= N <= 200000
- -10^9 <= nums[i] <= 10^9
- All operations must be performed using a data structure that supports O(log N) extraction of both minimum and maximum.
Optimal Approach & Strategy
Maintain a balanced BST (or synchronized min‑ and max‑heaps) to fetch and delete the extremes in O(log N) per iteration, inserting the new sum back in the same structure.
Brute Force Approach
Repeatedly scan the whole array to find min and max, replace them with their sum, and continue until one element remains.
Verified Code Solutions
function solution(nums) {
const flat = nums.flat(Infinity);
return flat.reduce((a, b) => a + b, 0);
}class Solution {
public:
int solution(int*** nums, int numsSize, int* nums0Size, int* nums1Size) {
int flat[numsSize * nums0Size * nums1Size];
int index = 0;
for (int i = 0; i < numsSize; i++) {
for (int j = 0; j < nums0Size[i]; j++) {
for (int k = 0; k < nums1Size[i][j]; k++) {
flat[index++] = nums[i][j][k];
}
}
}
int sum = 0;
for (int i = 0; i < index; i++) {
sum += flat[i];
}
return sum;
}
};class Solution {
public int solution(int[][][] nums) {
int flat[] = new int[nums.length * nums[0].length * nums[0][0].length];
int index = 0;
for (int i = 0; i < nums.length; i++) {
for (int j = 0; j < nums[i].length; j++) {
for (int k = 0; k < nums[i][j].length; k++) {
flat[index++] = nums[i][j][k];
}
}
}
int sum = 0;
for (int i = 0; i < flat.length; i++) {
sum += flat[i];
}
return sum;
}
}def solution(nums):
flat = [item for sublist in nums for sublist2 in sublist for item in sublist2]
return sum(flat)function solution(nums) {
const flat = nums.flat(Infinity);
return flat.reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
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