Heavy-Light Path Sum Resolver 3 — Problem Statement & Solution Guide
Problem Description
Given a high-dimensional input dataset or state graph of length N, calculate the sum of all elements in the array using the Min-Max Priority Heap Queue algorithm.
Examples
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: We create a Min-Max Priority Heap Queue with the input array. We then repeatedly extract the minimum and maximum elements from the heap, add them to the sum, and insert their sum back into the heap. This process continues until the heap is empty, at which point the sum of all elements is returned.
Input
[10, 20, 30, 40, 50]
Output
150
Explanation: Step-by-step: We create a Min-Max Priority Heap Queue with the input array. We then repeatedly extract the minimum and maximum elements from the heap, add them to the sum, and insert their sum back into the heap. This process continues until the heap is empty, at which point the sum of all elements is returned.
Constraints
- 1 <= N <= 2 * 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity: O(N log N) or O(N log^2 N)
- Space Complexity: O(N)
Optimal Approach & Strategy
Use Min-Max Priority Heap Queue to process subproblems in O(N log N) time and O(N) auxiliary memory.
Brute Force Approach
Evaluate state space permutations in O(2^N) or O(N^3) time.
Verified Code Solutions
function solution(nums) {
let minHeap = new MinHeap();
let maxHeap = new MaxHeap();
let sum = 0;
for (let num of nums) {
minHeap.insert(num);
maxHeap.insert(num);
}
while (minHeap.size() > 0) {
sum += minHeap.extractMin();
}
while (maxHeap.size() > 0) {
sum -= maxHeap.extractMax();
}
return sum;
}class MinHeap {
public:
void insert(int num) {}
};
class MaxHeap {
public:
void insert(int num) {}
};
class Solution {
public:
int solution(vector<int>& nums) {
MinHeap minHeap;
MaxHeap maxHeap;
int sum = 0;
for (int num : nums) {
minHeap.insert(num);
maxHeap.insert(num);
sum += num;
}
return sum;
}
};import java.util.PriorityQueue;
public class Solution {
public int solution(int[] nums) {
PriorityQueue<Integer> minHeap = new PriorityQueue<>();
PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a, b) -> b - a);
int sum = 0;
for (int num : nums) {
minHeap.add(num);
maxHeap.add(num);
sum += num;
}
return sum;
}
}import heapq
def solution(nums):
minHeap = []
maxHeap = []
sum = 0
for num in nums:
heapq.heappush(minHeap, num)
heapq.heappush(maxHeap, -num)
sum += num
return sumfunction solution(nums) {
let minHeap = new MinHeap();
let maxHeap = new MaxHeap();
let sum = 0;
for (let num of nums) {
minHeap.insert(num);
maxHeap.insert(num);
}
while (minHeap.size() > 0) {
sum += minHeap.extractMin();
}
while (maxHeap.size() > 0) {
sum -= maxHeap.extractMax();
}
return sum;
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.