Total Node Weight Calculation — Problem Statement & Solution Guide
Problem Description
You are provided with the head pointer of a singly linked list. Each node in this structure stores an integer value representing its specific weight. Your task is to compute the aggregate sum of these weights across the entire list. If the list is empty (i.e., the head is null), the function must return 0. The solution should traverse the list exactly once to ensure optimal time complexity.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Total Node Weight Calculation"
WHY DOES IT MATTER?
Single-pass traversal ensures linear time complexity and constant space, which is critical for large data sets and systems with limited memory. It also avoids the risk of stack overflows that can occur with recursive approaches.
OPTIMIZATION CHALLENGE
The key insight is that you don’t need to revisit any node; a simple accumulator suffices. This eliminates the need for auxiliary data structures or recursion.
REAL-WORLD CONNECTION
Think of a streaming data pipeline where each record arrives sequentially; you must compute a running total without storing all records. Similarly, in log aggregation, you sum metrics on the fly to avoid memory bloat.
When explaining this to an interviewer, emphasize that the algorithm’s simplicity is its strength—no extra space, no recursion, just a clean loop and a variable.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem asks for the sum of all node weights in a singly linked list. A naive approach might involve recursively summing each node’s value, which would still traverse the list once but would incur O(n) stack space due to recursion. For very large lists, this recursion depth can lead to stack overflow and is generally considered unsafe in production code.
To avoid these pitfalls, the optimal solution follows a classic linear traversal pattern: iterate through each node once, accumulating the weight in a running total. This guarantees O(n) time complexity while keeping auxiliary space constant, O(1). The algorithm is straightforward yet powerful, illustrating the importance of single-pass solutions for linked list problems.
By using an iterative loop and a simple accumulator, we eliminate the overhead of function calls and potential memory issues. This pattern is widely applicable to other aggregate operations on linked structures, such as computing averages, finding maximum/minimum values, or concatenating strings, making it a foundational technique in algorithm design.
Interview Questions on This Problem
Q1How would you handle a linked list that contains negative weights when computing the total sum?
The algorithm remains unchanged; the accumulator simply adds each node’s value, whether positive or negative. However, it’s important to ensure that the data type used for the sum can accommodate the range of possible totals, such as using a 64-bit integer if the list can be large or contain large magnitude values.
Q2What modifications would you make if the list were doubly linked and you needed to compute the sum in reverse order?
You could start from the tail node and traverse backwards, adding each weight to the accumulator. If the tail pointer isn’t provided, you’d first traverse to the end to find it, which would still be O(n) overall. The core idea of a single-pass accumulation remains the same.
Q3In a distributed system where each node’s weight is stored on a different server, how would you compute the total sum efficiently?
You would perform a parallel reduction: each server computes the sum of its local sublist, then a coordinator aggregates these partial sums. This reduces the overall time to O(log k) where k is the number of servers, assuming communication overhead is minimal.
Examples
Input
head = [4, 7, 2, 9]
Output
22
Explanation: The list contains four nodes with weights 4, 7, 2, and 9. Summing these values: 4 + 7 = 11; 11 + 2 = 13; 13 + 9 = 22. The final total is 22.
Input
head = []
Output
0
Explanation: The input list is empty, meaning there are no nodes to process. According to the problem specification, the sum of an empty set of weights is defined as 0.
Input
head = [-5, 10, -3, 15]
Output
17
Explanation: The weights are -5, 10, -3, and 15. Calculation: -5 + 10 = 5; 5 + (-3) = 2; 2 + 15 = 17. The negative values are included in the summation.
Input
head = [100]
Output
100
Explanation: The list contains a single node with a weight of 100. The sum is simply the value of that node, which is 100.
Constraints
- 0 <= number of nodes <= 10^5
- -10^9 <= node.val <= 10^9
- The linked list is singly linked (each node has only a 'next' pointer).
Optimal Approach & Strategy
Iterate through the list once, maintaining a single accumulator variable. Add each node’s weight to the accumulator and move to the next node until the list ends. This uses O(n) time and O(1) space.
Brute Force Approach
A naive solution might recursively call a function on the next node and add the current node’s weight to the result. This uses O(n) stack space and can cause stack overflow on large lists.
Verified Code Solutions
function solution(head) {
let total = 0;
while (head !== null) {
total += head.val;
head = head.next;
}
return total;
}class Solution {
public:
int solution(ListNode* head) {
int total = 0;
while (head != nullptr) {
total += head->val;
head = head->next;
}
return total;
}
};class Solution {
public int solution(ListNode head) {
int total = 0;
while (head != null) {
total += head.val;
head = head.next;
}
return total;
}
}def solution(head):
total = 0
while head:
total += head.val
head = head.next
return totalfunction solution(head) {
let total = 0;
while (head !== null) {
total += head.val;
head = head.next;
}
return total;
}Asked in Top Tech Interviews
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