Protocol Pipeline Consolidator 23 — Problem Statement & Solution Guide
Problem Description
You are given a singly linked list containing N integer nodes and an integer threshold K. Your task is to compute the sum of all node values that are strictly greater than K. The list must be processed using a recursive backtracking approach: the recursion should traverse to the end of the list, then unwind while accumulating the required sum. The program should read the number of nodes, the node values in order, and the threshold K, and output a single integer representing the computed sum.
Input format:
- The first line contains an integer N, the length of the linked list.
- The second line contains N space‑separated integers, the values of the nodes from head to tail.
- The third line contains the integer K.
Output format:
- A single integer, the sum of all node values that are greater than K. If no node satisfies the condition, output 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Pipeline Consolidator 23"
WHY DOES IT MATTER?
Recursive backtracking cleanly separates traversal from aggregation, reducing mutable state bugs.
OPTIMIZATION CHALLENGE
The key is to avoid repeated passes; a single recursive pass yields O(N) time.
REAL-WORLD CONNECTION
Similar to unwinding network packets where the payload is processed only after the header chain is fully parsed.
Always return the accumulated sum from the recursive call rather than using a global variable.
COMPLEXITY AT A GLANCE
O(N)O(N)Core Theory — Why This Approach?
Recursive backtracking on a singly linked list leverages the call stack to reach the terminal node before any computation, effectively performing a post‑order traversal. By unwinding the recursion, each frame can add the current node's value to a running total only if it exceeds the threshold K, guaranteeing that every node is visited exactly once.
Iterative scans are straightforward, but they miss the pedagogical value of recursion and can lead to off‑by‑one errors when handling tail conditions. The optimal paradigm uses recursion to simplify code, avoid explicit pointer manipulation for the sum, and naturally respects the O(N) time bound while using O(N) auxiliary space for the stack, which is acceptable for the given constraints.
Interview Questions on This Problem
Q1How does recursion achieve a post‑order traversal on a singly linked list?
The recursive call is made before processing the current node, so the function reaches the list's end first. As the stack unwinds, each call processes its node, achieving post‑order semantics.
Q2What is the space complexity of this recursive solution and why?
It is O(N) because each node adds a frame to the call stack until the base case is reached. The stack depth equals the list length.
Q3Can this problem be solved iteratively with the same space complexity?
An iterative solution uses O(1) extra space by maintaining a running sum and a pointer. However, it does not use recursion, so the stack space is eliminated.
Examples
Input
6 4 10 -2 7 15 1 5
Output
32
Explanation: The nodes greater than K=5 are 10, 7, and 15. Their sum is 10+7+15 = 32. The recursive function reaches the tail, then returns 0, and each unwind step adds the current node's value if it exceeds K.
Input
4 -5 -1 -3 -2 -3
Output
-3
Explanation: Values greater than K=-3 are -2 and -1. Their sum is -2 + (-1) = -3. The recursion visits nodes in order, and during the unwind phase adds -2 and -1 while ignoring -5 and -3.
Input
1 100 50
Output
100
Explanation: The single node value 100 is greater than K=50, so the sum equals 100. The recursive call reaches the null terminator, returns 0, and the unwind adds 100.
Constraints
- 1 <= N <= 10^5
- -10^9 <= node value <= 10^9
- -10^9 <= K <= 10^9
- The solution must run in O(N) time and O(N) recursion depth (or use tail recursion optimization if supported).
Optimal Approach & Strategy
Use a recursive function that reaches the end first, then adds qualifying node values while the call stack unwinds.
Brute Force Approach
Iterate through the list with a loop, checking each node and accumulating the sum in a variable.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
}def solution(nums, K):
sum = 0
for num in nums:
if num > K:
sum += num
return sumfunction solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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