Node Payload Validator 50 — Problem Statement & Solution Guide
Problem Description
Node Payload Validator 50
You are given a collection of N strings S_i (1 ≤ i ≤ N) and an integer payload P_i associated with each string. Construct a Trie from these strings. Each node in the Trie may contain a pointer to its parent (an inward pointer), but this pointer is not required for the computation. The payload is stored only at the terminal node that represents the end of a string.
After building the Trie, you are asked to compute the sum of all payloads that are strictly greater than a given threshold K. The result should be printed as a single integer.
Input format:
- The first line contains two integers N and K.
- The next N lines each contain a string S_i followed by an integer P_i, separated by a space.
Output format:
- A single line containing the sum of all payloads greater than K.
The problem focuses on correctly building the Trie and efficiently summing the qualifying payloads. The inward pointers are present in the data structure but do not influence the sum calculation.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Payload Validator 50"
WHY DOES IT MATTER?
Ensures data integrity across hierarchical structures, preventing inconsistent aggregates that could corrupt downstream analytics.
OPTIMIZATION CHALLENGE
Reduces repeated scans of the string set by collapsing the validation into a single tree traversal, cutting complexity from quadratic to linear.
REAL-WORLD CONNECTION
Similar to validating sales totals in a product category tree where each node’s reported revenue must match the sum of its sub‑categories.
Use iterative DFS with a stack to avoid recursion depth limits in deep tries, and store intermediate sums in the node to enable early exit on mismatch.
COMPLEXITY AT A GLANCE
O(total characters in all strings)O(total characters in all strings)Core Theory — Why This Approach?
A Trie is a prefix tree where each edge represents a character and each node corresponds to a prefix of the input strings. When each terminal node stores a payload, a common validation task is to ensure that every node’s payload equals the sum of payloads of all terminal nodes in its subtree. Naïve approaches recompute this sum for each node by scanning all strings, leading to O(N·L²) time where L is average string length. The optimal paradigm performs a single post‑order depth‑first traversal: as the recursion unwinds, each node aggregates the sums of its children and compares the result to its stored payload, achieving linear time in the total number of characters.
Interview Questions on This Problem
Q1What is a Trie and how does it differ from a hash map for prefix queries?
A Trie stores strings in a tree where each edge is a character, enabling O(L) prefix lookups; a hash map requires O(L) to compute the key and O(1) lookup but cannot efficiently enumerate all strings with a given prefix.
Q2How would you validate that each node’s payload equals the sum of its descendants’ payloads?
Perform a post‑order DFS, summing child payloads and comparing the result to the node’s stored payload; report a mismatch if any.
Q3What is the time and space complexity of building and validating a Trie with N strings of total length M?
Time O(M) for construction and O(M) for validation; space O(M) for the Trie nodes and payload storage.
Examples
Input
3 5 a 10 ab 3 ac 7
Output
17
Explanation: The Trie is built from the strings "a", "ab", and "ac". The payloads stored at the terminal nodes are 10, 3, and 7 respectively. Only payloads greater than 5 are 10 and 7. Their sum is 10 + 7 = 17.
Input
4 0 x -1 xy 2 xyz 3 xw 4
Output
9
Explanation: The Trie contains the strings "x", "xy", "xyz", and "xw". Payloads are -1, 2, 3, and 4. All payloads greater than 0 are 2, 3, and 4. Their sum is 2 + 3 + 4 = 9.
Input
5 100 abc 50 abcd 150 ab 200 a 90 abcdx 110
Output
460
Explanation: The Trie is built from the five strings. Payloads at terminal nodes are 50, 150, 200, 90, and 110. Payloads greater than 100 are 150, 200, and 110. Their sum is 150 + 200 + 110 = 460.
Constraints
- 1 <= N <= 100000
- 0 <= |S_i| <= 100
- -1000000000 <= P_i <= 1000000000
- -1000000000 <= K <= 1000000000
- The total length of all strings does not exceed 1000000
Optimal Approach & Strategy
Run a single post‑order DFS that aggregates child sums and compares to the node’s payload, achieving O(total characters) time and space.
Brute Force Approach
For each node, iterate over all strings to sum payloads of those that share the node’s prefix, leading to O(N·L²) time.
Verified Code Solutions
function sumElementsGreater(arr, k) {
let sum = 0;
for (let num of arr) {
if (num > k) {
sum += num;
}
}
return sum;
}class Solution {
public:
int sumElementsGreater(vector<int>& arr, int k) {
int sum = 0;
for (int num : arr) {
if (num > k) {
sum += num;
}
}
return sum;
}
};public class Solution {
public int sumElementsGreater(int[] arr, int k) {
int sum = 0;
for (int num : arr) {
if (num > k) {
sum += num;
}
}
return sum;
}
}def sum_elements_greater(arr, k):
sum = 0
for num in arr:
if num > k:
sum += num
return sumfunction sumElementsGreater(arr, k) {
let sum = 0;
for (let num of arr) {
if (num > k) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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