Payload Token Validator 29 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing payload and token metrics, construct an optimal algorithm to evaluate and compute the target validator value under given operational constraints. The algorithm should iterate over the array from left to right, adding the element to the result if it's less than K, and adding the maximum of the element and K to the result if it's greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Token Validator 29"
WHY DOES IT MATTER?
Trie enables fast prefix‑based aggregation, turning repeated linear scans into constant‑time lookups per character.
OPTIMIZATION CHALLENGE
The key is to propagate and update aggregate data (like maximum payload) during insertion, cutting the overall complexity from quadratic to linear‑ish.
REAL-WORLD CONNECTION
Autocomplete engines and IP routing tables use Tries to resolve longest‑prefix matches in milliseconds.
When implementing, pre‑allocate node arrays or use a memory pool to avoid costly dynamic allocations during massive insertions.
COMPLEXITY AT A GLANCE
O(N·L)O(N·L)Core Theory — Why This Approach?
A Trie (prefix tree) stores a dynamic set of strings where each node represents a common prefix, enabling O(L) lookup, insertion, and prefix‑aggregation where L is the length of the processed token. In the Payload Token Validator problem, each payload element can be treated as a token string; by inserting tokens into a Trie while traversing left‑to‑right, we can instantly query the maximum metric among all previously seen tokens that share a prefix, satisfying the “add element if less than K, otherwise add the maximum of matching prefixes” rule without rescanning the entire array. Naïve solutions recompute prefix maxima for every element, leading to O(N²) time on large N, which quickly exceeds limits. The optimal paradigm leverages the Trie’s hierarchical structure to propagate and update maximum values during insertion, achieving linear‑ish performance proportional to total characters processed.
Interview Questions on This Problem
Q1How does a Trie achieve O(L) search time compared to a hash map’s O(1) average case?
A Trie follows the characters of the key sequentially, guaranteeing worst‑case time proportional to key length L, independent of the number of stored keys. Hash maps may degrade to O(N) with collisions, while a Trie’s structure avoids hashing altogether.
Q2What modifications are needed to store the maximum payload value for each prefix in a Trie?
Each node should maintain an extra field (e.g., maxVal) that is updated during insertion with the larger of the current maxVal and the new token’s metric. This allows O(1) retrieval of the prefix’s maximum during queries.
Q3Why might a naïve array‑scan solution time out on the Payload Token Validator problem?
Scanning the entire prefix for every element yields O(N²) complexity, which is prohibitive for N up to 10⁵ or higher. The Trie aggregates prefix information incrementally, reducing repeated work.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Output
45
Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], we iterate over the array from left to right. For the first element 1, the result is 1 (1 < K), for the second element 2, the result is 3 (1 + 2), for the third element 3, the result is 6 (3 < K), for the fourth element 4, the result is 10 (4 < K), for the fifth element 5, the result is 15 (5 < K), for the sixth element 6, the result is 21 (6 < K), for the seventh element 7, the result is 28 (7 < K), for the eighth element 8, the result is 36 (8 < K), for the ninth element 9, the result is 45 (9 < K), for the tenth element 10, we add max(10, 3) = 10 to the result.
Input
[5, 5, 5, 5]
Output
20
Explanation: Step-by-step: with input [5, 5, 5, 5], we iterate over the array from left to right. For the first element 5, the result is 5 (5 < K), for the second element 5, the result is 10 (5 + 5), for the third element 5, the result is 15 (5 < K), for the fourth element 5, the result is 20 (5 < K).
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Build a Trie while traversing the array; at each insertion update node‑level maximums, then query the needed prefix in O(L) time.
Brute Force Approach
Iterate over each element and recompute the maximum among all earlier elements that share the required prefix, leading to O(N²) time.
Verified Code Solutions
function solution(nums, K) {
if (nums.length === 0) return 0;
let result = 0;
for (let num of nums) {
if (num < K) {
result += num;
} else {
result += Math.max(num, K);
}
}
return result;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
if (nums.size() == 0) return 0;
int result = 0;
for (int num : nums) {
if (num < K) {
result += num;
} else {
result += max(num, K);
}
}
return result;
}
};class Solution {
public int solution(int[] nums, int K) {
if (nums.length == 0) return 0;
int result = 0;
for (int num : nums) {
if (num < K) {
result += num;
} else {
result += Math.max(num, K);
}
}
return result;
}
}def solution(nums, K):
if not nums:
return 0
result = 0
for num in nums:
if num < K:
result += num
else:
result += max(num, K)
return resultfunction solution(nums, K) {
if (nums.length === 0) return 0;
let result = 0;
for (let num of nums) {
if (num < K) {
result += num;
} else {
result += Math.max(num, K);
}
}
return result;
}Asked in Top Tech Interviews
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