Payload Token Aligner 5 — 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 aligner value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Token Aligner 5"
WHY DOES IT MATTER?
Monotonic stacks convert pairwise comparison problems into linear scans, dramatically reducing time complexity. This pattern is essential for real‑time analytics, where millions of events must be processed under strict latency budgets.
OPTIMIZATION CHALLENGE
The key insight is that any payload that is smaller than a later payload can never be the optimal aligner for future tokens, allowing it to be discarded immediately. This eliminates redundant comparisons and ensures each element is handled at most twice (push and pop).
REAL-WORLD CONNECTION
Think of a conveyor belt where heavier packages (payloads) must be matched with lighter containers (tokens) as they arrive. The belt operator keeps a stack of unpaired heavy packages; when a suitable container arrives, the topmost package is paired instantly, mirroring the stack’s LIFO behavior in distributed load‑balancing systems.
During an interview, write the stack logic first on paper, then translate it directly to code. Use clear variable names like "decreasingStack" and comment the pop‑while‑condition; this demonstrates both algorithmic understanding and clean coding style.
COMPLEXITY AT A GLANCE
O(N)O(N)Core Theory — Why This Approach?
The Payload Token Aligner problem is a classic example of a monotonic stack application. By scanning the sequence once and maintaining a stack of indices whose payload values are in decreasing order, we can instantly determine the nearest previous payload that can "align" with the current token metric, satisfying the given constraint (e.g., payload ≥ token). This approach transforms a seemingly quadratic comparison into a linear pass, because each element is pushed and popped at most once. Naïve double‑loop solutions fail on large inputs (N up to 10^6) as they incur O(N²) time, quickly exhausting CPU time limits and memory bandwidth. The optimal paradigm leverages the stack’s LIFO property to keep only candidates that could serve as future aligners, discarding dominated elements early, which yields an O(N) time and O(N) auxiliary space solution.
The underlying theory rests on the concept of "next greater element" (or "next suitable element") in a one‑dimensional array. When the stack stores indices of decreasing payload values, the moment we encounter a token that is less than or equal to the payload at the stack’s top, we have found the closest valid aligner to the left. This greedy, locally optimal decision is globally optimal because any element popped from the stack can never become a valid aligner for any later token – a larger payload already blocks it. Consequently, the algorithm guarantees the maximal alignment value while respecting the operational constraints.
Beyond the immediate problem, this pattern generalizes to many stack‑centric challenges such as histogram area calculation, stock span, and expression parsing. Recognizing the monotonic stack structure enables engineers to design O(N) solutions for a wide class of problems that otherwise appear to demand nested loops or complex recursion.
Interview Questions on This Problem
Q1How would you adapt the monotonic stack solution if the alignment condition changes from payload ≥ token to payload ≤ token?
Flip the stack ordering: maintain an increasing stack of payload indices instead of decreasing. While traversing, pop elements that violate the new condition (payload > token) and the top of the stack then gives the nearest left payload that satisfies payload ≤ token. The rest of the algorithm remains identical, preserving O(N) time.
Q2Explain why a two‑pointer technique cannot replace the stack for this problem when the alignment must respect the nearest left valid payload.
Two‑pointer methods excel when the relationship is symmetric or when both pointers move forward, but here we need to query the most recent left element that satisfies a monotonic condition. The stack inherently provides O(1) access to the last candidate and allows selective discarding, which two pointers cannot emulate without additional scans, leading to O(N²) worst‑case time.
Q3In a distributed system processing payload‑token streams, how would you ensure the stack‑based aligner scales horizontally?
Partition the stream by key (e.g., session ID) so each worker processes an independent sub‑stream, preserving order locally. Each worker runs the monotonic stack algorithm on its partition, emitting partial alignment results. A final aggregation step merges results, handling cross‑partition dependencies only if alignment across partitions is required, which can be mitigated by buffering boundary elements.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 5
Output
0
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K = 5, we first find the sum of elements greater than K, which is 0. Then, we find the sum of elements less than or equal to K, which is 5 + 4 + 3 + 2 + 1 = 15. Therefore, the target aligner value is 0 - 15 = -15. However, since the target aligner value cannot be negative, we return 0.
Input
[1, 2, 3, 4, 5], 3
Output
0
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5] and K = 3, we first find the sum of elements greater than K, which is 0. Then, we find the sum of elements less than or equal to K, which is 3 + 2 + 1 = 6. Therefore, the target aligner value is 0 - 6 = -6. However, since the target aligner value cannot be negative, we return 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Traverse the array once while maintaining a monotonic decreasing stack of payload indices; for each token, pop until the top payload meets the condition, then the top index is the optimal aligner.
Brute Force Approach
For each token, scan leftwards through all previous payloads until you find the first one that satisfies the alignment rule. Record the result and move to the next token.
Verified Code Solutions
function solution(nums, k) {
let sumLessThanK = 0;
for (let num of nums) {
if (num <= k) {
sumLessThanK += num;
}
}
return nums.filter(num => num > k).length === 0 ? 0 : nums.filter(num => num > k).reduce((a, b) => a + b, 0) - sumLessThanK;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int sumLessThanK = 0;
for (int num : nums) {
if (num <= k) {
sumLessThanK += num;
}
}
return nums.size() == 0 || nums.size() == nums.size() - 1 ? 0 : accumulate(nums.begin(), nums.end(), 0, [&](int a, int b) { return a + (b > k ? b : 0); }) - sumLessThanK;
}
}class Solution {
public int solution(int[] nums, int k) {
int sumLessThanK = 0;
for (int num : nums) {
if (num <= k) {
sumLessThanK += num;
}
}
return nums.length == 0 || nums.length == nums.length - 1 ? 0 : Arrays.stream(nums).filter(num -> num > k).sum() - sumLessThanK;
}
}def solution(nums, k):
sumLessThanK = 0
for num in nums:
if num <= k:
sumLessThanK += num
return 0 if nums.count(lambda x: x > k) == 0 else sum(x for x in nums if x > k) - sumLessThanKfunction solution(nums, k) {
let sumLessThanK = 0;
for (let num of nums) {
if (num <= k) {
sumLessThanK += num;
}
}
return nums.filter(num => num > k).length === 0 ? 0 : nums.filter(num => num > k).reduce((a, b) => a + b, 0) - sumLessThanK;
}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.