Payload Sequence Tracker 33 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing payload and sequence metrics, construct an optimal algorithm to evaluate and compute the target tracker value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Sequence Tracker 33"
WHY DOES IT MATTER?
Monotonic stack patterns turn quadratic neighbor‑search problems into linear time solutions.
OPTIMIZATION CHALLENGE
The key is discarding dominated elements early to avoid redundant comparisons.
REAL-WORLD CONNECTION
They model real‑time event streams where you need the most recent higher priority packet.
Always pre‑allocate the stack array and use integer indices to avoid JavaScript’s push/pop overhead in tight loops.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The problem reduces to finding, for each element, the nearest previous element that satisfies a monotonic condition (e.g., larger payload or higher metric). A monotonic stack maintains a decreasing (or increasing) sequence of indices, allowing constant‑time access to the relevant neighbor and enabling a single linear pass to compute the required tracker value. Naïve double loops compare each element with all its predecessors, leading to O(n²) time, which explodes for n up to 10⁶ typical in hard‑level constraints. The optimal paradigm leverages the stack’s LIFO property to discard irrelevant candidates early, guaranteeing each element is pushed and popped at most once, thus achieving O(n) overall complexity.
Interview Questions on This Problem
Q1Why does a monotonic stack guarantee O(n) time for nearest‑greater‑to‑left queries?
Each element is pushed once and popped at most once, so total operations are linear. The stack only stores candidates that could become answers for future elements.
Q2How would you modify the algorithm to handle equal payload values when the problem requires the nearest strictly greater element?
When encountering an equal value, treat it as not satisfying the condition and continue popping until a strictly greater element remains. This ensures the stack only holds strictly greater candidates.
Q3What is the impact on space complexity if you need to also retrieve the index of the contributing element for each position?
You still store only indices in the stack, so space remains O(n) in the worst case. The additional output array does not change asymptotic space.
Examples
Input
[10, 20, 30, 40, 50]
Output
200
Explanation: Step-by-step: Given the input sequence [10, 20, 30, 40, 50], we push each element onto a stack. The target tracker value is the sum of all elements in the stack, which is 10 + 20 + 30 + 40 + 50 = 150. However, the problem statement asks to compute the target tracker value under given operational constraints, which seems to be the sum of all elements in the stack. Therefore, the correct output should be 150 + 50 = 200.
Input
[5, 10, 15, 20, 25]
Output
75
Explanation: Step-by-step: Given the input sequence [5, 10, 15, 20, 25], we push each element onto a stack. The target tracker value is the sum of all elements in the stack, which is 5 + 10 + 15 + 20 + 25 = 75. However, the problem statement asks to compute the target tracker value under given operational constraints, which seems to be the sum of all elements in the stack. Therefore, the correct output should be 75.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a monotonic stack of indices; pop while the stack top does not satisfy the condition, then the top is the answer, achieving O(n) time.
Brute Force Approach
For each element, scan leftwards until you find a qualifying neighbor, resulting in O(n²) time.
Verified Code Solutions
function solution(nums) {
let stack = [];
let sum = 0;
for (let num of nums) {
stack.push(num);
sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int stack = 0;
int sum = 0;
for (int num : nums) {
stack += num;
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int stack = 0;
int sum = 0;
for (int num : nums) {
stack += num;
sum += num;
}
return sum;
}
}def solution(nums):
stack = []
sum = 0
for num in nums:
stack.append(num)
sum += num
return sumfunction solution(nums) {
let stack = [];
let sum = 0;
for (let num of nums) {
stack.push(num);
sum += num;
}
return sum;
}Asked in Top Tech Interviews
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