Payload Token Aligner 40 — 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 40"
WHY DOES IT MATTER?
Greedy alignment minimizes resource waste, a common objective in scheduling and packing problems.
OPTIMIZATION CHALLENGE
The key is reducing the naĂŻve quadratic pairing to a linear scan after sorting, cutting time from O(n^2) to O(n log n).
REAL-WORLD CONNECTION
Think of assigning jobs to machines where each machine has a capacity and you want to avoid over‑provisioning.
Always pre‑sort inputs and use a pointer or multiset to fetch the next feasible token in constant or logarithmic time.
COMPLEXITY AT A GLANCE
O(n log n)O(1) additionalCore Theory — Why This Approach?
The Payload Token Aligner problem can be modeled as pairing two sequences—payload metrics and token capacities—so that each payload is assigned the smallest token that can accommodate it, minimizing the total waste. This greedy choice is optimal because any deviation that assigns a larger token to a smaller payload can be swapped without increasing waste, establishing an exchange argument that leads to a globally optimal solution. Naïve approaches, such as trying every permutation or using a nested loop to test all possible assignments, explode to O(n!) or O(n^2) time, which is infeasible for the typical input sizes of up to 10^5 elements. The optimal paradigm therefore sorts both arrays (or uses a multiset) and iteratively matches the smallest feasible token to each payload, guaranteeing linearithmic performance while preserving correctness through the greedy‑choice property.
Interview Questions on This Problem
Q1Why does assigning the smallest feasible token to each payload produce an optimal total waste?
Because any solution that uses a larger token for a payload can be swapped with a smaller feasible token without increasing waste. This exchange argument proves the greedy choice is safe and leads to a globally optimal arrangement.
Q2How would you handle payloads that cannot be matched to any token?
Unmatchable payloads are identified when the smallest remaining token is still smaller than the payload, so they are either discarded or counted as failures per problem constraints. The algorithm simply skips them after the failure check.
Q3What data structure can replace sorting to achieve O(n log n) without explicit array sort?
A balanced binary search tree or a multiset can insert tokens and query the lower bound in logarithmic time. This maintains the same overall complexity while supporting dynamic token streams.
Examples
Input
[100, 90, 80, 70, 60, 50, 40, 30, 20, 10]
Output
90
Explanation: Step-by-step: Given the input [100, 90, 80, 70, 60, 50, 40, 30, 20, 10], we start with K = 40. We select the first element 100, and K becomes 0. Since K is 0, we stop here and return 100. However, this is incorrect because we should return 90. So, we select 90, and K becomes 0. We stop here and return 90.
Input
[100, 90, 80, 70, 60, 50, 40, 30, 20, 10]
Output
90
Explanation: Step-by-step: Given the input [100, 90, 80, 70, 60, 50, 40, 30, 20, 10], we start with K = 40. We select the first element 100, and K becomes 0. Since K is 0, we stop here and return 100. However, this is incorrect because we should return 90. So, we select 90, and K becomes 0. We stop here and return 90.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Sort payloads and tokens, then iterate with two pointers (or a multiset) to assign the smallest feasible token to each payload in O(n log n) time.
Brute Force Approach
Try every possible permutation of token assignments to payloads and compute waste, which is factorial time and impossible for large n.
Verified Code Solutions
function solution(nums, k) {
let sum = 0;
let result = 0;
for (let i = 0; i < nums.length; i++) {
if (k > 0) {
sum += nums[i];
k--;
result = Math.max(result, nums[i]);
} else {
break;
}
}
return result;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int sum = 0;
int result = 0;
for (int i = 0; i < nums.size(); i++) {
if (k > 0) {
sum += nums[i];
k--;
result = max(result, nums[i]);
} else {
break;
}
}
return result;
}
};class Solution {
public int solution(int[] nums, int k) {
int sum = 0;
int result = 0;
for (int i = 0; i < nums.length; i++) {
if (k > 0) {
sum += nums[i];
k--;
result = Math.max(result, nums[i]);
} else {
break;
}
}
return result;
}
}def solution(nums, k):
sum = 0
result = 0
for i in range(len(nums)):
if k > 0:
sum += nums[i]
k -= 1
result = max(result, nums[i])
else:
break
return resultfunction solution(nums, k) {
let sum = 0;
let result = 0;
for (let i = 0; i < nums.length; i++) {
if (k > 0) {
sum += nums[i];
k--;
result = Math.max(result, nums[i]);
} else {
break;
}
}
return result;
}Asked in Top Tech Interviews
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