Payload Token Tracker 10 — 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 tracker value under given operational constraints, where the tracker value is the sum of values greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Token Tracker 10"
WHY DOES IT MATTER?
Efficient range queries on sorted data are a staple in high‑throughput services.
OPTIMIZATION CHALLENGE
Replace an O(N) scan with O(log N) search plus constant‑time arithmetic to meet latency SLAs.
REAL-WORLD CONNECTION
Databases use B‑tree indexes to binary‑search rows and then aggregate values for analytics.
Cache the prefix sum once, and reuse it across queries to avoid recomputation.
COMPLEXITY AT A GLANCE
O(log N)O(1) extra (O(N) for prefix sums if not counted as extra)Core Theory — Why This Approach?
Binary search exploits the monotonic property of sorted data to locate a boundary in logarithmic time, turning a linear scan into a divide‑and‑conquer process. By finding the first index where element > K, we can compute the sum of all larger values using a pre‑computed suffix or prefix sum array, achieving O(log n) query time.
A naive linear scan adds each element greater than K, which is O(n) and becomes prohibitive for massive streams or multiple queries. The optimal paradigm combines binary search for the cutoff point with O(1) range‑sum retrieval, dramatically reducing per‑query cost while preserving correctness even under tight time constraints.
Interview Questions on This Problem
Q1Why does binary search require the input array to be sorted?
Because it relies on the order to discard half the search space each step; unsorted data breaks the monotonic guarantee.
Q2How can you compute the sum of elements greater than K after locating the split index?
Maintain a prefix (or suffix) sum array; the answer is totalSum - prefixSum[splitIndex‑1] for a prefix sum.
Q3What is the time complexity of handling Q independent queries after preprocessing?
O(Q·log N) after O(N) preprocessing for the prefix sums; each query is a binary search plus O(1) arithmetic.
Examples
Input
[1, 2, 3, 4, 5]
Output
3
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we first sort the array in ascending order. Then, we initialize a variable to store the tracker value and iterate through the array from left to right. If the current element is greater than K, we add it to the tracker value. Finally, we return the tracker value.
Input
[10, 20, 30, 40, 50]
Output
5
Explanation: Step-by-step: with input [10, 20, 30, 40, 50], we first sort the array in ascending order. Then, we initialize a variable to store the tracker value and iterate through the array from left to right. If the current element is greater than K, we add it to the tracker value. Finally, we return the tracker value.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Pre‑compute a prefix sum array, binary‑search for the first >K, then compute the result with a single subtraction; O(log N) time per query.
Brute Force Approach
Iterate through the entire list, adding each element that exceeds K; O(N) time per query.
Verified Code Solutions
function solution(nums, K) {
nums.sort((a, b) => a - b);
let trackerValue = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) {
trackerValue += nums[i];
}
}
return trackerValue;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
sort(nums.begin(), nums.end());
int trackerValue = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] > K) {
trackerValue += nums[i];
}
}
return trackerValue;
}
};class Solution {
public int solution(int[] nums, int K) {
Arrays.sort(nums);
int trackerValue = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] > K) {
trackerValue += nums[i];
}
}
return trackerValue;
}
}def solution(nums, K):
nums.sort()
tracker_value = 0
for i in range(len(nums)):
if nums[i] > K:
tracker_value += nums[i]
return tracker_valuefunction solution(nums, K) {
nums.sort((a, b) => a - b);
let trackerValue = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) {
trackerValue += nums[i];
}
}
return trackerValue;
}Asked in Top Tech Interviews
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