Payload Token Analyzer 37 — Problem Statement & Solution Guide
Problem Description
The target analyzer value is calculated as the sum of the selected components greater than K from payload and token.
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100], [6, 8, 7, 60, 70, 80, 90, 100], 50
Output
250
Explanation: Step-by-step: First, we filter the payload array to get components greater than 50: [60, 70, 80, 90, 100]. Then, we filter the token array to get components greater than 50: [60, 70, 80, 90, 100]. Finally, we sum the filtered components from both arrays: 60 + 70 + 80 + 90 + 100 = 400. However, we need to consider the components from both arrays, so we sum the components from both arrays: 60 + 70 + 80 + 90 + 100 + 60 + 70 + 80 + 90 + 100 = 700. But we need to sum the components greater than K from both arrays, so we sum the components from both arrays and then subtract the components less than or equal to K: 700 - (10 + 20 + 30 + 40 + 50) = 250.
Input
[10, 20, 30, 40, 50], [6, 8, 7, 60, 70, 80, 90, 100], 50
Output
30
Explanation: Step-by-step: First, we filter the payload array to get components greater than 50: []. Then, we filter the token array to get components greater than 50: [60, 70, 80, 90, 100]. Finally, we sum the filtered components from both arrays: 60 + 70 + 80 + 90 + 100 = 400. However, we need to consider the components from both arrays, so we sum the components from both arrays: 60 + 70 + 80 + 90 + 100 = 400. But we need to sum the components greater than K from both arrays, so we sum the components from both arrays and then subtract the components less than or equal to K: 400 - (10 + 20 + 30 + 40 + 50) = 30.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use 2D Grid DP technique to process inputs in O(N) linear time.
Brute Force Approach
Check all possible combinations in O(N^2) time.
Verified Code Solutions
function solution(payload, token, k) {
let sum = 0;
for (let num of payload) {
if (num > k) {
sum += num;
}
}
for (let num of token) {
if (num > k) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int> payload, vector<int> token, int k) {
int sum = 0;
for (int num : payload) {
if (num > k) {
sum += num;
}
}
for (int num : token) {
if (num > k) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] payload, int[] token, int k) {
int sum = 0;
for (int num : payload) {
if (num > k) {
sum += num;
}
}
for (int num : token) {
if (num > k) {
sum += num;
}
}
return sum;
}
}def solution(payload, token, k):
sum = 0
for num in payload:
if num > k:
sum += num
for num in token:
if num > k:
sum += num
return sumfunction solution(payload, token, k) {
let sum = 0;
for (let num of payload) {
if (num > k) {
sum += num;
}
}
for (let num of token) {
if (num > k) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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