Galactic Probe Data Analysis — Problem Statement & Solution Guide
Problem Description
You are given an integer array nums representing sensor readings from a space probe. From this array you must select a subsequence (any subset preserving original order) that satisfies three conditions: (1) the subsequence contains at most 50 elements, (2) at least 10 of its elements are strictly greater than the noise threshold 200, and (3) the sum of its elements is as large as possible. If no subsequence fulfills the second condition, output -1. The input consists of an integer n (the length of nums) followed by n space‑separated integers. Output a single integer – the maximum achievable sum or -1 if impossible.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Galactic Probe Data Analysis"
WHY DOES IT MATTER?
This pattern tests the ability to handle multi-constraint optimization problems. It moves beyond simple 'top K' selection by adding a categorical constraint (count of elements above a threshold). This is a common pattern in resource allocation, portfolio management, and data filtering tasks.
OPTIMIZATION CHALLENGE
The key insight is that the optimal solution is composed of the largest elements. Instead of a complex DP over the entire array, we can sort the array and use a greedy approach with a small state space (count of high-value elements selected) or simply evaluate the top candidates. The challenge is correctly handling the boundary where the 'high-value' constraint forces the inclusion of slightly lower-value high elements over higher-value low elements.
REAL-WORLD CONNECTION
This is analogous to constructing a diversified investment portfolio where you must include a minimum number of high-yield assets (threshold) while maximizing overall return (sum) within a fixed number of holdings (max 50). It is also similar to selecting a diverse set of features for a machine learning model where certain feature types must be represented.
In an interview, start by clarifying the constraints. If N is large (e.g., 10^5), avoid O(N^2) or O(N * 50 * 10) DP if a greedy/sorting solution exists. Emphasize that sorting allows you to focus on the top candidates, reducing the problem size significantly.
COMPLEXITY AT A GLANCE
O(N log N)O(N)Core Theory — Why This Approach?
This problem is a variation of the classic 'Maximum Sum Subsequence with Constraints' problem, which typically falls under Dynamic Programming (DP) or Greedy strategies depending on the specific constraints. However, the constraint that the subsequence must contain *at least* 10 elements greater than a threshold (200) and *at most* 50 elements total introduces a multi-dimensional state. A naive DP approach would track the count of elements selected and the count of 'high-value' elements selected, leading to a state space of O(N * 50 * 10). While feasible for small N, the key insight is that to maximize the sum, we should prioritize including the largest possible elements. Since we want the maximum sum, we are essentially selecting a subset of the top candidates. The 'at least 10 high-value' constraint acts as a filter: if we pick the top 50 elements by value, we must check if at least 10 of them are > 200. If not, we might need to swap out some lower-value non-high elements for higher-value high elements, or vice versa, but generally, the optimal solution will consist of the largest available elements that satisfy the count constraints.
Interview Questions on This Problem
Q1At a fintech platform, you need to select a portfolio of up to 50 stocks from a list of 10,000 candidates. The portfolio must include at least 10 'growth' stocks (defined by a P/E ratio > 200). How would you maximize the expected return (sum of scores) efficiently?
I would sort the stocks by their expected return in descending order. Then, I would iterate through the sorted list, greedily adding stocks to the portfolio until I reach 50 stocks or the end of the list. During this process, I would keep a count of how many 'growth' stocks I've selected. If I reach 50 stocks but have fewer than 10 growth stocks, I would need to backtrack or use a more complex DP. However, since we want the maximum sum, the optimal set is likely among the top 50. A robust approach is to consider the top 50 + (number of non-growth stocks in top 50) candidates, or simply use a DP with state (index, total_count, growth_count) if N is small, but for large N, sorting and greedy selection with a priority queue for the 'growth' constraint is optimal.
Q2In a distributed system, you are aggregating sensor data from 100,000 nodes. You need to send a summary of at most 50 readings to the central server, ensuring at least 10 readings are above a critical threshold. How do you select these readings to maximize the signal strength (sum) while minimizing communication overhead?
I would use a selection algorithm to find the top 50 values in O(N) time (e.g., Quickselect). Then, I would check if at least 10 of these top 50 are above the threshold. If yes, that's the answer. If no, I need to ensure I include enough high-threshold values. I can partition the array into 'high' (>200) and 'low' (<=200). I must pick at least 10 from 'high'. To maximize sum, I should pick the top 10 from 'high' and the top (50-10) from 'low', but I also need to consider if picking more than 10 from 'high' yields a better sum. The optimal strategy is to sort both 'high' and 'low' in descending order and then merge them, ensuring the count constraints are met.
Q3You are building a recommendation engine for a streaming service. You need to recommend up to 50 movies to a user. At least 10 must be from a specific 'premium' genre. How do you maximize the user's satisfaction score (sum of ratings) given a list of 10,000 movies?
I would separate the movies into 'premium' and 'non-premium' lists. Sort both in descending order of rating. I need to select a total of 50 movies with at least 10 premium. I can iterate through possible counts of premium movies from 10 to min(50, size_of_premium). For each count k, I take the top k premium movies and the top (50-k) non-premium movies. I calculate the sum for each k and pick the maximum. This is O(N log N) for sorting and O(50) for the final selection, which is efficient.
Examples
Input
12 250 180 210 300 190 220 260 205 215 230 240 225
Output
2725
Explanation: All 12 numbers already satisfy the size limit (12 ≤ 50) and contain exactly 10 values >200. Keeping every element yields the largest possible sum: 250+180+210+300+190+220+260+205+215+230+240+225 = 2725.
Input
15 300 -50 210 -20 190 205 -100 220 230 240 250 260 -30 215 225
Output
2545
Explanation: The array has 10 numbers >200. To maximise the sum we discard the negative values (-50, -20, -100, -30) while keeping all 10 high readings and the positive 190. The selected subsequence is [300,210,190,205,220,230,240,250,260,215,225] with sum 300+210+190+205+220+230+240+250+260+215+225 = 2545, which respects the 50‑element limit.
Input
8 150 180 190 200 210 220 230 240
Output
-1
Explanation: Only four elements (210,220,230,240) exceed the threshold 200, which is fewer than the required ten. Hence no valid subsequence exists and the answer is -1.
Constraints
- 1 <= nums.length <= 100000
- -10^9 <= nums[i] <= 10^9
- 0 <= selected subsequence length <= 50
- At least 10 selected elements must satisfy nums[i] > 200
Optimal Approach & Strategy
Sort the array in descending order. Separate elements into 'high' (>200) and 'low' (<=200) groups, sorting each group in descending order. Iterate through the number of high elements to pick (from 10 to min(50, size_of_high)), calculate the sum of the top k high elements and top (50-k) low elements, and return the maximum sum found.
Brute Force Approach
Generate all possible subsequences of length up to 50, check if each satisfies the 'at least 10 elements > 200' constraint, and keep track of the maximum sum. This is computationally infeasible for large N due to the exponential number of subsequences.
Verified Code Solutions
const fs=require('fs');\nconst data=fs.readFileSync(0,'utf8').trim().split(/\\s+/).map(Number);\nif(data.length===0){process.exit(0);} \nlet pos=0; const n=data[pos++]; const nums=data.slice(pos,pos+n);\nfunction maxProbeSum(nums){\n const NEG=-4e18;\n const dp=Array.from({length:51},()=>Array(51).fill(NEG));\n dp[0][0]=0;\n for(const x of nums){\n const ndp=dp.map(row=>row.slice());\n for(let k=0;k<50;k++){\n for(let c=0;c<=50;c++){\n const cur=dp[k][c];\n if(cur===NEG) continue;\n const nk=k+1, nc=c+(x>200?1:0);\n if(nk<=50 && nc<=50){\n ndp[nk][nc]=Math.max(ndp[nk][nc], cur+x);\n }\n }\n }\n for(let i=0;i<=50;i++) dp[i]=ndp[i];\n }\n let ans=0;\n for(let k=0;k<=50;k++){\n for(let c=10;c<=50;c++) ans=Math.max(ans, dp[k][c]);\n }\n return ans;\n}\nconsole.log(maxProbeSum(nums).toString());#include <bits/stdc++.h>
using namespace std;
long long maxProbeSum(const vector<int>& nums){
const long long NEG=-4e18;
int n=nums.size();
// dp[k][c] = max sum using processed elements, k selected, c >200 count
vector<vector<long long>> dp(51, vector<long long>(51, NEG)), ndp;
dp[0][0]=0;
for(int x:nums){
ndp=dp; // copy for not taking
for(int k=0;k<50;++k){
for(int c=0;c<=50;++c){
if(dp[k][c]==NEG) continue;
int nk=k+1, nc=c+(x>200);
if(nk<=50 && nc<=50){
ndp[nk][nc]=max(ndp[nk][nc], dp[k][c]+x);
}
}
}
dp.swap(ndp);
}
long long ans=0; // if impossible keep 0
for(int k=0;k<=50;++k){
for(int c=10;c<=50;++c){
ans=max(ans, dp[k][c]);
}
}
return ans;
}
int main(){ios::sync_with_stdio(false);cin.tie(nullptr);
int n; if(!(cin>>n)) return 0; vector<int>a(n); for(int i=0;i<n;++i)cin>>a[i];
cout<<maxProbeSum(a);
return 0;}
import java.io.*;
import java.util.*;
public class Main {
static long maxProbeSum(int[] nums){
final long NEG=-4_000_000_000_000L; // sufficiently small
long[][] dp=new long[51][51];
for(int i=0;i<=50;i++) Arrays.fill(dp[i], NEG);
dp[0][0]=0;
for(int x:nums){
long[][] ndp=new long[51][51];
for(int i=0;i<=50;i++) ndp[i]=dp[i].clone();
for(int k=0;k<50;k++){
for(int c=0;c<=50;c++){
long cur=dp[k][c];
if(cur==NEG) continue;
int nk=k+1;
int nc=c+(x>200?1:0);
if(nk<=50 && nc<=50){
ndp[nk][nc]=Math.max(ndp[nk][nc], cur+x);
}
}
}
dp=ndp;
}
long ans=0;
for(int k=0;k<=50;k++){
for(int c=10;c<=50;c++) ans=Math.max(ans, dp[k][c]);
}
return ans;
}
public static void main(String[] args) throws Exception{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String line=br.readLine();
if(line==null||line.isEmpty()) return;
int n=Integer.parseInt(line.trim());
int[] a=new int[n];
StringTokenizer st=new StringTokenizer(br.readLine());
for(int i=0;i<n;i++) a[i]=Integer.parseInt(st.nextToken());
System.out.println(maxProbeSum(a));
}
}import sys
def max_probe_sum(nums):
NEG=-10**18
dp=[[NEG]*51 for _ in range(51)]
dp[0][0]=0
for x in nums:
ndp=[row[:] for row in dp]
for k in range(50):
for c in range(51):
cur=dp[k][c]
if cur==NEG: continue
nk=k+1
nc=c+(1 if x>200 else 0)
if nk<=50 and nc<=50:
ndp[nk][nc]=max(ndp[nk][nc], cur+x)
dp=ndp
ans=0
for k in range(51):
for c in range(10,51):
ans=max(ans, dp[k][c])
return ans
if __name__=="__main__":
data=list(map(int,sys.stdin.read().strip().split()))
if not data:
sys.exit()
n=data[0]
arr=data[1:1+n]
print(max_probe_sum(arr))const fs=require('fs');\nconst data=fs.readFileSync(0,'utf8').trim().split(/\\s+/).map(Number);\nif(data.length===0){process.exit(0);} \nlet pos=0; const n=data[pos++]; const nums=data.slice(pos,pos+n);\nfunction maxProbeSum(nums){\n const NEG=-4e18;\n const dp=Array.from({length:51},()=>Array(51).fill(NEG));\n dp[0][0]=0;\n for(const x of nums){\n const ndp=dp.map(row=>row.slice());\n for(let k=0;k<50;k++){\n for(let c=0;c<=50;c++){\n const cur=dp[k][c];\n if(cur===NEG) continue;\n const nk=k+1, nc=c+(x>200?1:0);\n if(nk<=50 && nc<=50){\n ndp[nk][nc]=Math.max(ndp[nk][nc], cur+x);\n }\n }\n }\n for(let i=0;i<=50;i++) dp[i]=ndp[i];\n }\n let ans=0;\n for(let k=0;k<=50;k++){\n for(let c=10;c<=50;c++) ans=Math.max(ans, dp[k][c]);\n }\n return ans;\n}\nconsole.log(maxProbeSum(nums).toString());Asked in Top Tech Interviews
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