The Hudson River Trading Online Assessment evaluates problem-solving and coding skills for roles like Software Engineer. Hosted on platforms such as CodeSignal, it consists of 3–4 algorithmic challenges with a total duration of 70–150 minutes. Languages commonly used include C++ and Python. Below we cover an overview, three sample problems with solutions, and key tips for success
1. Order Book Matching
Problem: Given a list of buy/sell orders [price, quantity, type], match buys with sells whenever buy price ≥ sell price. Compute the total executed volume.
Example:
Input: [[100,5,'B'], [90,3,'S'], [95,2,'S'], [105,4,'B']]
Output: 5
Approach: • Separate buy orders (sort descending) and sell orders (sort ascending). • Use two pointers to match and accumulate the min(quantity) for each pair.
#include
#include
using namespace std;
long long matchOrders(vector<vector>& orders) {
vector<pair> buy, sell;
for (auto &o : orders) {
if (o[2]=='B') buy.emplace_back(o[0],o[1]);
else sell.emplace_back(o[0],o[1]);
}
sort(buy.rbegin(), buy.rend());
sort(sell.begin(), sell.end());
long long volume=0;
int i=0, j=0;
while (i<buy.size() && j= sell[j].first) {
auto qty = min(buy[i].second, sell[j].second);
volume += qty;
buy[i].second -= qty;
sell[j].second -= qty;
if (buy[i].second==0) i++;
if (sell[j].second==0) j++;
} else {
j++;
}
}
return volume;
}2. Minimum Transformations for Unique Substrings
Problem: For a string s and window length k, find the minimum character changes so every substring of length k has all unique letters.
Example:
Input: s="aabbcc", k=3 → Output: 2
Approach: • Slide a window of size k over s. • Count duplicates in each window and record the minimum needed replacements.
from collections import Counter
def minTransformations(s: str, k: int) -> int:
def needed(w):
cnt = Counter(w)
return sum(v-1 for v in cnt.values() if v>1)
return min(needed(s[i:i+k]) for i in range(len(s)-k+1))3. Optimal Trade Execution
Problem: Given stock prices array and max k transactions, compute the maximum profit. You cannot hold more than one share at a time.
Example:
Input: prices=[3,2,6,5,0,3], k=2 → Output: 7
Approach: • If k ≥ n/2, sum all positive price jumps. • Otherwise use DP with states for transaction count and holding/not-holding.
#include
using namespace std;
int maxProfit(int k, vector& prices) {
int n = prices.size();
if (k >= n/2) {
int profit=0;
for (int i=1;iprices[i-1])
profit += prices[i]-prices[i-1];
return profit;
}
vector<vector> dp(k+1, vector(2, -1e9));
dp[0][0]=0; dp[0][1]=-prices[0];
for (int i=1;i0) dp[j][1] = max(dp[j][1], dp[j-1][0] - prices[i]);
}
}
return dp[k][0];
}ProgramHelp: Beyond Interview Prep
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