https://leetcode.com/discuss/general-discussion/886274/online-assessment-questions-matrix-summation

Matrix Summation ✅

long getMaxProfit(vector<int> pnl, int k) {
    int n = pnl.size();
    long maxprofit = 0;
    int i;
    long profit = 0;
    int start = 0;
    for(i=0;i<n;i++) {
        profit += pnl[i];
        if (profit <= 0) {
            profit = 0;
            start = i+1;
        } else {
            maxprofit = max(maxprofit , profit);
            if (i-start+1 == k) {
               
                profit -= pnl[start];
                start++;
            }
            while(pnl[start] <= 0 && start <=i) {
                profit -= pnl[start];
                maxprofit = max(maxprofit , profit);
                start++;
            }
        }
    }
    return maxprofit;
}

Profit Analysis ✅

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#include <iostream>
#include <vector>
#include <unordered_map>
#include <set>
#include <algorithm>
using namespace std;
long long minimumCost(vector<int>& arr) {
    unordered_map<int, int> freq;
    for (int num : arr) {
        freq[num]++;
    }
    vector<pair<int, int>> elements;
    for (const auto& entry : freq) {
        elements.push_back(entry);
    }
    sort(elements.begin(), elements.end(), [](pair<int, int>& a, pair<int, int>& b) {
        return a.second > b.second; 
    });
    set<int> distinct;
    long long totalCost = 0;
    for (const auto& element : elements) {
        for (int i = 0; i < element.second; ++i) {
            distinct.insert(element.first);
            totalCost += distinct.size(); 
        }
    }
    return totalCost;
}

IBM ✅

SELECT product_name, cost AS product_cost, category AS product_category
FROM products AS p1
WHERE cost = (
    SELECT MAX(cost)
    FROM products AS p2
    WHERE p1.category = p2.category
)
ORDER BY product_cost DESC, product_name ASC;

Fractal✅