3155. Maximum Number of Upgradable Servers

Intuitive:

For each data center, can upgrade 1, 2, 3,... up to K servers, using O(N) loop over all the server to calculate each's profit.

But we can do it using bin search

Solution

• For each data center, set up a binary search,
  • • low = 0, high = count[i] - for a data center, the minimum servers to upgrade is 0, and the maximum number is all servers, which is count[i]
  • • Use the common mid = (low + high) // 2 which will function as the number of servers to upgrade. Calculate net money
  • • That net money metric drives which half of the space to search. If net money < 0, the test number of servers needs to be reduced, so set high = mid - 1. Otherwise, set low = mid + 1.

class Solution {
    public int[] maxUpgrades(int[] count, int[] upgrade, int[] sell, int[] money) {
        int[] res = new int[count.length];

        // loop over all data centers
        for (int i = 0; i < res.length; i++) {
            int low = 0, high = count[i];

            while (low <= high) {
                int mid = low + (high - low) / 2;

                // if int, might be overflow!
                long cost = (long) upgrade[i] * mid; // cost of upgrade [mid] servers
                long profit = (long) sell[i] * (count[i] - mid); // profit by selling rest servers
                
                // net revenue by doing so(upgrading mid servers)
                long net = money[i] + profit - cost;

                if (net >= 0) {
                    low = mid + 1; // could upgrade more!
                } else {
                    high = mid - 1; // losing money, cant do so
                }
            }
            res[i] = low - 1;
            //When the loop terminates, low is one greater than the last valid (affordable) mid.
        }
        return res;
    }
}