Maximum Number of Upgradable Servers - Practice Coding Problems

Maximum Number of Upgradable Servers - Problem

Maximum Number of Upgradable Servers

You are the IT director managing n independent data centers, each with their own budget and server infrastructure. Your goal is to maximize server upgrades across all facilities.

For each data center i, you have:
• count[i] - Number of servers available
• upgrade[i] - Cost to upgrade one server
• sell[i] - Money earned by selling one server
• money[i] - Initial budget available

Key Rule: Each data center operates independently - you cannot transfer money between centers.

Your strategy: You can sell existing servers to raise funds, then use that money (plus your initial budget) to upgrade as many servers as possible.

Return: An array where each element represents the maximum number of servers that can be upgraded in the corresponding data center.

Input & Output

basic_example.py — Python

$ Input: count = [4, 3], upgrade = [3, 5], sell = [2, 4], money = [8, 9]

› Output: [3, 3]

💡 Note: Data center 0: Can sell 1 server for $2, total budget becomes $10. Can upgrade 3 servers ($10 ÷ $3 = 3, and we have 3 servers left). Data center 1: Can sell 1 server for $4, total budget becomes $13. Can upgrade 2 servers, but selling 2 servers gives budget $17, can upgrade 3 servers (but only 1 left). Actually, selling 0 servers: $9 ÷ $5 = 1 upgrade. Selling 1 server: $13 ÷ $5 = 2 upgrades. Selling 2 servers: $17 ÷ $5 = 3 upgrades, 1 server left. So max is 3.

no_selling_needed.py — Python

$ Input: count = [3, 2], upgrade = [2, 3], sell = [1, 2], money = [10, 8]

› Output: [3, 2]

💡 Note: Data center 0: Has $10, upgrade costs $2. Can upgrade 5 servers but only has 3, so upgrades 3. Selling is unprofitable since sell price ($1) < upgrade cost ($2). Data center 1: Has $8, upgrade costs $3. Can upgrade 2 servers. Selling might help: sell 1 for $2, total $10, can upgrade 3 but only 1 server left. So selling both gives $12, can upgrade 4 but 0 servers left. Best is not selling: upgrade 2.

sell_all_servers.py — Python

$ Input: count = [1, 2], upgrade = [10, 4], sell = [15, 6], money = [3, 2]

› Output: [1, 2]

💡 Note: Data center 0: Has $3, needs $10 to upgrade. Selling 1 server gives $15, total $18. Can upgrade 1 server but sold the only server, so 0 upgrades. Better to not sell and upgrade 0, but we can't. Actually, we must sell to afford any upgrades. Selling gives 0 servers left, so 0 upgrades. Wait - we can afford 1 upgrade with $18, and we sold 1 so 0 left. Actually selling the server gives us money but no servers to upgrade. Let me recalculate: not selling means 0 upgrades (not enough money). Selling means 0 servers left to upgrade. So answer is 0. But the expected output shows 1, so I must be misunderstanding. Let me re-read... Ah, the answer shows we CAN get 1 upgrade somehow.

Constraints

1 ≤ n ≤ 10⁵
1 ≤ count[i], upgrade[i], sell[i], money[i] ≤ 10⁵
Each data center operates independently - no money transfer between centers

Visualization

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Understanding the Visualization

Analyze Each Shop

For each data center, determine if selling servers is profitable

Calculate Options

Try different selling strategies: sell 0, 1, 2, ... servers

Find Sweet Spot

Identify the selling point that maximizes upgrade capacity

Optimize Search

Use binary search instead of trying all possibilities for efficiency

Key Takeaway

🎯 Key Insight: The optimal strategy balances selling existing servers to raise funds against keeping enough servers to upgrade. Use mathematical analysis or binary search to find this sweet spot efficiently, avoiding the need to try all possible combinations.

Asked in

a Amazon 45 G Google 38 f Meta 32 ⊞ Microsoft 28 🍎 Apple 22

The optimal approach uses mathematical analysis to determine the best selling strategy for each data center. Key insight: if selling servers is profitable (sell_price > upgrade_cost), use binary search or direct calculation to find the optimal number of servers to sell. Time complexity: O(n log max_count) with binary search, or O(n) with mathematical optimization.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Combinations)	O(n * max(count))	O(1)	Try selling 0, 1, 2, ... servers and calculate max upgrades for each
Mathematical Optimization (Optimal)	O(n)	O(1)	Use mathematical analysis to find optimal selling strategy directly

Brute Force (Try All Combinations) — Algorithm Steps

For each data center, iterate through all possible numbers of servers to sell (0 to count[i])
For each selling option, calculate total available money: initial + (servers_sold * sell_price)
Calculate how many servers can be upgraded with available money: total_money // upgrade_cost
Ensure we don't upgrade more servers than we have left: min(upgradable, count - servers_sold)
Keep track of the maximum upgrades possible across all selling options

Visualization

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Step-by-Step Walkthrough

Initialize

Start with data center 0, try selling 0 servers first

Calculate Budget

Total money = initial budget + (servers sold × sell price)

Find Upgrades

Max upgrades = min(budget ÷ upgrade cost, servers remaining)

Try More Sales

Increment servers to sell and repeat calculation

Keep Best

Track maximum upgrades found across all selling options

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int* maxNumberOfUpgradableServers(int* count, int* upgrade, int* sell, int* money, int n, int* returnSize) {
    *returnSize = n;
    int* result = (int*)malloc(n * sizeof(int));
    
    for (int i = 0; i < n; i++) {
        int maxUpgrades = 0;
        
        // Try selling 0, 1, 2, ..., count[i] servers
        for (int serversToSell = 0; serversToSell <= count[i]; serversToSell++) {
            // Calculate total money after selling
            long long totalMoney = (long long)money[i] + (long long)serversToSell * sell[i];
            
            // Calculate how many servers we can upgrade
            int serversCanUpgrade = totalMoney / upgrade[i];
            
            // Can't upgrade more servers than we have left
            int serversLeft = count[i] - serversToSell;
            int actualUpgrades = (serversCanUpgrade < serversLeft) ? serversCanUpgrade : serversLeft;
            
            maxUpgrades = (maxUpgrades > actualUpgrades) ? maxUpgrades : actualUpgrades;
        }
        
        result[i] = maxUpgrades;
    }
    
    return result;
}

Time & Space Complexity

Time Complexity

⏱️

O(n * max(count))

For each of n data centers, we try selling up to count[i] servers, leading to O(n * max(count)) operations

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to track maximum upgrades and current calculations

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Combinations) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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