LeetCode 2412 - Minimum Money Required Before Transactions

The problem asks us to determine the minimum initial money required to complete all transactions in any order. Each transaction is defined by [costi, cashbacki], meaning that performing the transaction requires at least costi money, and after completing it, we receive…

leetcodehardarraygreedysorting

LeetCode Problem 2412

Difficulty: 🔴 Hard
Topics: Array, Greedy, Sorting

Solution

Problem Understanding

The problem asks us to determine the minimum initial money required to complete all transactions in any order. Each transaction is defined by [costi, cashbacki], meaning that performing the transaction requires at least costi money, and after completing it, we receive cashbacki back.

In other words, we must choose a starting amount of money such that, no matter how we order the transactions, we never run out of money during any transaction. The output is a single integer representing this minimum initial money.

The input constraints indicate that we could have up to 10^5 transactions, and each cost or cashback can be as large as 10^9. This rules out brute-force approaches that try all permutations of transactions, as factorial-time solutions are infeasible. We also know that all numbers are non-negative, so we do not have to handle negative costs or cashback values.

Important edge cases include transactions where costi equals cashbacki, transactions with zero cost, and transactions where cashbacki exceeds costi.

Approaches

Brute Force

The brute-force approach would attempt every possible permutation of transactions, calculating the running balance to see if a given starting money is sufficient. For each permutation, we check if the current money is enough for each transaction, updating the money after applying the cost and cashback. The minimum starting money would be the smallest amount that succeeds across all permutations.

This approach is correct in theory but impractical. With up to 10^5 transactions, the number of permutations is 10^5!, which is astronomically large. Even checking a single permutation takes O(n), so the brute-force approach is completely infeasible.

Optimal Approach

The key insight is to consider each transaction individually and classify them based on whether they lose money (costi > cashbacki) or gain money (costi <= cashbacki).

Transactions where costi > cashbacki are risky because performing them reduces your money, so you must ensure you have enough money to cover the cost upfront. The critical value for each transaction is the maximum of all transactions' min(costi, costi - cashbacki) because this represents the worst-case scenario for each transaction.

In simpler terms, the minimum initial money required can be computed as the sum of costi - cashbacki for all transactions where costi > cashbacki plus the maximum cost among transactions where costi <= cashbacki. This guarantees you can complete all transactions regardless of order.

Approach Time Complexity Space Complexity Notes
Brute Force O(n!) O(n) Check all permutations; infeasible for large n
Optimal O(n) O(1) Linear scan; compute sums and max values; uses greedy classification

Algorithm Walkthrough

  1. Initialize two variables: sum_loss to store the total "loss" from transactions where costi > cashbacki, and max_gain_cost to store the maximum cost of transactions where costi <= cashbacki.
  2. Iterate through each transaction [c, r]. If c > r, add (c - r) to sum_loss. Otherwise, update max_gain_cost to be the maximum of its current value and c.
  3. After processing all transactions, the minimum starting money is sum_loss + max_gain_cost.
  4. Return this value.

Why it works: The algorithm works because it separates transactions into "risky" (lose money) and "safe" (gain or neutral money). By summing all risky losses and ensuring you have enough for the largest safe transaction, you guarantee enough initial money to complete every transaction, in any order.

Python Solution

from typing import List

class Solution:
    def minimumMoney(self, transactions: List[List[int]]) -> int:
        sum_loss = 0
        max_gain_cost = 0
        
        for cost, cashback in transactions:
            if cost > cashback:
                sum_loss += cost - cashback
            else:
                max_gain_cost = max(max_gain_cost, cost)
        
        return sum_loss + max_gain_cost

This implementation first classifies each transaction into either a "loss" or "gain" type. It accumulates the net loss of all risky transactions and separately tracks the largest cost among gain transactions. The final result guarantees completion of all transactions in any order.

Go Solution

func minimumMoney(transactions [][]int) int64 {
    var sumLoss int64 = 0
    var maxGainCost int64 = 0
    
    for _, t := range transactions {
        cost := int64(t[0])
        cashback := int64(t[1])
        
        if cost > cashback {
            sumLoss += cost - cashback
        } else {
            if cost > maxGainCost {
                maxGainCost = cost
            }
        }
    }
    
    return sumLoss + maxGainCost
}

In Go, we explicitly use int64 to handle large values up to 10^9 and to avoid integer overflow during summation. The logic mirrors the Python solution.

Worked Examples

Example 1: transactions = [[2,1],[5,0],[4,2]]

Transaction cost > cashback? sum_loss max_gain_cost
[2,1] Yes 1 0
[5,0] Yes 6 0
[4,2] Yes 8 0

Minimum money = sum_loss + max_gain_cost = 8 + 0 = 10

Example 2: transactions = [[3,0],[0,3]]

Transaction cost > cashback? sum_loss max_gain_cost
[3,0] Yes 3 0
[0,3] No 3 0

Minimum money = sum_loss + max_gain_cost = 3 + 0 = 3

Complexity Analysis

Measure Complexity Explanation
Time O(n) Single pass through the transactions array
Space O(1) Only two variables are maintained; no additional structures needed

Since the algorithm only iterates through the array once and uses a constant amount of extra space, it is highly efficient even for the maximum constraint of 10^5 transactions.

Test Cases

# Provided examples
assert Solution().minimumMoney([[2,1],[5,0],[4,2]]) == 10  # all losses
assert Solution().minimumMoney([[3,0],[0,3]]) == 3  # one loss, one gain

# Boundary cases
assert Solution().minimumMoney([[0,0]]) == 0  # zero cost and cashback
assert Solution().minimumMoney([[5,5]]) == 5  # neutral transaction, should be max cost

# Mixed transactions
assert Solution().minimumMoney([[1,5],[7,2],[3,3]]) == 7  # sum loss = 5 (7-2), max gain cost = 2

# Large numbers
assert Solution().minimumMoney([[10**9, 0],[0, 10**9]]) == 10**9  # stress test large numbers
Test Why
[[2,1],[5,0],[4,2]] all transactions reduce money; tests sum of losses
[[3,0],[0,3]] mix of gain and loss; tests correct combination
[[0,0]] minimum values; zero cost edge case
[[5,5]] neutral transaction; ensures max cost handling
[[1,5],[7,2],[3,3]] mix of gain/loss; checks proper classification
[[10^9,0],[0,10^9]] stress test with large values

Edge Cases

Edge Case 1: All transactions are gains or neutral

If costi <= cashbacki for all transactions, the initial money only needs to cover the maximum cost among them. The algorithm handles this by tracking max_gain_cost.

Edge Case 2: All transactions are losses

If every transaction reduces money, the algorithm must sum all net losses to guarantee that the starting money is enough. This is done through sum_loss.

Edge Case 3: Transactions with zero cost

Transactions with zero cost and non-zero cashback are safe and do not increase the required starting money. The algorithm correctly ignores these for sum_loss but considers their cost in max_gain_cost if applicable.

This approach is robust to all combinations of cost and cashback values and guarantees correctness.