README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Hard	https://github.com/doocs/leetcode/edit/main/solution/2400-2499/2412.Minimum%20Money%20Required%20Before%20Transactions/README_EN.md	2092	Biweekly Contest 87 Q4	Greedy Array Sorting

2412. Minimum Money Required Before Transactions

中文文档

Description

You are given a 0-indexed 2D integer array transactions, where transactions[i] = [costi, cashbacki].

The array describes transactions, where each transaction must be completed exactly once in some order. At any given moment, you have a certain amount of money. In order to complete transaction i, money >= costi must hold true. After performing a transaction, money becomes money - costi + cashbacki.

Return the minimum amount of money required before any transaction so that all of the transactions can be completed regardless of the order of the transactions.

 

Example 1:

Input: transactions = [[2,1],[5,0],[4,2]]
Output: 10
Explanation:
Starting with money = 10, the transactions can be performed in any order.
It can be shown that starting with money < 10 will fail to complete all transactions in some order.

Example 2:

Input: transactions = [[3,0],[0,3]]
Output: 3
Explanation:
- If transactions are in the order [[3,0],[0,3]], the minimum money required to complete the transactions is 3.
- If transactions are in the order [[0,3],[3,0]], the minimum money required to complete the transactions is 0.
Thus, starting with money = 3, the transactions can be performed in any order.

 

Constraints:

• 1 <= transactions.length <= 105
• transactions[i].length == 2
• 0 <= costi, cashbacki <= 109

Solutions

Solution 1: Greedy

First, we accumulate all the negative profits, denoted as $s$. Then we enumerate each transaction as the last transaction. If transactions[i].x > transactions[i].y, it means the current transaction is losing money, and this transaction has been calculated when we previously accumulated negative profits, so we update the answer with s + transactions[i].y; otherwise, we update the answer with s + transactions[i].x.

The time complexity is $O(n)$, where $n$ is the number of transactions. The space complexity is $O(1)$.

Python3

class Solution:
    def minimumMoney(self, transactions: List[List[int]]) -> int:
        s = sum(max(0, a - b) for a, b in transactions)
        ans = 0
        for a, b in transactions:
            if a > b:
                ans = max(ans, s + b)
            else:
                ans = max(ans, s + a)
        return ans

Java

class Solution {
    public long minimumMoney(int[][] transactions) {
        long s = 0;
        for (var e : transactions) {
            s += Math.max(0, e[0] - e[1]);
        }
        long ans = 0;
        for (var e : transactions) {
            if (e[0] > e[1]) {
                ans = Math.max(ans, s + e[1]);
            } else {
                ans = Math.max(ans, s + e[0]);
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    long long minimumMoney(vector<vector<int>>& transactions) {
        long long s = 0, ans = 0;
        for (auto& e : transactions) {
            s += max(0, e[0] - e[1]);
        }
        for (auto& e : transactions) {
            if (e[0] > e[1]) {
                ans = max(ans, s + e[1]);
            } else {
                ans = max(ans, s + e[0]);
            }
        }
        return ans;
    }
};

Go

func minimumMoney(transactions [][]int) int64 {
	s, ans := 0, 0
	for _, e := range transactions {
		s += max(0, e[0]-e[1])
	}
	for _, e := range transactions {
		if e[0] > e[1] {
			ans = max(ans, s+e[1])
		} else {
			ans = max(ans, s+e[0])
		}
	}
	return int64(ans)
}