2412. Minimum Money Required Before Transactions

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 negative profits, denoted as \(s\). Then, we enumerate each transaction \(\text{transactions}[i] = [a, b]\) as the last transaction. If \(a > b\), it means the current transaction is a loss, and this transaction has already been included when we accumulated the negative profits earlier. Therefore, we update the answer with \(s + b\). Otherwise, we update the answer with \(s + a\).

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

Python3JavaC++GoTypeScriptRustJavaScript

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

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;
    }
}

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;
    }
};

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)
}

function minimumMoney(transactions: number[][]): number {
    const s = transactions.reduce((acc, [a, b]) => acc + Math.max(0, a - b), 0);
    let ans = 0;
    for (const [a, b] of transactions) {
        if (a > b) {
            ans = Math.max(ans, s + b);
        } else {
            ans = Math.max(ans, s + a);
        }
    }
    return ans;
}

impl Solution {
    pub fn minimum_money(transactions: Vec<Vec<i32>>) -> i64 {
        let mut s: i64 = 0;
        for transaction in &transactions {
            let (a, b) = (transaction[0], transaction[1]);
            s += (a - b).max(0) as i64;
        }
        let mut ans = 0;
        for transaction in &transactions {
            let (a, b) = (transaction[0], transaction[1]);
            if a > b {
                ans = ans.max(s + b as i64);
            } else {
                ans = ans.max(s + a as i64);
            }
        }
        ans
    }
}

/**
 * @param {number[][]} transactions
 * @return {number}
 */
var minimumMoney = function (transactions) {
    const s = transactions.reduce((acc, [a, b]) => acc + Math.max(0, a - b), 0);
    let ans = 0;
    for (const [a, b] of transactions) {
        if (a > b) {
            ans = Math.max(ans, s + b);
        } else {
            ans = Math.max(ans, s + a);
        }
    }
    return ans;
};

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