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Formatted question description: https://leetcode.ca/all/2412.html

2412. Minimum Money Required Before Transactions

  • Difficulty: Hard.
  • Related Topics: .
  • Similar Questions: .

Problem

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

Solution (Java, C++, Python)

  • class Solution {
    public long minimumMoney(int[][] transactions) {
        Arrays.sort(transactions,(int a[],int b[])->(a[1]-b[1]));

        long max=0,ans=0,ab=0;
        for(int a[]:transactions){
            if(a[0]>a[1]){
                max+=a[0];
                ans=Math.max(ans,max);
                max-=a[1];
            }
            else ab=Math.max(ab,a[0]);
        }
        ans=Math.max(ans,max+ab);
        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:
    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

############

# 2412. Minimum Money Required Before Transactions
# https://leetcode.com/problems/minimum-money-required-before-transactions

class Solution:
    def minimumMoney(self, A: List[List[int]]) -> int:
        return sum(max(0, i - j) for i, j in A) + max(map(min, A))



  • 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)
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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