LeetCode in Kotlin

2412. Minimum Money Required Before Transactions

Hard

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

class Solution {
    fun minimumMoney(transactions: Array<IntArray>): Long {
        var max = 0
        var sum: Long = 0
        for (transaction in transactions) {
            val diff = transaction[1] - transaction[0]
            if (diff < 0) {
                sum -= diff.toLong()
                transaction[0] += diff
            }
            max = max.coerceAtLeast(transaction[0])
        }
        return max + sum
    }
}

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