LeetCode #2412 — HARD

Minimum Money Required Before Transactions

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

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 >= cost_i must hold true. After performing a transaction, money becomes money - cost_i + cashback_i.

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 <= cost_i, cashback_i <= 10⁹

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[[2,1],[5,0],[4,2]]

Example 2

[[3,0],[0,3]]

Step 02

Core Insight

What unlocks the optimal approach

Split transactions that have cashback greater or equal to cost apart from transactions that have cashback less than cost. You will always <strong>earn</strong> money in the first scenario.
For transactions that have cashback greater or equal to cost, sort them by cost in descending order.
For transactions that have cashback less than cost, sort them by cashback in ascending order.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2412: Minimum Money Required Before Transactions
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;
    }
}

// Accepted solution for LeetCode #2412: Minimum Money Required Before Transactions
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)
}

# Accepted solution for LeetCode #2412: Minimum Money Required Before Transactions
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

// Accepted solution for LeetCode #2412: Minimum Money Required Before Transactions
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
    }
}

// Accepted solution for LeetCode #2412: Minimum Money Required Before Transactions
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.