LeetCode #2706 — EASY

Buy Two Chocolates

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given an integer array prices representing the prices of various chocolates in a store. You are also given a single integer money, which represents your initial amount of money.

You must buy exactly two chocolates in such a way that you still have some non-negative leftover money. You would like to minimize the sum of the prices of the two chocolates you buy.

Return the amount of money you will have leftover after buying the two chocolates. If there is no way for you to buy two chocolates without ending up in debt, return money. Note that the leftover must be non-negative.

Example 1:

Input: prices = [1,2,2], money = 3
Output: 0
Explanation: Purchase the chocolates priced at 1 and 2 units respectively. You will have 3 - 3 = 0 units of money afterwards. Thus, we return 0.

Example 2:

Input: prices = [3,2,3], money = 3
Output: 3
Explanation: You cannot buy 2 chocolates without going in debt, so we return 3.

Constraints:

2 <= prices.length <= 50
1 <= prices[i] <= 100
1 <= money <= 100

Step 01

Brute Force Baseline

Problem summary: You are given an integer array prices representing the prices of various chocolates in a store. You are also given a single integer money, which represents your initial amount of money. You must buy exactly two chocolates in such a way that you still have some non-negative leftover money. You would like to minimize the sum of the prices of the two chocolates you buy. Return the amount of money you will have leftover after buying the two chocolates. If there is no way for you to buy two chocolates without ending up in debt, return money. Note that the leftover must be non-negative.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[1,2,2]
3

Example 2

[3,2,3]
3

Step 02

Core Insight

What unlocks the optimal approach

Sort the array and check if the money is more than or equal to the sum of the two cheapest elements.

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

Upper-end input sizes

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 #2706: Buy Two Chocolates
class Solution {
    public int buyChoco(int[] prices, int money) {
        Arrays.sort(prices);
        int cost = prices[0] + prices[1];
        return money < cost ? money : money - cost;
    }
}

// Accepted solution for LeetCode #2706: Buy Two Chocolates
func buyChoco(prices []int, money int) int {
	sort.Ints(prices)
	cost := prices[0] + prices[1]
	if money < cost {
		return money
	}
	return money - cost
}

# Accepted solution for LeetCode #2706: Buy Two Chocolates
class Solution:
    def buyChoco(self, prices: List[int], money: int) -> int:
        prices.sort()
        cost = prices[0] + prices[1]
        return money if money < cost else money - cost

// Accepted solution for LeetCode #2706: Buy Two Chocolates
impl Solution {
    pub fn buy_choco(mut prices: Vec<i32>, money: i32) -> i32 {
        prices.sort();
        let cost = prices[0] + prices[1];
        if cost > money {
            return money;
        }
        money - cost
    }
}

// Accepted solution for LeetCode #2706: Buy Two Chocolates
function buyChoco(prices: number[], money: number): number {
    prices.sort((a, b) => a - b);
    const cost = prices[0] + prices[1];
    return money < cost ? money : money - cost;
}

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(log n)

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.