LeetCode #1561 — MEDIUM

Maximum Number of Coins You Can Get

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

There are 3n piles of coins of varying size, you and your friends will take piles of coins as follows:

In each step, you will choose any 3 piles of coins (not necessarily consecutive).
Of your choice, Alice will pick the pile with the maximum number of coins.
You will pick the next pile with the maximum number of coins.
Your friend Bob will pick the last pile.
Repeat until there are no more piles of coins.

Given an array of integers piles where piles[i] is the number of coins in the i^th pile.

Return the maximum number of coins that you can have.

Example 1:

Input: piles = [2,4,1,2,7,8]
Output: 9
Explanation: Choose the triplet (2, 7, 8), Alice Pick the pile with 8 coins, you the pile with 7 coins and Bob the last one.
Choose the triplet (1, 2, 4), Alice Pick the pile with 4 coins, you the pile with 2 coins and Bob the last one.
The maximum number of coins which you can have are: 7 + 2 = 9.
On the other hand if we choose this arrangement (1, 2, 8), (2, 4, 7) you only get 2 + 4 = 6 coins which is not optimal.

Example 2:

Input: piles = [2,4,5]
Output: 4

Example 3:

Input: piles = [9,8,7,6,5,1,2,3,4]
Output: 18

Constraints:

3 <= piles.length <= 10⁵
piles.length % 3 == 0
1 <= piles[i] <= 10⁴

Step 01

Brute Force Baseline

Problem summary: There are 3n piles of coins of varying size, you and your friends will take piles of coins as follows: In each step, you will choose any 3 piles of coins (not necessarily consecutive). Of your choice, Alice will pick the pile with the maximum number of coins. You will pick the next pile with the maximum number of coins. Your friend Bob will pick the last pile. Repeat until there are no more piles of coins. Given an array of integers piles where piles[i] is the number of coins in the ith pile. Return the maximum number of coins that you can have.

Baseline thinking

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

Pattern signal: Array · Math · Greedy

Example 1

[2,4,1,2,7,8]

Example 2

[2,4,5]

Example 3

[9,8,7,6,5,1,2,3,4]

Step 02

Core Insight

What unlocks the optimal approach

Which pile of coins will you never be able to pick up?
Bob is forced to take the last pile of coins, no matter what it is. Which pile should you give to him?

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 #1561: Maximum Number of Coins You Can Get
class Solution {
    public int maxCoins(int[] piles) {
        Arrays.sort(piles);
        int ans = 0;
        for (int i = piles.length / 3; i < piles.length; i += 2) {
            ans += piles[i];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1561: Maximum Number of Coins You Can Get
func maxCoins(piles []int) (ans int) {
	sort.Ints(piles)
	for i := len(piles) / 3; i < len(piles); i += 2 {
		ans += piles[i]
	}
	return
}

# Accepted solution for LeetCode #1561: Maximum Number of Coins You Can Get
class Solution:
    def maxCoins(self, piles: List[int]) -> int:
        piles.sort()
        return sum(piles[len(piles) // 3 :][::2])

// Accepted solution for LeetCode #1561: Maximum Number of Coins You Can Get
impl Solution {
    pub fn max_coins(mut piles: Vec<i32>) -> i32 {
        piles.sort();
        let mut ans = 0;
        for i in (piles.len() / 3..piles.len()).step_by(2) {
            ans += piles[i];
        }
        ans
    }
}

// Accepted solution for LeetCode #1561: Maximum Number of Coins You Can Get
function maxCoins(piles: number[]): number {
    piles.sort((a, b) => a - b);
    let ans = 0;
    for (let i = piles.length / 3; i < piles.length; i += 2) {
        ans += piles[i];
    }
    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(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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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.