LeetCode #2029 — MEDIUM

Stone Game IX

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

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The Problem

Problem Statement

Alice and Bob continue their games with stones. There is a row of n stones, and each stone has an associated value. You are given an integer array stones, where stones[i] is the value of the ith stone.

Alice and Bob take turns, with Alice starting first. On each turn, the player may remove any stone from stones. The player who removes a stone loses if the sum of the values of all removed stones is divisible by 3. Bob will win automatically if there are no remaining stones (even if it is Alice's turn).

Assuming both players play optimally, return true if Alice wins and false if Bob wins.

Example 1:

Input: stones = [2,1]
Output: true
Explanation: The game will be played as follows:
- Turn 1: Alice can remove either stone.
- Turn 2: Bob removes the remaining stone. 
The sum of the removed stones is 1 + 2 = 3 and is divisible by 3. Therefore, Bob loses and Alice wins the game.

Example 2:

Input: stones = [2]
Output: false
Explanation: Alice will remove the only stone, and the sum of the values on the removed stones is 2. 
Since all the stones are removed and the sum of values is not divisible by 3, Bob wins the game.

Example 3:

Input: stones = [5,1,2,4,3]
Output: false
Explanation: Bob will always win. One possible way for Bob to win is shown below:
- Turn 1: Alice can remove the second stone with value 1. Sum of removed stones = 1.
- Turn 2: Bob removes the fifth stone with value 3. Sum of removed stones = 1 + 3 = 4.
- Turn 3: Alices removes the fourth stone with value 4. Sum of removed stones = 1 + 3 + 4 = 8.
- Turn 4: Bob removes the third stone with value 2. Sum of removed stones = 1 + 3 + 4 + 2 = 10.
- Turn 5: Alice removes the first stone with value 5. Sum of removed stones = 1 + 3 + 4 + 2 + 5 = 15.
Alice loses the game because the sum of the removed stones (15) is divisible by 3. Bob wins the game.

Constraints:

  • 1 <= stones.length <= 105
  • 1 <= stones[i] <= 104
Patterns Used

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Alice and Bob continue their games with stones. There is a row of n stones, and each stone has an associated value. You are given an integer array stones, where stones[i] is the value of the ith stone. Alice and Bob take turns, with Alice starting first. On each turn, the player may remove any stone from stones. The player who removes a stone loses if the sum of the values of all removed stones is divisible by 3. Bob will win automatically if there are no remaining stones (even if it is Alice's turn). Assuming both players play optimally, return true if Alice wins and false if Bob wins.

Baseline thinking

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

Pattern signal: Array · Math · Greedy

Example 1

[2,1]

Example 2

[2]

Example 3

[5,1,2,4,3]

Related Problems

  • Stone Game (stone-game)
  • Stone Game II (stone-game-ii)
  • Stone Game III (stone-game-iii)
  • Stone Game IV (stone-game-iv)
  • Stone Game V (stone-game-v)
Step 02

Core Insight

What unlocks the optimal approach

  • There are limited outcomes given the current sum and the stones remaining.
  • Can we greedily simulate starting with taking a stone with remainder 1 or 2 divided by 3?
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #2029: Stone Game IX
class Solution {
    public boolean stoneGameIX(int[] stones) {
        int[] c1 = new int[3];
        for (int x : stones) {
            c1[x % 3]++;
        }
        int[] c2 = {c1[0], c1[2], c1[1]};
        return check(c1) || check(c2);
    }

    private boolean check(int[] cnt) {
        if (--cnt[1] < 0) {
            return false;
        }
        int r = 1 + Math.min(cnt[1], cnt[2]) * 2 + cnt[0];
        if (cnt[1] > cnt[2]) {
            --cnt[1];
            ++r;
        }
        return r % 2 == 1 && cnt[1] != cnt[2];
    }
}
Step 06

Interactive Study Demo

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

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.

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.