How to Solve Stone Game I X Leetcode Problem - Interview Coder Guide
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How to Solve Stone Game I X Problem

Master the Stone Game I X LeetCode problem with undetectable real-time assistance. Get instant solutions and explanations during your coding interviews.

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Medium#2156
LeetCode Problem

Stone Game IX

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...

ArrayMathGreedy+2 more

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Problem Breakdown

Understanding the Stone Game I X Problem

Let's break down this LeetCode problem and understand what makes it challenging in interview settings.

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.

MediumProblem #2156
LeetCode

Stone Game IX

Related Topics
ArrayMathGreedyCountingGame Theory
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Examples

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

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