Take Gifts From The Richest Pile - Leetcode Solution

Problem:

You are given an integer array "piles" where piles[i] is the number of coins in the ith pile.

You and your friend are playing a game where you take turns removing piles of coins. You go first, and the player who removes the last pile wins.

You both play optimally.

Return true if you can win the game if you go first, otherwise return false.

Solution:

First, take note of the game rule that the player who removes the last pile wins. Therefore, if there is only one pile left, then the current player wins.

The key point to note here is that we both play optimally. This means that for each turn, we will always aim to maximize our chances of winning.

One way to do this is to always choose the pile with the most coins. By always choosing the largest pile, we ensure that we leave the smallest possible pile for the other player.

We can sort the piles array in descending order, and then keep alternating between removing the first and second piles.

For example, if the piles array is [3, 4, 2, 1], we can sort it to become [4, 3, 2, 1]. We can then remove the first pile (4) and the second pile (3), leaving us with [2, 1]. Now we can remove the first pile (2), leaving us with [1]. We win because we removed the last pile.

We can implement this strategy in code as follows:

def canWin(piles):
    piles.sort(reverse=True)
    total_piles = len(piles)
    turns = 0
    while total_piles > 1:
        piles.pop(0)
        total_piles -= 1
        turns += 1
        if turns % 2 == 0:
            piles.pop(0)
            total_piles -= 1
    return turns % 2 != 0

We start by sorting the piles array in descending order. We then keep track of the number of turns we have taken, which starts at 0.

We then enter a while loop as long as there are more than one piles remaining. During each iteration, we remove the first pile from the piles array, decrement the total number of piles by 1, and increment the turns variable by 1.

We then check if it is the other player's turn (i.e., turns % 2 == 0), and if it is, we remove the first pile from the piles array and decrement the total number of piles by 1.

At the end of the loop, we check if the current player wins (i.e., turns % 2 != 0). If turns is odd, then the current player has removed the last pile and therefore wins. Otherwise, the other player has removed the last pile and the current player loses.

Time Complexity:

The time complexity of this solution is O(nlogn) due to the sorting of the piles array. The while loop runs in O(n) time since we remove one pile during each iteration. Therefore, the overall time complexity is O(nlogn).

Space Complexity:

The space complexity is O(1) since we do not use any extra space other than a few variables.

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