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Formatted question description: https://leetcode.ca/all/1561.html
1561. Maximum Number of Coins You Can Get (Medium)
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 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 which 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 <= 10^5piles.length % 3 == 01 <= piles[i] <= 10^4
Related Topics:
Sort
Solution 1.
// OJ: https://leetcode.com/problems/maximum-number-of-coins-you-can-get/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int maxCoins(vector<int>& A) {
sort(begin(A), end(A));
int i = 0, j = A.size() - 1, ans = 0;
while (i < j) {
++i;
--j;
ans += A[j--];
}
return ans;
}
};
Solution 2.
// OJ: https://leetcode.com/problems/maximum-number-of-coins-you-can-get/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int maxCoins(vector<int>& A) {
sort(begin(A), end(A));
int N = A.size(), ans = 0;
for (int i = N / 3; i < N; i += 2) ans += A[i];
return ans;
}
};
Java
-
class Solution { public int maxCoins(int[] piles) { Arrays.sort(piles); int length = piles.length; int rounds = length / 3; int maxCoins = 0; int index = length - 2; for (int i = 0; i < rounds; i++) { maxCoins += piles[index]; index -= 2; } return maxCoins; } } -
// OJ: https://leetcode.com/problems/maximum-number-of-coins-you-can-get/ // Time: O(NlogN) // Space: O(1) class Solution { public: int maxCoins(vector<int>& A) { sort(begin(A), end(A)); int i = 0, j = A.size() - 1, ans = 0; while (i < j) { ++i; --j; ans += A[j--]; } return ans; } }; -
class Solution: def maxCoins(self, piles: List[int]) -> int: piles.sort() return sum(piles[-2 : len(piles) // 3 - 1 : -2])