LeetCode LCP 06. Get coins

【LetMeFly】LCP 06. Get coins

Leetcode question link: https://leetcode.cn/problems/na-ying-bi/

There are npiles of coins on the table, and the number of each pile is stored in an array coins. We can choose any pile each time, take away one or two of them, and find the minimum number of times to deduct coins after taking all the coins.

Example 1:

enter:[4,2,1]

Output:4

Explanation: The first pile of force coins needs to be taken at least 2 times, the second pile needs to be taken at least 1 time, the third pile needs to be taken at least 1 time, and a total of 4 times can be obtained.

Example 2:

enter:[2,3,10]

Output:8

limit:

• 1 <= n <= 4
• 1 <= coins[i] <= 10

Method 1: Traverse

If you can get 1 or 2 coins each time, you must get as many as possible. For a bunch of $n$ coins, the minimum number of times required is$\lceil \frac{n}{2}\rceil$。

Tip: $\lfloor \frac{n+1}{2}\rfloor =\lceil \frac{n}{2}\rceil$

• Time complexity $O(len(coins))$
• Space complexity $O(1)$

AC code

C++

class Solution {
    
    
public:
    int minCount(vector<int>& coins) {
    
    
        int ans = 0;
        for (int t : coins) {
    
    
            ans += (t + 1)/ 2;
        }
        return ans;
    }
};

Python

# from typing import List

class Solution:
    def minCount(self, coins: List[int]) -> int:
        return sum((i + 1) // 2 for i in coins)

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