# LC 135. Candy

### [Problem link](https://leetcode.com/problems/candy/)

###### tags: `leedcode` `python` `c++` `hard` `Greedy`

There are <code>n</code> children standing in a line. Each child is assigned a rating value given in the integer array <code>ratings</code>.

You are giving candies to these children subjected to the following requirements:

- Each child must have at least one candy.
- Children with a higher rating get more candies than their neighbors.

Return the minimum number of candies you need to have to distribute the candies to the children.

**Example 1:**

Input: ratings = [1,0,2]
Output: 5
Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.

**Example 2:**

Input: ratings = [1,2,2]
Output: 4
Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
The third child gets 1 candy because it satisfies the above two conditions.

**Constraints:**

- <code>n == ratings.length</code>
- <code>1 <= n <= 2 * 10⁴</code>
- <code>0 <= ratings[i] <= 2 * 10⁴</code>

## Solution 1 - Greedy
#### Python

class Solution:
    def candy(self, ratings: List[int]) -> int:
        c = [1] * len(ratings)

        # Compare with the left child
        for i in range(1, len(ratings)):
            if ratings[i - 1] < ratings[i]:
                c[i] = c[i - 1] + 1

        # Compare with the right child, then compare the size.
        for i in range(len(ratings) - 2, -1, -1):
            if ratings[i] > ratings[i + 1]:
                c[i] = max(c[i], c[i + 1] + 1)

        return sum(c)

#### C++

class Solution {
public:
    int candy(vector<int>& ratings) {
        int n = ratings.size();
        vector<int> cnt(n, 0);

        for (int i = 1; i < n; i++) {
            if (ratings[i] > ratings[i - 1]) {
                cnt[i] = cnt[i - 1] + 1;
            }
        }
        for (int i = n - 2; i >= 0; i--) {
            if (ratings[i] > ratings[i + 1]) {
                cnt[i] = max(cnt[i], cnt[i + 1] + 1);
            }
        }
        int res = accumulate(cnt.begin(), cnt.end(), 0) + n;
        return res;
    }
};

>### Complexity
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(n) | O(n) |

## Note
sol1:
先從左遍歷到右, 再從右遍歷到左
[reference](https://github.com/youngyangyang04/leetcode-master/blob/master/problems/0135.%E5%88%86%E5%8F%91%E7%B3%96%E6%9E%9C.md)