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1. Description

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

2. Example

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.

3. Constraints

• n == ratings.length
• 1 <= n <= 2 * $10^4$
• 0 <= ratings[i] <= 2 * $10^4$

4. Solutions

Two Iterations

n = ratings.size()
Time complexity: O(n)
Space complexity: O(n)

class Solution {
public:
    int candy(const vector<int>& ratings) {
        vector<int> candy_to_right(ratings.size(), 1);
        for (int i = 1; i < candy_to_right.size(); ++i) {
            if (ratings[i] > ratings[i - 1]) {
                candy_to_right[i] = candy_to_right[i - 1] + 1;
            }
        }
        
        int candy_count = *candy_to_right.rbegin();
        for (int i = candy_to_right.size() - 2, right_to_left = 1; i >= 0; --i) {
            if(ratings[i] > ratings[i + 1]) {
                ++right_to_left;
            } else {
                right_to_left = 1;
            }

            candy_count += max(right_to_left, candy_to_right[i]);
        }   

        return candy_count;
    }
};