15.Candy

There are N children standing in a line. Each child is assigned a rating value. 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.

Example 1:
Input: [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: [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.

Solution : (Using two arrays)

Approach: We maintain 2 arrays left neighbor and right neighbor. Left Arr: (Assuming Right neighbour gets more candy than left) Right Arr: (Assuming Left neighbour gets more candy than right) Find the max of both the left and right.

class Solution
{
public:
    int candy(vector<int> &ratings)
    {

        int n = ratings.size();
        vector<int> ln(n, 1);
        vector<int> rn(n, 1);
        int sum = 0;
        for (int i = 0; i < n - 1; i++)
        {
            if (ratings[i + 1] > ratings[i])
            {
                rn[i + 1] = rn[i] + 1;
            }
        }

        for (int i = n - 1; i > 0; i--)
        {
            if (ratings[i - 1] > ratings[i])
            {
                ln[i - 1] = ln[i] + 1;
            }
        }

        for (int i = 0; i < n; i++)
        {
            sum = sum + (max(ln[i], rn[i]));
        }

        return sum;
    }
};