> For the complete documentation index, see [llms.txt](https://soumyajit4419.gitbook.io/ds-algo/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://soumyajit4419.gitbook.io/ds-algo/arrays/16.candy.md).

# 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.**

```cpp
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;
    }
};
```