# Solving Leetcode Interviews in Seconds with AI: Candy


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "135" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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.   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:  n == ratings.length 1 <= n <= 2 * 104 0 <= ratings[i] <= 2 * 104  

	# Explanation
	Here's the solution to the candy distribution problem:

*   **Greedy Approach:** The solution uses a greedy approach to distribute candies. First, it assigns each child one candy.
*   **Left to Right Pass:** It then iterates from left to right, ensuring that if a child has a higher rating than their left neighbor, they receive one more candy than their neighbor.
*   **Right to Left Pass:** Finally, it iterates from right to left, ensuring that if a child has a higher rating than their right neighbor, and they have fewer candies than their neighbor, they receive one more candy than their neighbor.

*   **Runtime Complexity:** O(n), where n is the number of children. **Storage Complexity:** O(n)

	
	# Code
	```python
	def candy(ratings):
    n = len(ratings)
    candies = [1] * n

    # Left to right pass
    for i in range(1, n):
        if ratings[i] > ratings[i - 1]:
            candies[i] = candies[i - 1] + 1

    # Right to left pass
    for i in range(n - 2, -1, -1):
        if ratings[i] > ratings[i + 1]:
            candies[i] = max(candies[i], candies[i + 1] + 1)

    return sum(candies)
	```