How do you solve Candy on LeetCode?

Candy (LeetCode #135) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Each child must have at least one candy, which is the base case.

What is the brute force solution for Candy?

The brute force approach for Candy is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we can start by giving each child one candy and then iteratively adjust the candies based on the ratings. This method is straightforward but inefficient as it may require multiple passes through the list. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Candy?

Candy uses the following concepts: Array, Greedy. The recommended patterns to study are: Greedy Algorithm, Array Manipulation.

What are the interview tips for Candy?

Clearly explain your thought process while solving the problem, especially when transitioning between different approaches. Use examples to illustrate your points during the interview, as this can help clarify your reasoning. Practice coding the solution on a whiteboard or in a shared coding environment to simulate the interview experience.

What are common mistakes when solving Candy?

Failing to account for both left-to-right and right-to-left passes, which can lead to incorrect distributions. Not initializing the candies array correctly, which can cause index errors or incorrect results.

#135

Candy

Hard

ArrayGreedyGreedy AlgorithmArray Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses two passes through the ratings array to ensure that the candy distribution meets the requirements efficiently. This method is more efficient because it only requires linear time while ensuring that all conditions are met.

⚙️

Algorithm

4 steps

1Step 1: Initialize an array 'candies' of the same length as 'ratings', with each element set to 1.
2Step 2: Traverse the 'ratings' array from left to right. If a child has a higher rating than the previous child, give them one more candy than the previous child.
3Step 3: Traverse the 'ratings' array from right to left. If a child has a higher rating than the next child, ensure they have more candies than the next child, adjusting if necessary.
4Step 4: Sum all the values in the 'candies' array to get the total number of candies needed.

solution.py10 lines

1def candy(ratings):
2    n = len(ratings)
3    candies = [1] * n
4    for i in range(1, n):
5        if ratings[i] > ratings[i - 1]:
6            candies[i] = candies[i - 1] + 1
7    for i in range(n - 2, -1, -1):
8        if ratings[i] > ratings[i + 1]:
9            candies[i] = max(candies[i], candies[i + 1] + 1)
10    return sum(candies)

ℹ

Complexity note: The time complexity is O(n) because we traverse the ratings array twice. The space complexity is O(n) due to the additional candies array used to store the candy distribution.

1Each child must have at least one candy, which is the base case.
2The relationship between ratings and candies is directional; we need to check both left and right neighbors.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.