135. Candy LeetCode Solution

Last updated: February 14, 2025

In this guide, you will get 135. Candy LeetCode Solution with the best time and space complexity. The solution to Candy problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Candy solution in C++
Candy solution in Java
Candy solution in Python
Additional Resources

Problem Statement of Candy

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

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

135. Candy LeetCode Solution in C++

class Solution {
 public:
  int candy(vector<int>& ratings) {
    const int n = ratings.size();
    int ans = 0;
    vector<int> l(n, 1);
    vector<int> r(n, 1);

    for (int i = 1; i < n; ++i)
      if (ratings[i] > ratings[i - 1])
        l[i] = l[i - 1] + 1;

    for (int i = n - 2; i >= 0; --i)
      if (ratings[i] > ratings[i + 1])
        r[i] = r[i + 1] + 1;

    for (int i = 0; i < n; ++i)
      ans += max(l[i], r[i]);

    return ans;
  }
};
/* code provided by PROGIEZ */

135. Candy LeetCode Solution in Java

class Solution {
  public int candy(int[] ratings) {
    final int n = ratings.length;

    int ans = 0;
    int[] l = new int[n];
    int[] r = new int[n];
    Arrays.fill(l, 1);
    Arrays.fill(r, 1);

    for (int i = 1; i < n; ++i)
      if (ratings[i] > ratings[i - 1])
        l[i] = l[i - 1] + 1;

    for (int i = n - 2; i >= 0; --i)
      if (ratings[i] > ratings[i + 1])
        r[i] = r[i + 1] + 1;

    for (int i = 0; i < n; ++i)
      ans += Math.max(l[i], r[i]);

    return ans;
  }
}
// code provided by PROGIEZ

135. Candy LeetCode Solution in Python

class Solution:
  def candy(self, ratings: list[int]) -> int:
    n = len(ratings)

    ans = 0
    l = [1] * n
    r = [1] * n

    for i in range(1, n):
      if ratings[i] > ratings[i - 1]:
        l[i] = l[i - 1] + 1

    for i in range(n - 2, -1, -1):
      if ratings[i] > ratings[i + 1]:
        r[i] = r[i + 1] + 1

    for a, b in zip(l, r):
      ans += max(a, b)

    return ans
 # code by PROGIEZ

Additional Resources

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