Candy LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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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

Candy Solutions

Time: O(n)
Space: O(n)

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;
  }
};

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;
  }
}

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

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