Question

135. Candy

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Description

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

Solutions

Solution 1: Two traversals

We initialize two arrays $left$ and $right$, where $left[i]$ represents the minimum number of candies the current child should get when the current child's score is higher than the left child's score, and $right[i]$ represents the minimum number of candies the current child should get when the current child's score is higher than the right child's score. Initially, $left[i]=1$, $right[i]=1$.

We traverse the array from left to right once, and if the current child's score is higher than the left child's score, then $left[i]=left[i-1]+1$; similarly, we traverse the array from right to left once, and if the current child's score is higher than the right child's score, then $right[i]=right[i+1]+1$.

Finally, we traverse the array of scores once, and the minimum number of candies each child should get is the maximum of $left[i]$ and $right[i]$, and we add them up to get the answer.

Time complexity $O(n)$, space complexity $O(n)$. Where $n$ is the length of the array of scores.

class Solution:
  def candy(self, ratings: List[int]) -> int:
    n = len(ratings)
    left = [1] * n
    right = [1] * n
    for i in range(1, n):
      if ratings[i] > ratings[i - 1]:
        left[i] = left[i - 1] + 1
    for i in range(n - 2, -1, -1):
      if ratings[i] > ratings[i + 1]:
        right[i] = right[i + 1] + 1
    return sum(max(a, b) for a, b in zip(left, right))

class Solution {
  public int candy(int[] ratings) {
    int n = ratings.length;
    int[] left = new int[n];
    int[] right = new int[n];
    Arrays.fill(left, 1);
    Arrays.fill(right, 1);
    for (int i = 1; i < n; ++i) {
      if (ratings[i] > ratings[i - 1]) {
        left[i] = left[i - 1] + 1;
      }
    }
    for (int i = n - 2; i >= 0; --i) {
      if (ratings[i] > ratings[i + 1]) {
        right[i] = right[i + 1] + 1;
      }
    }
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      ans += Math.max(left[i], right[i]);
    }
    return ans;
  }
}

class Solution {
public:
  int candy(vector<int>& ratings) {
    int n = ratings.size();
    vector<int> left(n, 1);
    vector<int> right(n, 1);
    for (int i = 1; i < n; ++i) {
      if (ratings[i] > ratings[i - 1]) {
        left[i] = left[i - 1] + 1;
      }
    }
    for (int i = n - 2; ~i; --i) {
      if (ratings[i] > ratings[i + 1]) {
        right[i] = right[i + 1] + 1;
      }
    }
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      ans += max(left[i], right[i]);
    }
    return ans;
  }
};

func candy(ratings []int) int {
  n := len(ratings)
  left := make([]int, n)
  right := make([]int, n)
  for i := range left {
    left[i] = 1
    right[i] = 1
  }
  for i := 1; i < n; i++ {
    if ratings[i] > ratings[i-1] {
      left[i] = left[i-1] + 1
    }
  }
  for i := n - 2; i >= 0; i-- {
    if ratings[i] > ratings[i+1] {
      right[i] = right[i+1] + 1
    }
  }
  ans := 0
  for i, a := range left {
    b := right[i]
    ans += max(a, b)
  }
  return ans
}

function candy(ratings: number[]): number {
  const n = ratings.length;
  const left = new Array(n).fill(1);
  const right = new Array(n).fill(1);
  for (let i = 1; i < n; ++i) {
    if (ratings[i] > ratings[i - 1]) {
      left[i] = left[i - 1] + 1;
    }
  }
  for (let i = n - 2; i >= 0; --i) {
    if (ratings[i] > ratings[i + 1]) {
      right[i] = right[i + 1] + 1;
    }
  }
  let ans = 0;
  for (let i = 0; i < n; ++i) {
    ans += Math.max(left[i], right[i]);
  }
  return ans;
}

public class Solution {
  public int Candy(int[] ratings) {
    int n = ratings.Length;
    int[] left = new int[n];
    int[] right = new int[n];
    Array.Fill(left, 1);
    Array.Fill(right, 1);
    for (int i = 1; i < n; ++i) {
      if (ratings[i] > ratings[i - 1]) {
        left[i] = left[i - 1] + 1;
      }
    }
    for (int i = n - 2; i >= 0; --i) {
      if (ratings[i] > ratings[i + 1]) {
        right[i] = right[i + 1] + 1;
      }
    }
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      ans += Math.Max(left[i], right[i]);
    }
    return ans;
  }
}

Solution 2

class Solution {
  public int candy(int[] ratings) {
    int n = ratings.length;
    int up = 0;
    int down = 0;
    int peak = 0;
    int candies = 1;
    for (int i = 1; i < n; i++) {
      if (ratings[i - 1] < ratings[i]) {
        up++;
        peak = up + 1;
        down = 0;
        candies += peak;
      } else if (ratings[i] == ratings[i - 1]) {
        peak = 0;
        up = 0;
        down = 0;
        candies++;
      } else {
        down++;
        up = 0;
        candies += down + (peak > down ? 0 : 1);
      }
    }
    return candies;
  }
}