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Hard
Greedy
Array

135. Candy

中文文档

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.

Python3

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))

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

C#

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

Java

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;
    }
}