135. Candy

There are N children standing in a line. Each child is assigned a rating value.

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.

What is the minimum candies you must give?

Example 1:

Input: [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: [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.

 

Approach #1: C++. [Using two array]

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

　　

Approach #2: Java. [Using one array]

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

　　

Approach #3: Java. [Single Pass Approach with Constant Space]

class Solution {
    public int count(int n) {
        return (n * (n + 1)) / 2;
    }
    
    public int candy(int[] ratings) {
        if (ratings.length <= 1) return ratings.length;
        int candies = 0;
        int up = 0;
        int down = 0;
        int old_slope = 0;
        
        for (int i = 1; i < ratings.length; ++i) {
            int new_slope = ratings[i] > ratings[i-1] ? 1 : ratings[i] < ratings[i-1] ? -1 : 0;
            if (old_slope > 0 && new_slope == 0 || old_slope < 0 && new_slope >= 0) {
                candies += count(up) + count(down) + Math.max(up, down);
                up = 0;
                down = 0;
            }
            if (new_slope > 0) up++;
            if (new_slope < 0) down++;
            if (new_slope == 0) candies++;
            old_slope = new_slope;
        }
        
        candies += count(up) + count(down) + Math.max(up, down) + 1;
        
        return candies;
    }
}

　　

reference:

https://leetcode.com/problems/candy/solution/