135

135. Candy (Hard)

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

Companies:
Amazon, Apple

Related Topics:
Array, Greedy

Solution 1. BFS

// OJ: https://leetcode.com/problems/candy/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int candy(vector<int>& A) {
        int N = A.size();
        vector<int> v(N);
        queue<int> q;
        for (int i = 0; i < N; ++i) {
            if ((i - 1 < 0 || A[i] <= A[i - 1]) && (i + 1 >= N || A[i] <= A[i + 1])) {
                q.push(i);
                v[i] = 1;
            }
        }
        while (q.size()) {
            int i = q.front();
            q.pop();
            for (int j = -1; j <= 1; j += 2) {
                int k = i + j;
                if (k < 0 || k >= N) continue;
                if (A[k] > A[i] && v[k] < 1 + v[i]) {
                    v[k] = 1 + v[i];
                    q.push(k);
                } 
            }
        }
        return accumulate(begin(v), end(v), 0);
    }
};

TODO

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