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.

// Two Pass Method
int candy(vector<int>& ratings) { // time: O(n); space: O(n)
    int n = ratings.size(), res = 0;
    vector<int> nums(n, 1);
    for (int i = 0; i < n - 1; ++i) {
        if (ratings[i + 1] > ratings[i]) nums[i + 1] = nums[i] + 1;
    }
    for (int i = n - 1; i > 0; --i) {
        if (ratings[i - 1] > ratings[i]) nums[i - 1] = max(nums[i - 1], nums[i] + 1);
    }
    // for (int num : nums) res += num;
    res = accumulate(nums.begin(), nums.end(), 0);
    return res;
}

一開始給第一個人1個糖果，然後對於下一個人有三種情形。 1. 下一個人的rating == 前一個人的rating，給1個糖果就可以 2. 下一個人的rating > 前一個人的rating，比前一個人多給1個糖果 3. 下一個人的rating < 前一個人的rating，我們得計算連續遞減的ratings有多少個，最後可能需要再開始遞減的前一個人補糖果，因為假設後面遞減很多，那麼每個人減少一個糖果，可是題目條件限制每個人至少拿一個，也就是糖果數量不可以遞減至少於0。可以用例子:[1, 3, 2, 1]來想想，一開始給第一個人1個，第二個人給2個的話，這樣會不夠，所以最後需要補上一些糖果。

// One Pass Method
int candy(vector<int>& ratings) { // time: O(n); space: O(1)
    if (ratings.empty()) return 0;
    // pre: the candies for the student before this descending sequences of students
    // cnt: the number of descending ratings except the first one in the descending sequence
    int res = 1, pre = 1, cnt = 0;
    for (int i = 1; i < ratings.size(); ++i) {
        if (ratings[i] >= ratings[i - 1]) {
            if (cnt > 0) {
                res += cnt * (cnt + 1) / 2;
                if (cnt >= pre) res += cnt - pre + 1;
                cnt = 0;
                pre = 1;
            }
            pre = (ratings[i] == ratings[i - 1]) ? 1 : pre + 1;
            res += pre;
        } else {
            ++cnt;
        }
    }
    if (cnt > 0) {
        res += cnt * (cnt + 1) / 2;
        if (cnt >= pre) res += cnt - pre + 1;
    }
    return res;
}