Leetcode 0135. distribute candy

【LetMeFly】135. Distribute candy

Force button topic link ：https://leetcode.cn/problems/candy/

n Two kids in a row . Give you an array of integers ratings Indicates the score of each child .

You need to follow these requirements , Distribute candy to these children ：

• Each child is assigned at least 1 A candy .
• Children with higher scores from two adjacent children will get more candy .

Please distribute candy to each child , Calculate and return what needs to be prepared Minimum number of sweets .

 

Example  1：

 Input ：ratings = [1,0,2]
 Output ：5
 explain ： You can give the first one separately 、 the second 、 The third child distributed  2、1、2  Candy .

Example  2：

 Input ：ratings = [1,2,2]
 Output ：4
 explain ： You can give the first one separately 、 the second 、 The third child distributed  1、2、1  Candy .
      The third child only got  1  Candy , This satisfies two conditions in the problem surface .

 

Tips ：

• n == ratings.length
• 1 <= n <= 2 * 104
• 0 <= ratings[i] <= 2 * 104

Method 1 ： Find the smallest

The idea is simple ： Find all of them first “ Minima ”（ Previous rating And the latter rating Are greater than this rating）

Then for each minimum , from $1$ Start distributing candy , And continue to extend to both sides , Until it no longer increases .

During extension , The amount of candy distributed is accumulated .

after , Go through it again “ Candy allocation array ”, If you encounter adjacent The unqualified situation , Correct the allocation of non call and condition ：

• Time complexity $O(n)$, among $n$ It's the number of children
• Spatial complexity $O(n)$

AC Code

C++

class Solution {
    
public:
    int candy(vector<int>& ratings) {
    
        vector<int> mins;
        vector<int> candies(ratings.size());
		//  look for “ Minima ”
        for (int i = 0; i < ratings.size(); i++) {
    
            if ((i - 1 >= 0 && ratings[i - 1] < ratings[i]) || (i + 1 < ratings.size() && ratings[i + 1] < ratings[i])) {
    
                continue;
            }
            mins.push_back(i);
        }

		//  Start from the minimum point and expand to both sides 
        for (int thisMin : mins) {
    
            int thisCandy = 1;
            int i = thisMin;
            while (true) {
    
                candies[i] = thisCandy;
                thisCandy++;
                if (i - 1 >= 0 && ratings[i - 1] > ratings[i]) {
    
                    i--;
                }
                else {
    
                    break;
                }
            }
            i = thisMin;
            thisCandy = 1;
            while (true) {
    
                candies[i] = thisCandy;
                thisCandy++;
                if (i + 1 < ratings.size() && ratings[i + 1] > ratings[i]) {
    
                    i++;
                }
                else {
    
                    break;
                }
            }
        }
        
		//  Update the adjacent conditions that do not meet 
        for (int i = 1; i < candies.size(); i++) {
    
            if (ratings[i - 1] > ratings[i] && candies[i - 1] <= candies[i]) {
    
                candies[i - 1] = candies[i] + 1;
            }
            if (ratings[i] > ratings[i - 1] && candies[i] <= candies[i - 1]) {
    
                candies[i] = candies[i - 1] + 1;
            }
        }
		
		//  Sum by accumulation 
        int ans = 0;
        for (int& t : candies)
            ans += t;
        return ans;
    }
};

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