Candy

Problem Statement:

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:

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.

✅ Solution: Two-Pass Greedy (Left-to-Right & Right-to-Left)

class Solution {
public:
    int candy(vector<int>& ratings) {
        int n = ratings.size();
        vector<int> candies(n, 1); // Step 1: Give 1 candy to each child

        // Step 2: Left to Right - if current rating > previous, give more candy
        for(int i = 1; i < n; i++){
            if(ratings[i] > ratings[i - 1]){
                candies[i] = candies[i - 1] + 1;
            }
        }

        // Step 3: Right to Left - if current rating > next, ensure correct count
        for(int i = n - 2; i >= 0; i--){
            if(ratings[i] > ratings[i + 1]){
                candies[i] = max(candies[i], candies[i + 1] + 1);
            }
        }

        // Step 4: Sum all candies
        return accumulate(candies.begin(), candies.end(), 0);
    }
};

📝 How It Works

Start by giving 1 candy to every child.
First pass (left → right): If the next child has a higher rating, they get more candies than the previous one.
Second pass (right → left): If the current child has a higher rating than the next one, and doesn’t have more candies, update accordingly.
This ensures both neighbors conditions are satisfied.

🧩 Key Formula / Recurrence

candies[i] = candies[i-1] + 1 if ratings[i] > ratings[i-1] (left to right)
candies[i] = max(candies[i], candies[i+1] + 1) if ratings[i] > ratings[i+1] (right to left)

⏱️ Time & Space Complexity

Metric	Complexity
Time	O(N)
Space	O(N)

⚠️ Edge Cases

All ratings are equal → everyone gets 1 candy.
Strictly increasing or decreasing ratings → forms arithmetic progression of candies.
Single child → gets 1 candy.

💡 Other Approaches

Approach	Time	Space	Notes
Brute Force	O(N²)	O(N)	Update until stable (TLE) ❌
Two-Pass Greedy ✅	O(N)	O(N)	Optimal and simple
Priority Queue	O(N log N)	O(N)	Overkill, not needed

💬

Problem Statement:​

✅ Solution: Two-Pass Greedy (Left-to-Right & Right-to-Left)​

📝 How It Works​

🧩 Key Formula / Recurrence​

⏱️ Time & Space Complexity​

⚠️ Edge Cases​

💡 Other Approaches​

🔁 Related Problems​

Discussion & Doubts