Last updated on October 9th, 2024 at 06:10 pm

Here, We see Candy LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Candy LeetCode Solution

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 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.

1. Candy LeetCode Solution C++

class Solution {
public:
    int candy(vector<int>& ratings) {
        int n = ratings.size();
        int candy = n, i=1;
        while(i<n){
            if(ratings[i] == ratings[i-1]){
                i++;
                continue;
            }
            int peak = 0;
            while(ratings[i] > ratings [i-1]){
                peak++;
                candy += peak;
                i++;
                if(i == n) return candy;
            }
            int valley = 0;
            while(i<n && ratings[i] < ratings[i-1]){
                valley++;
                candy += valley;
                i++;
            }
            candy -= min(peak, valley); //Keep only the higher peak
        }
        return candy;
    }
};

2. Candy LeetCode Solution Java

class Solution {
    public int candy(int[] ratings) {
        int n = ratings.length;
        int[] candies = new int[n];
        Arrays.fill(candies, 1);

        for (int i = 1; i < n; i++) {
            if (ratings[i] > ratings[i - 1]) {
                candies[i] = candies[i - 1] + 1;
            }
        }
        for (int i = n - 2; i >= 0; i--) {
            if (ratings[i] > ratings[i + 1]) {
                candies[i] = Math.max(candies[i], candies[i + 1] + 1);
            }
        }
        int totalCandies = 0;
        for (int candy : candies) {
            totalCandies += candy;
        }
        return totalCandies;
    }
}

3. Candy Solution JavaScript

var candy = function(ratings) {
    const n = ratings.length;
    const candies = new Array(n).fill(1);

    for (let i = 1; i < n; i++) {
        if (ratings[i] > ratings[i - 1]) {
            candies[i] = candies[i - 1] + 1;
        }
    }
    for (let i = n - 2; i >= 0; i--) {
        if (ratings[i] > ratings[i + 1]) {
            candies[i] = Math.max(candies[i], candies[i + 1] + 1);
        }
    }
    return candies.reduce((a, b) => a + b, 0);
};

4. Candy Solution Python

class Solution(object):
    def candy(self, ratings):
        n = len(ratings)
        candies = [1] * n 
        for i in range(1, n):
            if ratings[i] > ratings[i-1]:
                candies[i] = candies[i-1] + 1
        for i in range(n-2, -1, -1):
            if ratings[i] > ratings[i+1]:
                candies[i] = max(candies[i], candies[i+1] + 1)
        return sum(candies)