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2 changes: 1 addition & 1 deletion dsa-problems/leetcode-problems/0100-0199.md
Original file line number Diff line number Diff line change
Expand Up @@ -224,7 +224,7 @@ export const problems = [
"problemName": "135. Candy",
"difficulty": "Hard",
"leetCodeLink": "https://leetcode.com/problems/candy/",
"solutionLink": "#"
"solutionLink": "/dsa-solutions/lc-solutions/0100-0199/Candy"
},
{
"problemName": "136. Single Number",
Expand Down
194 changes: 194 additions & 0 deletions dsa-solutions/lc-solutions/0100-0199/0135-Candy.md
Original file line number Diff line number Diff line change
@@ -0,0 +1,194 @@
---
id: candy
title: Candy
sidebar_label: 0135 Candy
tags:
- Greedy
- LeetCode
- Java
- Python
- C++
description: "This is a solution to the Candy problem on LeetCode."
---

## Problem Description

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.

### Examples

**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.

```
Input: root = [1,2,2,null,3,null,3]
Output: false
```

### Constraints

- n == ratings.length
- 1 <= n <= 2 * 10^4
- 0 <= ratings[i] <= 2 * 10^4

---

## Solution for Candy Distribution Problem

### Intuition And Approach

To solve this problem, we can use a greedy approach. We can start by assigning each child 1 candy. Then, we iterate over the ratings array twice, once from left to right and once from right to left, adjusting the number of candies as needed to satisfy the given conditions.



#### Code in Different Languages

<Tabs>
<TabItem value="Java" label="Java">
<SolutionAuthor name="@YourName"/>
```java
public class Solution {
public int candy(int[] ratings) {
int n = ratings.length;
int[] candies = new int[n];

// Initialize all candies to 1
for (int i = 0; i < n; i++) {
candies[i] = 1;
}

// Left to right pass
for (int i = 1; i < n; i++) {
if (ratings[i] > ratings[i - 1]) {
candies[i] = candies[i - 1] + 1;
}
}

// Right to left pass
for (int i = n - 2; i >= 0; i--) {
if (ratings[i] > ratings[i + 1] && candies[i] <= candies[i + 1]) {
candies[i] = candies[i + 1] + 1;
}
}

int totalCandies = 0;
for (int candy : candies) {
totalCandies += candy;
}

return totalCandies;
}

public static void main(String[] args) {
Solution solution = new Solution();
int[] ratings1 = {1, 0, 2};
int[] ratings2 = {1, 2, 2};

System.out.println(solution.candy(ratings1)); // Output: 5
System.out.println(solution.candy(ratings2)); // Output: 4
}
}
```
</TabItem>
<TabItem value="Python" label="Python">
<SolutionAuthor name="@YourName"/>
```python
class Solution(object):
def candy(self, ratings):
"""
:type ratings: List[int]
:rtype: int
"""
n = len(ratings)
candies = [1] * n

# Left to right pass
for i in range(1, n):
if ratings[i] > ratings[i - 1]:
candies[i] = candies[i - 1] + 1

# Right to left pass
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)
```
</TabItem>
<TabItem value="C++" label="C++">
<SolutionAuthor name="@mahek0620"/>
```cpp
#include <vector>
#include <algorithm>

using namespace std;

class Solution {
public:
int candy(vector<int>& ratings) {
int n = ratings.size();
vector<int> candies(n, 1);

// Left to right pass
for (int i = 1; i < n; i++) {
if (ratings[i] > ratings[i - 1]) {
candies[i] = candies[i - 1] + 1;
}
}

// Right to left pass
for (int i = n - 2; i >= 0; i--) {
if (ratings[i] > ratings[i + 1]) {
candies[i] = max(candies[i], candies[i + 1] + 1);
}
}

// Sum up all candies
int totalCandies = 0;
for (int candy : candies) {
totalCandies += candy;
}

return totalCandies;
}
};
```
</TabItem>
</Tabs>

#### Complexity Analysis

- Time Complexity: O(n) where n is the number of children (length of the ratings array).
- Space Complexity: O(n) for storing the candies array.

</tabItem>



## References

- **LeetCode Problem:** [Candy Problem](https://leetcode.com/problems/candy/)
- **Solution Link:** [Candy Solution on LeetCode](https://leetcode.com/problems/candy/solutions/5273312/candy-solution)
- **Authors GeeksforGeeks Profile:** [Mahek Patel](https://leetcode.com/u/mahekrpatel611/)

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