Skip to content

Instantly share code, notes, and snippets.

Show Gist options
  • Save Ojasvi2005/4693ba0775dc9a5e5616678fb9dbd81e to your computer and use it in GitHub Desktop.
Save Ojasvi2005/4693ba0775dc9a5e5616678fb9dbd81e to your computer and use it in GitHub Desktop.
LEETCODE CANDY QUESTION
QUES NO. 135
Question-
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.
Mistake I did initially - earlier I was overwriting some values in the second pass. which I fixed later.
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.
Constraints:
n == ratings.length
1 <= n <= 2 * 104
0 <= ratings[i] <= 2 * 104
Solution-
class Solution {
public:
//two pass greedy solution.
int candy(vector<int>& ratings) {
int n=ratings.size();
if(n==1) return 1;
int sum=0;
vector<int> temp(n,1);
for(int i=1;i<n;i++){
if(ratings[i]>ratings[i-1]){
temp[i]=temp[i-1]+1;
}
}
for(int i=n-2;i>=0;i--){
if(ratings[i]>ratings[i+1]){
temp[i]=max(temp[i],temp[i+1]+1);
}
sum+=temp[i];
}
sum+=temp[n-1];
return sum;
}
};
@Ojasvi2005
Copy link
Author

this is a C++ code.

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment