Candy in C++

Suppose there are N children, they are standing in a line. Here each child is assigned a rating value. We are supplying candies to these children subjected to the following requirements −

• Each child must have at least one candy.

• Children whose rating is high will get more candies than their neighbors.

We have to find the minimum number of candies we must give?

So if the input is like [1,1,3], then the output will be 4. So they will get 1, 1 and 2 candies respectively.

To solve this, we will follow these steps −

• n := size of the array ratings, create array called dp of size n, fill this using 1

• ret := 0

• for i in range 1 to n – 1

  • if ratings[i] > ratings[i – 1], then dp[i] := dp[i - 1] + 1

• for i in range n - 2 down to 0

  • if ratings[i] > ratings[i + 1], then dp[i] := max of dp[i] and dp[i + 1] + 1

• ret := sum of the elements of dp

• return ret

Example (C++)

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   int candy(vector<int>& ratings) {
      int n = ratings.size();
      vector <int> dp(n, 1);
      int ret = 0;
      for(int i = 1; i < n; i++){
         if(ratings[i] > ratings[i - 1]){
            dp[i] = dp[i - 1] + 1;
         }
      }
      for(int i = n - 2; i >= 0; i--){
         if(ratings[i] > ratings[i + 1]){
            dp[i] = max(dp[i], dp[i + 1] + 1);
         }
      }
      for(int i = 0; i < n; i+=1){
         ret += dp[i];
      }
      return ret;
   }
};
main(){
   Solution ob;
   vector<int> v = {1,1,3};
   cout << (ob.candy(v));
}

Input

[1,1,3]

Output

4