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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 −
#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
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