# C++ solution, 45ms, O(n) time and O(n) space, passed at the first attempt

• ``````class Solution {
public:
int candy(vector<int>& ratings) {
int n=ratings.size();
if(n<1) return 0;
if(n==1) return 1;

vector<int> flag;
vector<int> pos;
vector<int> results(n,1);
int count=0;

if(ratings[0]<ratings[1]){
flag.push_back(1);
pos.push_back(0);
count++;
}
if(ratings[n-2]>ratings[n-1]){
flag.push_back(1);
pos.push_back(n-1);
count++;
}

for(int i=1;i<n-1;i++){
if(ratings[i]<ratings[i-1] && ratings[i]<ratings[i+1]){
flag.push_back(1);
pos.push_back(i);
count++;
}
if(ratings[i]<ratings[i-1] && ratings[i]==ratings[i+1]){
flag.push_back(2);
pos.push_back(i);
count++;
}
if(ratings[i]==ratings[i-1] && ratings[i]<ratings[i+1]){
flag.push_back(3);
pos.push_back(i);
count++;
}
}

if(count>0){
for(int j=0;j<count;j++){
int i=pos[j];
if(flag[j]==1){
while(ratings[i+1]>ratings[i] && results[i]+1>results[i+1]){
results[i+1]=results[i]+1;
i++;
}

i=pos[j];
while(ratings[i-1]>ratings[i] && results[i]+1>results[i-1]){
results[i-1]=results[i]+1;
i--;
}
}

if(flag[j]==2){
while(ratings[i-1]>ratings[i] && results[i]+1>results[i-1]){
results[i-1]=results[i]+1;
i--;
}
}

if(flag[j]==3){
while(ratings[i+1]>ratings[i] && results[i]+1>results[i+1]){
results[i+1]=results[i]+1;
i++;
}
}
}
}

int sum=0;
for(int i=0;i<n;i++)
sum+=results[i];
return sum;
}
};
``````

Although the code seems long, the idea is pretty simple. I just record three types of minimum points and do some calculations from the minimum points.

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