# Share an ugly but correct O(1) space O(n) time (1 pass) solution.

• Not sure if it can be improved to look elegant.

``````class Solution {
public:
int candy(vector<int>& ratings) {
if (ratings.empty()) return 0;
int gd_cnt = 0, cur = 1, sum = 1; // gd_cnt: going down count -- how many steps of going down so far. cur: candy for current kid. Not accurate.
for (int i = 1; i < ratings.size(); ++i)
{
if (ratings[i-1] <= ratings[i])
{
if (gd_cnt > 0)
{
//this part adjusts the sum by comparing how deep it went. If it went below 1, we will increase sum. Or we will decrease sum.
if (cur > 1)
{
sum += gd_cnt * (1 - cur);
}else{
sum += (gd_cnt + 1) * (1 - cur);
}

cur = 2;
gd_cnt = 0;
}else{
cur++; //r(i-1) < r(i), increase candy
}
if (ratings[i - 1] == ratings[i]) //r(i-1) = r(i), reset candy
{
cur = 1;
}
}else {
cur--;  //r(i-1) > r(i), decrease candy
gd_cnt ++;

}
sum += cur;
}
if (gd_cnt > 0)
{
if (cur > 1)
{
sum += gd_cnt * (1 - cur);
}else{
sum += (gd_cnt + 1) * (1 - cur);
}

}
return sum;
}
};``````

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