# Diet Plan Performance

A dieter consumes calories[i] calories on the i-th day. For every consecutive sequence of k days, they look at T, the total calories consumed during that sequence of k days:

• If T < lower, they performed poorly on their diet and lose 1 point;
• If T > upper, they performed well on their diet and gain 1 point;
• Otherwise, they performed normally and there is no change in points.

Return the total number of points the dieter has after all calories.length days.

Note that: The total points could be negative.

Example 1:

Input: calories = [1,2,3,4,5], k = 1, lower = 3, upper = 3
Output: 0
Explaination: calories[0], calories[1] < lower and calories[3], calories[4] > upper, total points = 0.


Example 2:

Input: calories = [3,2], k = 2, lower = 0, upper = 1
Output: 1
Explaination: calories[0] + calories[1] > upper, total points = 1.


Example 3:

Input: calories = [6,5,0,0], k = 2, lower = 1, upper = 5
Output: 0
Explaination: calories[0] + calories[1] > upper, lower <= calories[1] + calories[2] <= upper, calories[2] + calories[3] < lower, total points = 0.


Constraints:

• 1 <= k <= calories.length <= 10^5
• 0 <= calories[i] <= 20000
• 0 <= lower <= upper

Solution:

Sliding window.

class Solution {
public:
int dietPlanPerformance(vector<int>& c, int k, int lower, int upper) {
int sum = 0;
for(int i = 0;i < k-1;i++){
sum += c[i];
}
int ans = 0;
int i = k-1;
while(i < c.size()){
if(i > k-1) sum -= c[i-k];
sum += c[i];
if(sum < lower){
ans--;
}
if(sum > upper){
ans++;
}
i++;
}
return ans;
}
};