### Problem Statement

A dieter consumes

`calories[i]`

calories on the `i`

-th day.
Given an integer

`k`

, for **every**consecutive sequence of`k`

days (`calories[i], calories[i+1], ..., calories[i+k-1]`

for all `0 <= i <= n-k`

), they look at *T*, the total calories consumed during that sequence of`k`

days (`calories[i] + calories[i+1] + ... + calories[i+k-1]`

):- 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.

Initially, the dieter has zero points. Return the total number of points the dieter has after dieting for

`calories.length`

days.
Note that the total points can be negative.

**Example 1:**

Input:calories = [1,2,3,4,5], k = 1, lower = 3, upper = 3Output:0Explanation: Since k = 1, we consider each element of the array separately and compare it to lower and upper. calories[0] and calories[1] are less than lower so 2 points are lost. calories[3] and calories[4] are greater than upper so 2 points are gained.

**Example 2:**

Input:calories = [3,2], k = 2, lower = 0, upper = 1Output:1Explanation: Since k = 2, we consider subarrays of length 2. calories[0] + calories[1] > upper so 1 point is gained.

**Example 3:**

Input:calories = [6,5,0,0], k = 2, lower = 1, upper = 5Output:0Explanation: calories[0] + calories[1] > upper so 1 point is gained. lower <= calories[1] + calories[2] <= upper so no change in points. calories[2] + calories[3] < lower so 1 point is lost.

**Constraints:**

`1 <= k <= calories.length <= 10^5`

`0 <= calories[i] <= 20000`

`0 <= lower <= upper`

### Video Tutorial

You can find the detailed video tutorial here### Thought Process

This is an easy problem but might be a bit hard to code it up correctly in one pass. We can use a sliding window to keep a rolling sum and compare it with upper and lower bound.A few caveats in implementation:

- We can simplify using two pointers start and end by keeping an index i (start) and i - k (end)
- We can only have one loop instead of having two separate loops

sliding window |

### Solutions

#### Sliding window

Time Complexity: O(N) visited the array once

Space Complexity: O(1) No extra space is needed

### References

- None