Diet Plan Performance

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ArraysSliding Windows

A dieter consumes calories[i] calories on the i^th day.

Given an integer array calories representing the number of calories consumed each day, and an integer k representing the rolling average days, and two integers lower and upper representing the acceptable range of average calories, you need to calculate the diet plan performance.

Your diet plan performance is calculated as follows:

If the average calories consumed for every k consecutive days is strictly less than lower, award 1 point.
If the average calories consumed for every k consecutive days is strictly greater than upper, award 1 point.
Otherwise, award 0 points.

Return the total number of points you get.

Note that the average is calculated as the sum of the k consecutive days divided by k, using integer division.

Example 1:

Input: calories = [1,2,3,4,5], k = 1, lower = 3, upper = 3
Output: 0
Explanation:
Calories Consumed in each day
day 1 (1) < lower (3) => 1 point
day 2 (2) < lower (3) => 1 point
day 3 (3) == lower (3) => 0 points
day 4 (4) > upper (3) => 1 point
day 5 (5) > upper (3) => 1 point
Total points = 1 + 1 + 0 + 1 + 1 = 4

Example 2:

Input: calories = [3,2,1], k = 2, lower = 2, upper = 3
Output: 1
Explanation:
The calories consumed for each consecutive 2 days are (3+2)/2 = 2, (2+1)/2 = 1.
The first day (2) == lower (2) => 0 points
The second day (1) < lower (2) => 1 point
Total points = 0 + 1 = 1

Example 3:

Input: calories = [6,5,4,3,2,1], k = 2, lower = 2, upper = 3
Output: 0
Explanation:
The calories consumed for each consecutive 2 days are (6+5)/2 = 5, (5+4)/2 = 4, (4+3)/2 = 3, (3+2)/2 = 2, (2+1)/2 = 1.
The first day (5) > upper (3) => 1 point
The second day (4) > upper (3) => 1 point
The third day (3) == upper (3) => 0 points
The fourth day (2) == lower (2) => 0 points
The fifth day (1) < lower (2) => 1 point
Total points = 1 + 1 + 0 + 0 + 1 = 3

Constraints:

1 <= k <= calories.length <= 10⁵
0 <= calories[i] <= 20000
0 <= lower <= upper

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible values for the calories in each food item and the lower/upper limits k?
If no such window exists that satisfies the diet plan, what should the function return?
Can the input calories array be empty or null?
Is 'consecutively' strictly referring to directly adjacent elements in the input array, or could there be gaps?
What happens if the lower and upper limits (k) are equal?

Brute Force Solution

Approach

The brute force approach for this diet problem means we're going to try every single possible combination of days to see if they meet the weight loss/gain criteria. We will examine every possible consecutive sequence of days within the provided data.

Here's how the algorithm would work step-by-step:

Start with the first day's weight.
Consider only the first day. Calculate if this single day's calorie intake results in a weight gain, loss, or neither.
Now, consider the first two days. Calculate the total calorie intake for those two days and check if it causes a weight gain, loss, or neither.
Continue adding one more day at a time, calculating the total calorie intake for each consecutive group of days, and checking if it results in a weight gain, loss, or neither.
After covering all consecutive sequences starting from the first day, move to the second day and repeat the process. Consider only the second day, then the second and third days, then the second, third, and fourth days, and so on.
Keep doing this until you have considered every possible consecutive sequence of days in the given range.
Count how many times the total calorie intake falls outside the specified range (less than the lower limit or greater than the upper limit). This count represents the number of days the diet performed poorly.
Based on the problem's specific rules (weight loss, gain, or neither), determine the final output based on the comparison with the lower and upper limits.

Code Implementation

def diet_plan_performance_brute_force(calories, lower_limit, upper_limit):
    poor_performance_days = 0

    for start_day_index in range(len(calories)):
        current_calories_sum = 0
        for end_day_index in range(start_day_index, len(calories)):
            current_calories_sum += calories[end_day_index]

            # Accumulate the total to check against thresholds
            if current_calories_sum < lower_limit or current_calories_sum > upper_limit:
                poor_performance_days += 1

    return poor_performance_days

Big(O) Analysis

Time Complexity

O(n²) – The described brute force approach iterates through all possible subarrays of the input array. The outer loop iterates n times, where n is the number of days, representing the starting day of a potential subarray. The inner loop, for each starting day, iterates up to n times to consider all consecutive days, calculating the sum of calories for each subarray. This nested loop structure results in a time complexity proportional to n * n, or O(n²).

Space Complexity

O(1) – The brute force approach described iterates through subarrays of the input array. It calculates the sum of each subarray but doesn't store these sums or any other significant data structures. The only extra memory used consists of a few integer variables to keep track of the current window and the counts of performance outcomes. Therefore, the space complexity is constant and independent of the input size N.

Optimal Solution

Approach

We need to efficiently figure out how well someone is sticking to their diet. The key is to look at chunks of days at a time to see if they exceeded the limit or fell short. This allows us to calculate the performance score without checking every single day individually.

Here's how the algorithm would work step-by-step:

First, think of each consecutive set of days as a sliding window, like looking through a moving frame.
Calculate the total calories consumed in the first set of days (the size of the window).
Check if the total calories are less than the lower limit or more than the upper limit to adjust the score accordingly.
Slide the window forward by one day. This means we remove the calorie count of the first day from the previous window and add the calorie count of the new day at the end.
Repeat the calorie calculation and score adjustment for the new set of days.
Continue sliding the window and checking until you've covered all the days.
The final score after checking all windows represents the person's diet performance.

Code Implementation

def diet_plan_performance(calories, day_window, lower_limit, upper_limit):
    performance_score = 0
    current_calories_sum = sum(calories[:day_window])

    # Check performance for initial window
    if current_calories_sum < lower_limit:
        performance_score -= 1
elif current_calories_sum > upper_limit:
        performance_score += 1

    # Iterate through remaining windows
    for i in range(day_window, len(calories)):
        # Slide the window by subtracting the oldest day and adding the newest day.
        current_calories_sum += calories[i] - calories[i - day_window]

        # Update performance score based on calorie consumption.
        if current_calories_sum < lower_limit:
            performance_score -= 1
elif current_calories_sum > upper_limit:
            performance_score += 1

    return performance_score

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the calorie array using a sliding window of fixed size k. The initial sum of the first window of size k takes O(k) time, but subsequent window sums are calculated by subtracting the calorie count of the outgoing day and adding the calorie count of the incoming day, which takes O(1) time per window. Since the window slides n-k+1 times and each slide takes O(1) time, the dominant factor is the single loop that iterates through the calorie array, resulting in a time complexity of O(n).

Space Complexity

O(1) – The algorithm uses a sliding window approach to calculate the diet plan performance. It maintains a fixed-size window and updates the sum of calories within that window. No auxiliary data structures that scale with the input size (N, the number of days) are created. The space used for storing the current window's sum and the final score remains constant regardless of the input size.

Edge Cases

Case	How to Handle
Empty calories array	Return 0 for all windows since no calories can be summed.
Calories array size smaller than k	Return 0 since no window of size k exists.
k is zero or negative	Treat k as invalid and return 0 as no meaningful window exists.
Calories array contains negative values	The problem statement is valid with negative numbers, so the sliding window approach should correctly account for weight loss and gains.
Calories array contains very large values leading to integer overflow when summed	Use a larger integer data type like long to store the sum of calories.
All windows have calories sum greater than upper threshold	The solution will consistently decrement the performance score for all windows.
All windows have calories sum less than lower threshold	The solution will consistently increment the performance score for all windows.
Upper threshold equals lower threshold	The solution correctly compares window sums and only modifies performance if window sum != thresholds.

Empty calories array

How to Handle:

Return 0 for all windows since no calories can be summed.

Calories array size smaller than k

How to Handle:

Return 0 since no window of size k exists.

k is zero or negative

How to Handle:

Treat k as invalid and return 0 as no meaningful window exists.

Calories array contains negative values

How to Handle:

The problem statement is valid with negative numbers, so the sliding window approach should correctly account for weight loss and gains.

Calories array contains very large values leading to integer overflow when summed

How to Handle:

Use a larger integer data type like long to store the sum of calories.

All windows have calories sum greater than upper threshold

How to Handle:

The solution will consistently decrement the performance score for all windows.

All windows have calories sum less than lower threshold

How to Handle:

The solution will consistently increment the performance score for all windows.

Upper threshold equals lower threshold

How to Handle:

The solution correctly compares window sums and only modifies performance if window sum != thresholds.