1176. Diet Plan Performance

Problem Description

In this problem, a person is tracking their diet over a number of days, where each day they consume a certain number of calories. The caloric intake for each day is stored in an array where calories[i] represents the calories consumed on the i-th day. The person evaluates their diet every k consecutive days, assessing whether they have gained or lost points based on their total caloric intake over those days. If the total caloric intake is less than a lower limit, the person loses a point; if it's more than an upper limit, they gain a point; otherwise, their score does not change. The task is to calculate the total number of points the person has after they have finished their diet, which can be a positive or negative number.


The solution is based on a sliding window technique where the window size is k. A sliding window is helpful in tracking a subset of data in a larger dataset, in this case, the caloric intake over k consecutive days. The solution starts by calculating the initial sum of the first k days. Then, as the window moves forward by one day at a time, instead of recalculating the sum from scratch, the solution simply subtracts the calories from the day that is no longer in the window and adds the calories from the new day. This ensures that the sum is maintained for the current window efficiently.

After each shift of the window, the new sum is checked against the lower and upper bounds to determine if a point should be lost, gained, or if the score remains the same. This check is encapsulated in a function called check(s), which returns -1, 1, or 0, respectively. By summing these up, we can find the total score after all days have been accounted for.

Learn more about Sliding Window patterns.

Solution Approach

The provided Python solution uses an efficient sliding window approach to solve the diet plan performance problem. The sliding window concept is applicable when it is required to calculate something among all contiguous subarrays of a certain length in an efficient manner.

Here is a breakdown of the implementation steps:

  1. Initial Sum Calculation: The initial sum s of the calories for the first k days is calculated using the built-in sum() function on the slice of the first k elements of the calories list.

  2. Initial Points Calculation: The initial sum s is then passed to the check(s) function, which compares s with the lower and upper bounds. Depending on the comparison, the function returns -1, 1, or 0, representing losing a point, gaining a point, or no change in points, respectively. This initial points value is stored in the ans variable.

  3. Sliding the Window: The code enters a loop that will iterate starting from the k-th day to the last day. For each iteration, it adjusts the sum by subtracting the calorie count of the day that's exiting the window (calories[i - k]) and adding the count of the new day entering the window (calories[i]). This keeps the sum s up-to-date with the calorie count of the current window without having to re-calculate the sum from scratch.

  4. Update Points: After updating the sum for the new window position, the check(s) function is used again to determine if points should be gained or lost based on the new sum. The result is added to the ans variable, which accumulates the total points.

  5. Returning the Result: After all windows have been processed, the function returns ans, which contains the total points the dieter has after completing the diet.

This approach has a time complexity of O(n), where n is the number of days, and a space complexity of O(1), which makes it a very efficient solution.

The main tools used in the provided solution are the sliding window pattern and a helper function to calculate the point changes. The calories list serves as the main data structure, with simple arithmetic used to manage the sum and points.

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Example Walkthrough

Let's consider a simple example where a person is tracking their calories over 5 days:

1Calories = [1200, 1300, 1250, 1500, 1100], k = 2, lower = 2000, upper = 3000

The person evaluates their diet every 2 consecutive days and loses or gains points if their caloric intake is below 2000 or above 3000, respectively.

  1. Initial Sum Calculation: Start by summing the first k days (2 days in this case), so sum the first and second day's calories: s = 1200 + 1300 = 2500.

  2. Initial Points Calculation: Pass the initial sum s to the check(s) function. Since 2500 is between the lower and upper limits (2000 and 3000, respectively), they neither gain nor lose a point: ans = 0.

  3. Sliding the Window: Now, for each day from the 3rd day to the 5th:

    • Day 3 (New window: 1300, 1250): Adjust the sum to only include day 2 and 3 by subtracting day 1's calories and adding day 3's calories: s = 2500 - 1200 + 1250 = 2550. After that, check the s value with check(s). Since 2550 is still between the lower and upper limits, no points are gained or lost, and ans = 0.
    • Day 4 (New window: 1250, 1500): Adjust the sum to include the calories from day 3 and 4: s = 2550 - 1300 + 1500 = 2750. Pass s to check(s). Since 2750 is also between the limits, ans remains at 0.
    • Day 5 (New window: 1500, 1100): Adjust the sum for day 4 and 5: s = 2750 - 1250 + 1100 = 2600. Check s with check(s). The sum is still between the limits, so the points stay the same, ans = 0.
  4. Update Points: At each step of the sliding window, we updated the ans based on the check(s) function, which led to no change in this example.

  5. Returning the Result: The loop ends as there are no more windows to slide. The total points the dieter has after completing the diet are ans = 0.

The individual did not gain or lose any points during their diet according to the rules provided because their caloric intake stayed within the specified range for every assessment period.

Solution Implementation

1from typing import List
3class Solution:
4    def dietPlanPerformance(self, calories: List[int], k: int, lower: int, upper: int) -> int:
5        # Helper function to evaluate the points for the given sum of calories
6        def evaluate_points(calories_sum):
7            # If the sum is less than the lower bound, return -1 point
8            if calories_sum < lower:
9                return -1
10            # If the sum is greater than the upper bound, return 1 point
11            elif calories_sum > upper:
12                return 1
13            # Otherwise, no points are awarded or deducted
14            else:
15                return 0
17        # Calculate the initial sum of the first 'k' elements
18        sliding_window_sum = sum(calories[:k])
19        # Initialize score with the points from initial sum
20        score = evaluate_points(sliding_window_sum)
21        # Length of the calories list
22        n = len(calories)
24        # Iterate over the remaining elements, updating the sum and score
25        for i in range(k, n):
26            # Update the sliding window sum by adding the current element and removing the oldest one
27            sliding_window_sum += calories[i] - calories[i - k]
28            # Update the score based on the updated sum
29            score += evaluate_points(sliding_window_sum)
31        # Return the total score after evaluating all sliding windows
32        return score
1class Solution {
2    public int dietPlanPerformance(int[] calories, int k, int lower, int upper) {
4        // Initialize the sum of the first 'k' elements.
5        int windowSum = 0;
7        // Calculate the sum of the first 'k' calories.
8        for (int i = 0; i < k; ++i) {
9            windowSum += calories[i];
10        }
12        // Initialize the performance points.
13        int points = 0;
15        // Check if the initial 'k' day period is below or above the threshold.
16        if (windowSum < lower) {
17            points--;
18        } else if (windowSum > upper) {
19            points++;
20        }
22        // Iterate through the array starting from the 'k'th day.
23        for (int i = k; i < calories.length; ++i) {
24            // Slide the window by 1: remove the first element and add the new one.
25            windowSum += calories[i] - calories[i - k];
27            // Adjust points based on the new sum.
28            if (windowSum < lower) {
29                points--;
30            } else if (windowSum > upper) {
31                points++;
32            }
33        }
35        // Return the calculated points.
36        return points;
37    }
1#include <vector>
2#include <numeric>
4class Solution {
6    int dietPlanPerformance(std::vector<int>& calories, int k, int lower, int upper) {
7        int totalDays = calories.size();
9        // Calculate the total calories for the initial 'k' day period.
10        int currentCalories = std::accumulate(calories.begin(), calories.begin() + k, 0);
12        int performanceScore = 0; // Initialize the performance score.
14        // Update the performance score based on the initial 'k' day period.
15        if (currentCalories < lower) {
16            --performanceScore; // Decrease score when below lower limit.
17        } else if (currentCalories > upper) {
18            ++performanceScore; // Increase score when above upper limit.
19        }
21        // Slide the 'k' day window through the remaining days and update the performance score.
22        for (int i = k; i < totalDays; ++i) {
23            // Add the calorie of the new day and subtract the calorie of the day exiting the 'k' day window.
24            currentCalories += calories[i] - calories[i - k];
26            // Update the score based on the new 'k' day period's calorie count.
27            if (currentCalories < lower) {
28                --performanceScore;
29            } else if (currentCalories > upper) {
30                ++performanceScore;
31            }
32        }
34        return performanceScore; // Return the final performance score.
35    }
1// Evaluates the diet plan performance by calculating points based on calorie intake over a sliding window.
3// Parameters:
4// calories: An array representing daily calorie intake.
5// k: An integer representing the width of the sliding window of days.
6// lower: An integer representing the lower threshold of calories.
7// upper: An integer representing the upper threshold of calories.
9// Returns:
10// The total points calculated by comparing the total calories in each window against the lower and upper thresholds.
11function dietPlanPerformance(calories: number[], k: number, lower: number, upper: number): number {
12    const totalDays = calories.length; // Total number of days represented in the calories array.
13    let windowSum = calories.slice(0, k).reduce((sum, current) => sum + current, 0); // Sum of first 'k' days.
14    let points = 0; // Initialize points to judge the performance.
16    // Award or deduct points based on the first 'k' days' total calories.
17    if (windowSum < lower) {
18        points--; 
19    } else if (windowSum > upper) {
20        points++;
21    }
23    // Slide the window one day at a time, updating the sum and points accordingly.
24    for (let i = k; i < totalDays; ++i) {
25        windowSum += calories[i] - calories[i - k]; // Update the window sum by adding the new day's calories and subtracting the first day's calories from the previous window.
27        // Award or deduct points based on the current window's calories.
28        if (windowSum < lower) {
29            points--; 
30        } else if (windowSum > upper) {
31            points++;
32        }
33    }
35    return points; // Return the calculated points.

Time and Space Complexity

The provided code defines a function dietPlanPerformance that calculates a diet plan performance score based on calories intake over a sliding window of size k, and compares each window's sum with given lower and upper bounds.

Time Complexity

The time complexity of the code is O(n), where n is the total number of days (length of the calories array).

Let's break down the operation:

  • Summation of the first k elements is done in O(k) time.
  • Following this initialization, the function iterates over the remaining elements, from k to n-1. For each iteration, it takes constant time to update the sum s (due to the subtraction of the oldest calorie value and the addition of the new calorie value) and to evaluate the check function. The loop, therefore, runs (n-k) times.
  • Since k is a constant with respect to n, the time taken for initial summation is negligible for large n. Thus, the dominant term is (n-k), which simplifies to O(n) for large n.
1O(k) + O(n-k) = O(n)

Space Complexity

The space complexity of the code is O(1).

Here's why:

  • The extra space used by the program is constant, as it only requires a fixed number of single-value variables (s, n, and ans) regardless of the input size.
  • No additional data structures that scale with the input size are used; the input list itself is not modified.
  • The check function is defined within the method and does not consume extra space that depends on the input size.

Learn more about how to find time and space complexity quickly using problem constraints.

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